Exercises Attempts

Exercise after §13

1. Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x \in A$ there is an open set $U$ containing $x$ such that $U \subseteq A$. Show that $A$ is open in $X$.

Since for any $x \in A$, there exists an open set $U_x$ such that $x \in U_x \subseteq A$, it follows that $A \subseteq \bigcup_{x \in A}U_x \subseteq A \implies A=\bigcup_{x \in A}U_x$, so by the definition of a topology, $A$ is open in $X$.

3 (reworded). Show that the countable complement topology $\mathcal{T}_c$ is indeed a topology on the set $X$.

Is the collection

$$\mathcal{T}_\infty := \{U : X \backslash U \text{ is infinite or empty or all of } X\}$$

a topology on $X$?

$\mathcal{T}_c$ case.

Indeed, $X$ and $\emptyset$ are in $\mathcal{T}_c$ because $X = X-\emptyset$ and $\emptyset=X-X$, and $\emptyset$ is countable.

Consider an indexed family of sets $\{U_\alpha\}$ in $\mathcal{T}_c$, then the (relative) component of its union $\bigcup U_\alpha$ with respect to $X$ has the following equivalence (due to de Morgan's law):

$$X \backslash \bigcup U_\alpha = \bigcap \left(X \backslash U_\alpha \right).$$

Note that since each $U_\alpha$ is in $\mathcal{T}_c$, each $X \backslash U_\alpha$ is countable. Due to the fact that an arbitrary intersection of countable sets is countable, $X \backslash \bigcup U_\alpha$ is thus countable, so $\bigcup U_\alpha$ is in $\mathcal{T}_c$.

Now, consider a finite family of sets $\{U_i\}_{i=1,\ldots,n}$ in $\mathcal{T}_c$, then by a similar argument, the (relative) component of its intersection $\bigcap_{i=1}^{n}U_i$ with respect to $X$ has the following equivalence:

$$X \backslash \bigcap_{i=1}^{n}U_i=\bigcup_{i=1}^{n}(X \backslash U_i).$$

Similarly, each $X \backslash U_i$ is countable. Since a finite union of countable sets is countable¹, $X \backslash \bigcap_{i=1}^{n}U_i$ is thus countable, so $\bigcap_{i=1}^{n}U_i$ is in $\mathcal{T}_c$.

Therefore, by the definition of a topology, $\mathcal{T}_c$ is indeed a topology.

$\mathcal{T}_\infty$ case.

By the similar reason as above, $X$ and $\emptyset$ are in $\mathcal{T}_\infty$.

Since an arbitrary, in particular finite, union of infinite sets is infinite, we can see from a similar argument as above that a finite intersection of sets in $\mathcal{T}_\infty$ is in $\mathcal{T}_\infty$.

However, an arbitrary intersection of infinite sets may not be infinite nor empty. A relatively straightforward example is when $X = \mathbb{R}$ and a family of sets $U_\alpha \subseteq X$ is defined as sets in the form of $\{ x \in \mathbb{R} : |x| > r \}$ where $r$ is an arbitrary positive real number. Note that each $X \backslash U_\alpha$, which is given by $\{ x \in \mathbb{R} : |x| \leqslant r \}$, is infinite, so each $U_\alpha$ is in $\mathcal{T}_\infty$ and thus open.

It follows that the intersection of this family of sets is $\{0\}$, which is finite and non-empty. This fact causes $\bigcup U_\alpha$, i.e. an arbitrary union of sets in $\mathcal{T}_\infty$, not to be in $\mathcal{T}_\infty$, due to the similar argument as mentioned in the $\mathcal{T}_c$ case.

This is enough to conclude that $\mathcal{T}_\infty$ is not a topology.

4. (a) If $\{\mathcal{T}_\alpha \}$ is a family of topologies on $X$, show that $\bigcap\mathcal{T}_\alpha$ is a topology on $X$. Is $\bigcup\mathcal{T}_\alpha$ a topology on $X$?

(b) Let $\{\mathcal{T}_\alpha \}$ be a family of topologies on $X$. Show that there is a unique smallest topology on $X$ containing all the collections $\mathcal{T}_\alpha$, and a unique largest topology contained in all $\mathcal{T}_\alpha$.

(c) If $X = \{a,b,c\}$, let

$$\mathcal{T}_1 = \{\emptyset, X, \{a\}, \{a,b\}\} \quad \text{and} \quad \mathcal{T}_2=\{\emptyset, X, \{a\},\{b,c\}\}.$$

Find the smallest topology containing $\mathcal{T}_1$ and $\mathcal{T}_2$, and the largest topology contained in $\mathcal{T}_1$ and $\mathcal{T}_2$.

4. (a) Indeed, $\emptyset$ and $X$ are in the intersection as all topologies are required to contain them. For any set that we take from the intersection, they belong to all the topologies that form the intersection, so any arbitrary union and finite intersection of them also belong to all the topologies that form the intersection and thus are in the intersection, so the intersection is a topology.

   The union operation does not preserve topologies in general. Consider $X=\{a,b,c\}$ and two of its topologies $\mathcal{T}_1=\{\emptyset,X,\{a\},\{b,c\}\}$ and $\mathcal{T}_2=\{\emptyset,X,\{b\}\}$, then $\mathcal{T}_1 \cup \mathcal{T}_2 = \{\emptyset,X,\{a\},\{b\},\{b,c\}\}$ and we can see that it is not a topology since $\{a\} \cup \{b\}=\{a,b\}$ is not in $\mathcal{T}_1 \cup \mathcal{T}_2$.

(b) We know from (a) that $\bigcap \mathcal{T}_\alpha$ is a topology on $X$ contained in all $\mathcal{T}_\alpha$. We claim that it is the unique largest such topology. It is the largest because a topology that is contained in all $\mathcal{T}_\alpha$ must contain $\bigcap \mathcal{T}_\alpha$, and if any $x \notin \bigcap \mathcal{T}_\alpha$ is added in, there must be some $\mathcal{T}_\alpha$ that does not contain $x$. The uniqueness comes from the fact that if $\mathcal{T}_1$ and $\mathcal{T}_2$ are the largest topologies, then $\mathcal{T}_1 \subseteq \mathcal{T}_2$ and $\mathcal{T}_2 \subseteq \mathcal{T}_1$ , so $\mathcal{T}_1=\mathcal{T}_2$.

(c) The smallest topology containing $\mathcal{T}_1$ and $\mathcal{T}_2$ is given by $\{\emptyset,X,\{a\},\{b\},\{a,b\},\{b,c\}\}$. The largest topology contained in $\mathcal{T}_1$ and $\mathcal{T}_2$ is given by $\{\emptyset, X , \{a\}\}$.

5. Show that if $\mathcal{A}$ is a basis for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\mathcal{A}$. Prove the same if $\mathcal{A}$ is a subbasis.

note

Exercise 13.5 also means that the topology generated by a basis $\mathcal{A}$ is the 'smallest' topology of a set $X$ that contains $\mathcal{A}$.

Basis case.

Suppose that $U \in \mathcal{T}_\mathcal{A}$, then by Lemma 13.1, $U$ is a union of some elements of $\mathcal{A}$, i.e. $U = \bigcup\mathcal{C}$ for some $\mathcal{C} \subseteq \mathcal{A}$. Since $\mathcal{C} \subseteq \mathcal{A} \subseteq \bigcap_{\mathcal{A} \subseteq \mathcal{T}}\mathcal{T}$, $\bigcap_{\mathcal{A} \subseteq \mathcal{T}}\mathcal{T}$ is a topology as previously shown in 4(a), and any arbitrary union of open sets is open, this implies that $U \in \bigcap_{\mathcal{A} \subseteq \mathcal{T}}\mathcal{T}$.

Conversely, suppose that $U' \in \bigcap_{\mathcal{A} \subseteq \mathcal{T}}\mathcal{T}$. It then follows immediately that $U' \in \mathcal{T}_\mathcal{A}$, because $\mathcal{T}_\mathcal{A}$ is one of the topologies that contain $\mathcal{A}$, as we know that each basis element is in $\mathcal{T}_\mathcal{A}$ and any arbitrary union of open sets is open, which results in the union of the singletons of each basis element, which is exactly $\mathcal{A}$ itself, to be in $\mathcal{T}_\mathcal{A}$ as well.

Therefore, $\mathcal{T}_\mathcal{A} = \bigcap_{\mathcal{A} \subseteq \mathcal{T}}\mathcal{T}$.

Subbasis case.

If $\mathcal{A}$ is a subbasis, then there exists a corresponding basis $\mathcal{A}'$ in $X$ such that each of its basis elements is a finite intersection of subbasis elements in $\mathcal{A}$, and the resulting topology generated respectively by $\mathcal{A}$ and $\mathcal{A}'$ are equal.

Since a finite intersection of elements of a topology is still in the topology, for all topologies on $X$ that contain $\mathcal{A}$ (which is a subbasis this time), they also contain $\mathcal{A}'$. This allows us to revert back to the basis case, and argue similarly to reach the desired conclusion.

Reference.

6. Show that the topologies of $\mathbb{R}_l$ and $\mathbb{R}_K$ are not comparable.

Note that there is an abuse of notation to denote the topologies, instead of the corresponding sets, as $\mathbb{R}_l$ and $\mathbb{R}_K$ respectively.

It suffices to find a set (element) in $\mathbb{R}_l$ that is not in $\mathbb{R}_K$ and vice versa. We recall that $K := \left\{\frac{1}{n} : n \in \mathbb{Z}_+ \right\}$ here.

Take $[0,1) \in \mathbb{R}_l$ and $(0,1) \backslash K \in \mathbb{R}_K$. Since $K \subset [0,1)$, $[0,1) \notin \mathbb{R}_K$. We also have that $(0,1) \backslash K \notin \mathbb{R}_l$ because any arbitrary union and finite intersection of elements (sets) in $\mathbb{R}_l$ (which are in the form of $[a,b)$ with $a,b \in \mathbb{R}$) that contains $(0,1) \backslash K$ will contain a non-empty subset of $K$.

8. (a) Apply Lemma 13.2 to show that the countable collection

$$\mathcal{B} = \{ (a,b) : a<b,\; a \text{ and } b \text{ rational}\}$$

is a basis that generates the standard topology on $\mathbb{R}$.

(b) Show that the collection

$$\mathcal{C} = \{[a,b) : a<b,\; a \text{ and } b \text{ rational}\}$$

is a basis that generates a topology different from the lower limit topology on $\mathbb{R}$.

(a) Let $U$ be an open set in the standard topology on $\mathbb{R}$.

We know that rational numbers are dense in real numbers, i.e. for any $x,y \in \mathbb{R}$ with $x<y$, there exists a rational number $r$ such that $x<r<y$. Let $u \in U$, then $u$ is contained in some open interval $(a,b) \subseteq U$ where $a,b \in \mathbb{R}$ and $a < b$, so that $a<u<b$. By the density of rational numbers, there exists rational numbers $r_1$ and $r_2$ such that $a < r_1 < u$ and $u < r_2 < b$.

This implies that $r_1 < u < r_2$, so $u$ is contained in the open interval $(r_1,r_2)$, which is in $\mathcal{B}$ as well as contained in the interval $(a,b)$, and thus in $U$. Therefore, by Lemma 13.2, $\mathcal{B}$ is a basis that generates the standard topology.

(b) We first show that $\mathcal{C}$ satisfies the two conditions for a basis.

For the first condition, let $x \in \mathbb{R}$ be arbitrary. If $x$ is rational, then we see that $x$ is contained in $[x,y)$, where $y$ is rational $y > x$, and the interval is in $\mathcal{C}$. If $x$ is irrational, then the density of irrational numbers in the real numbers, and thus rational numbers in particular, gives rise to the existence of an interval $[a',b')$ in $\mathcal{C}$ that contains $x$.

For the second condition, let $[c,d)$ and $[c',d')$ be in $\mathcal{C}$, then their intersection is given by $[\max\{c,c'\},\min\{d,d'\})$, which is in $\mathcal{C}$, so the second condition is satisfied.

Consider the set $U := [\sqrt{2},3) \subseteq \mathbb{R}$. It is open in the lower limit topology on $\mathbb{R}$ but it is not open in the topology generated by $\mathcal{C}$ as there is no interval $C$ in $\mathcal{C}$ such that $\sqrt{2} \in C$ and $C \subseteq U$, as any interval in $\mathcal{C}$ that contains $\sqrt{2}$ must contain some numbers strictly less than $\sqrt{2}$. Thus, the topology generated by $\mathcal{C}$ is different from the lower limit topology.

Exercise after §16

1. Show that if $Y$ is a subspace of $X$, and $A$ is a subset of $Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$.

Suppose that $X$ is a set endowed with a topology $\mathcal{T}$, $Y$ is a subspace of $X$ endowed with the topology $\mathcal{T}' := \{ Y \cap U : U \in \mathcal{T}\}$, and $A \subseteq Y$.

Let $\mathcal{T}_Y := \{A \cap V : V \in \mathcal{T}' \}$ be the topology $A$ inherits as a subspace of $Y$ and $\mathcal{T}_X := \{ A \cap W : W \in \mathcal{T} \}$ as the topology $A$ inherits as a subspace of $X$.

Suppose that a set $S$ lies in $\mathcal{T}_Y$, then $S = A \cap V$ where $V$ is in $\mathcal{T}'$. This also means that $S = A \cap (Y \cap W)$ where $W$ is in $\mathcal{T}$. Now, note that $A \cap (Y \cap W) = (A \cap Y) \cap W$ and since $A$ is a subset of $Y$, $A \cap Y$ is just $A$. Therefore, $S = A \cap W$ where $W$ is in $\mathcal{T}$. Thus, $\mathcal{T}_Y \subseteq \mathcal{T}_X$.

Since it is valid to reverse the arguments above, we have that $\mathcal{T}_X \subseteq \mathcal{T}_Y$ as well. Thus, $\mathcal{T}_Y = \mathcal{T}_X$, so the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$.

2. If $\mathcal{T}$ and $\mathcal{T}'$ are topologies on $X$ and $\mathcal{T}'$ is strictly finer than $\mathcal{T}$, what can you say about the corresponding subspace topologies on the subset $Y$ of $X$?

We can say that the subspace topologies on the subset $Y$ of $X$ inherited from $\mathcal{T}'$ is strictly finer than that inherited from $\mathcal{T}$.

To see this, let's look into the definitions of the respective subspace topologies inherited from $\mathcal{T}$ and $\mathcal{T}'$; we denote them by $\mathcal{T}_Y := \{ Y \cap U : U \in \mathcal{T} \}$ and $\mathcal{T}'_Y := \{ Y \cap V : V \in \mathcal{T}' \}$.

Since $\mathcal{T}'$ is strictly finer than $\mathcal{T}$, $V \in \mathcal{T}' \implies V \in \mathcal{T}$ but the converse is in general not true. It follows that $Y \cap V \in \mathcal{T}'_Y \implies Y \cap V \in \mathcal{T}_Y$ but the converse is also not true in general. This shows that $\mathcal{T}'_Y$ is strictly finer than $\mathcal{T}_Y$.

4. A map $f : X \to Y$ is said to be an open map if for every open set $U$ of $X$, the set $f(U)$ is open in $Y$. Show that $\pi_1 : X \times Y \to X$ and $\pi_2 : X \times Y \to Y$ are open maps.

(Note: $\pi_1$ and $\pi_2$ are projection maps of $X \times Y$ onto its first and second factors respectively.)

For every open set of $X \times Y$ (under the product topology), it is generated by a basis given by sets in the form of $U \times V$, where $U$ and $V$ are open respectively in $X$ and $Y$.

Consider the images $\pi_1(U \times V)$ and $\pi_2(U \times V)$. Indeed, by the definition of function images as well as $\pi_1$ and $\pi_2$,

$$\pi_1(U \times V) = \{x \in X : (x, y) \in U \times V \} = U, \text{ and } \pi_2(U \times V) = \{y \in Y : (x, y) \in U \times V \} = V.$$

Since for both $\pi_1$ and $\pi_2$, the images of the basis elements of $X \times Y$ are open (respectively in $X$ and $Y$), the same applies for every open set of $X \times Y$ (since images preserve arbitrary unions; as images do not preserve intersections in general, for finite intersections, some work needs to be done: note that if $U_1,\ldots,U_n \subseteq X$ and $V_1,\ldots,V_n \subseteq Y$ are open, then $\bigcap_{i=1}^{n}(U_i \times V_i)=\left(\bigcap_{i=1}^{n}U_i\right) \times \left(\bigcap_{i=1}^{n}V_i\right)$, with first and second factors being open as they both are finite intersections of open sets, and this is enough to apply the same argument as above to show that its images under $\pi_1$ and $\pi_2$ are open), so we can conclude that they are open maps.

6. Show that the countable collection

$$\{(a,b) \times (c,d) : a < b \text{ and } c < d \text{ and } a,b,c,d \text{ are rational}\}$$

is a basis for $\mathbb{R}^2$.

(Note: $(a,b)$ denotes $\{x \in \mathbb{R} : a < x < b\}$, not Cartesian coordinates.)

Using Theorem 15.1, it suffices to prove that the open intervals $(x,y)$, where $x < y$ and $x$ and $y$ are rational numbers, form a basis for $\mathbb{R}$, which is already shown in Exercise 13.8(a).

Exercise after §17

1. Let $\mathcal{C}$ be a collection of subsets of the set $X$. Suppose that $\emptyset$ and $X$ are in $\mathcal{C}$, and that finite unions and arbitrary intersections of elements of $\mathcal{C}$ are in $\mathcal{C}$. Show that

$$\mathcal{T} = \{ X \backslash C : C \in \mathcal{C} \}$$

is a topology on $X$.

Indeed, since $X \backslash X = \emptyset$ and $X \backslash \emptyset = X$, whence $\emptyset$ and $X$ are in $\mathcal{C}$, so $\emptyset$ and $X$ are in $\mathcal{T}$.

Let $\{C_\alpha\}$ be an arbitrarily indexed family of sets in $\mathcal{C}$, then since

$$\bigcup_{\alpha}(X \backslash C_\alpha)=X \backslash \bigcap_{\alpha}(C_\alpha),$$

and arbitary intersections of $\mathcal{C}$ are in $\mathcal{C}$, it follows that arbitrary unions of elements in $\mathcal{T}$ are in $\mathcal{T}$.

Let $C_1,\ldots,C_n$ be a finite number of elements in $\mathcal{C}$, then since

$$\bigcap_{i=1}^{n}(X \backslash C_i) = X \backslash \bigcup_{i=1}^{n}(C_i),$$

and finite unions of elements of $\mathcal{C}$ are in $\mathcal{C}$, it follows that finite intersections of elements in $\mathcal{T}$ are in $\mathcal{T}$.

Thus, $\mathcal{T}$ is a topology on $X$.

2. Show that if $A$ is closed in $Y$ and $Y$ is closed in $X$, then $A$ is closed in $X$.

Suppose that $A$ is closed in $Y$ and $Y$ is closed in $X$, then $Y \backslash A$ and $X \backslash Y$ are open.

Note that $(X \backslash Y) \cup (Y \backslash A) = (X \cup Y) \backslash (Y \cap A)$ (why?). As $A$ is a subset of $Y$ and $Y$ is a subset of $X$, it follows that $Y \cap A = A$ and $X \cup Y = X$, so we can say that $X \backslash A = (X \backslash Y) \cup (Y \backslash A)$. Since $X \backslash A$ is a union of open sets, it is also open. Therefore, $A$ is closed in $X$ by definition.

Alternatively, using Theorem 17.2, since $A$ is closed in $Y$, it equals the intersection of a closed set of $X$ with $Y$. Meanwhile, $Y$ itself is closed in $X$, so this means that $A$ is an intersection of closed sets in $X$. It follows from Theorem 17.1 that $A$ is closed in $X$.

3. Show that if $A$ is closed in $X$ and $B$ is closed in $Y$, then $A \times B$ is closed in $X \times Y$.

Suppose that $A$ is closed in $X$ and $B$ is closed in $Y$, then $X \backslash A$ is open in $X$ and $Y \backslash B$ is open in $Y$. This implies that $(X \backslash A) \times Y$ and $X \times (Y \backslash B)$ are open in $X \times Y$.

Note that $(X \times Y) \backslash (A \times B) = ((X \backslash A) \times Y) \cup (X \times (Y \backslash B))$ (why?), so it is open. Therefore, $A \times B$ is closed in $X \times Y$.

4. Show that if $U$ is open in $X$ and $A$ is closed in $X$, then $U \backslash A$ is open in $X$, and $A \backslash U$ is closed in $X$.

Suppose that $U$ is open in $X$ and $A$ is closed in $X$, then $X \backslash A$ is open in $X$.

Note that $U \backslash A = (X \backslash A) \cap U$, so it is open in $X$.

Note that $X \backslash (A \backslash U) = (X \backslash A) \cup U$, so it is open in $X$, and thus $A \backslash U$ is closed in $X$.

5. Let $X$ be an ordered set in the order topology. Show that $\overline{(a,b)} \subseteq [a,b]$. Under what conditions does equality hold?

Suppose that $x \in \overline{(a,b)}$, we show that $x \in [a,b] \iff a \leqslant x \leqslant b$. From Theorem 17.5, we know that $x \in \overline{(a,b)}$ if and only if every open set that contains $x$ intersects $(a,b)$.

By contrapositivity, we prove that if $x \notin [a,b] \iff x < a \text{ or } x > b$, then $x \notin \overline{(a,b)}$. If $x < a$, then we obtain an open set, specifically $\{y \in X : y < a \}$, that contains $x$ but does not intersect $(a,b)$. This implies that $x \notin \overline{(a,b)}$. If $x > b$, then we obtain an open set, specifically $\{y \in X : y > b \}$, that contains $x$ but does not intersect $(a,b)$, which again implies that $x \notin \overline{(a,b)}$. This thus concludes the proof.

For the equality to hold, by Theorem 17.6, $a$ and $b$ need to be the only (two) points in $X$ that are limit points of $(a,b)$ but not in $(a,b)$.

7. Criticise the following 'proof' that $\overline{\bigcup A_\alpha} \subseteq \bigcup \overline{A}_\alpha$: if $\{A_\alpha\}$ is a collection of sets in $X$ and if $x \in \overline{\bigcup A_\alpha}$, then every neighbourhood $U$ of $x$ intersects $\bigcup A_\alpha$. Thus $U$ must intersect some $A_\alpha$, so that $x$ must belong to the closure of some $A_\alpha$. Therefore, $x \in \bigcup \overline{A}_\alpha$.

Let $U$ be a neighbourhood of $x \in \overline{\bigcup A_\alpha}$. $U$ intersecting some $A_\alpha$ does not mean that $x$ must belong to the closure of some $A_\alpha$.

For a counterexample, consider the standard topology on $\mathbb{R}$. Define $A_n$ by $\left(\frac{1}{n+1},\frac{1}{n}\right)$ where $n \in \mathbb{Z}_+$. It follows that $\overline{\bigcup A_n} =[0,1]$. Take $0 \in \overline{\bigcup A_n}$. Indeed, every neighbourhood $V$ of $0$ intersects some $A_n$ by Theorem 17.5, but the closure of any $A_n$ is given by $\left[\frac{1}{n+1},\frac{1}{n}\right]$, and $0$ is not in any of them.

10. Show that every order topology is Hausdorff.

Let $X$ be an ordered set in the order topology, and $x_1, x_2 \in X$ such that $x_1 < x_2$ without loss of generality.

If there exists some $y \in X$ such that $x_1 < y < x_2$, then let $U = \{x \in X : x < y\}$ and $V = \{x \in X : x > y\}$. We can see that $U$ is a neighbourhood of $x_1$, $V$ is a neighbourhood of $x_2$ and $U \cap V = \emptyset$.

If there is no such $y$, then let $U = \{ x \in X : x < x_2 \}$ and $V = \{ x \in X : x > x_1 \}$. Similarly, $U$ is a neighbourhood of $x_1$, $V$ is a neighbourhood of $x_2$ and $U \cap V = \emptyset$.

11. Show that the product of two Hausdorff spaces is Hausdorff.

Let $X \times Y$ be the product topology constructed from two Hausdorff spaces $X$ and $Y$. Let $(x_1,y_1) , (x_2,y_2) \in X \times Y$ be distinct, then three cases need to be considered.

Case I: $x_1=x_2 := x$ and $y_1 \neq y_2$. Let $U_x$ be a neighbourhood of $x$ in $X$. Let $U_{y_1}$ be a neighbourhood of $y_1$ and $U_{y_2}$ be a neighbourhood of $y_2$ in $Y$ such that $U_{y_1} \cap U_{y_2} = \emptyset$ due to $Y$ being Hausdorff. It follows that $(U_x \times U_{y_1}) \cap (U_x \times U_{y_2}) = U_x \times (U_{y_1} \cap U_{y_2}) = \emptyset$, whence $U_x \times U_{y_1}$ and $U_x \times U_{y_2}$ are respectively neighbourhoods of $(x_1,y_1)$ and $(x_2,y_2)$.

Case II: $x_1 \neq x_2$ and $y_1=y_2 := y$. Let $U_{x_1}$ and $U_{x_2}$ be respectively neighbourhoods of $x_1$ and $x_2$ in $X$ such that $U_{x_1} \cap U_{x_2}=\emptyset$ due to $X$ being Hausdorff. Let $U_y$ be a neighbourhood of $y$ in $Y$. It follows that $(U_{x_1} \times U_y) \cap (U_{x_2} \times U_y) = (U_{x_1} \cap U_{x_2}) \times U_y = \emptyset$, whence $U_{x_1} \times U_y$ and $U_{x_2} \times U_y$ are respectively neighbourhoods of $(x_1,y_1)$ and $(x_2,y_2)$.

Case III: $x_1 \neq x_2$ and $y_1 \neq y_2$. Let $U_{x_1}$ and $U_{x_2}$ be respectively neighbourhoods of $x_1$ and $x_2$ in $X$ such that $U_{x_1} \cap U_{x_2} = \emptyset$ due to $X$ being Hausdorff. Let $U_{y_1}$ and $U_{y_2}$ be respectively neighbourhoods of $y_1$ and $y_2$ in $Y$ such that $U_{y_1} \cap U_{y_2} = \emptyset$ due to $Y$ being Hausdorff. It follows that $(U_{x_1} \times U_{y_1}) \cap (U_{x_2} \times U_{y_2}) = (U_{x_1} \cap U_{x_2}) \times (U_{y_1} \cap U_{y_2}) = \emptyset$, whence $U_{x_1} \times U_{y_1}$ and $U_{x_2} \times U_{y_2}$ are respectively neighbourhoods of $(x_1,y_1)$ and $(x_2,y_2)$.

Combining all the cases above tells us that the product of two Hausdorff spaces is Hausdorff.

12. Show that a subspace of a Hausdorff space is Hausdorff.

Let $X$ be a Hausdorff space and $Y$ be a subset of $X$ equipped with the subspace topology inherited from $X$.

Let $y_1, y_2 \in Y$ such that $y_1 \neq y_2$. Note that $y_1, y_2 \in X$ as well and $X$ is Hausdorff, so there exists a neighbourhood $U$ of $y_1$ and $V$ of $y_2$ in $X$ such that $U \cap V = \emptyset$. It follows that $Y \cap U$ and $Y \cap V$ are neighbourhoods of $y_1$ and $y_2$ respectively in $Y$ as a subspace such that $(Y \cap U) \cap (Y \cap V)= Y \cap (U \cap V) = \emptyset$.

Thus, a subspace of a Hausdorff space is Hausdorff.

13. Show that $X$ is Hausdorff if and only if the diagonal $\Delta = \{ (x,x) : x \in X\}$ is closed in $X \times X$.

Suppose that $X$ is Hausdorff. We claim that $(X \times X) \backslash \Delta$ is open in $X \times X$, so that $\Delta$ is closed in $X \times X$. Note that for any element in $(X \times X) \backslash \Delta$, the first and second coordinates are not equal.

Since for $x_1, x_2 \in X$ such that $x_1 \neq x_2$, there exists a neighbourhood $U$ of $x_1$ and a neighbourhood $V$ of $x_2$ such that $U \cap V = \emptyset$. It follows that $U \times V$ is open and does not contain $(x_1,x_1)$ nor $(x_2,x_2)$. Taking unions of such $U \times V$ over all distinct pairs of elements in $X$, it follows that $(X \times X) \backslash \Delta$ can be expressed as a union of open sets, so it is open.

Conversely, if $\Delta$ is closed in $X \times X$, then $(X \times X) \backslash \Delta$ is open in $X \times X$. Take a point $(x_1,x_2) \in (X \times X) \backslash \Delta$, then $x_1 \neq x_2$. Since $(X \times X) \backslash \Delta$ is open, there exists a neighbourhood $U \subseteq X$ of $x_1$ and a neighbourhood $V \subseteq X$ of $x_2$ such that $U \times V \subseteq (X \times X) \backslash \Delta$. This implies that $U$ and $V$ do not have common elements, i.e. $U \cap V = \emptyset$; if it did have one, say, $x'$, then $(x',x') \in (X \times X) \backslash \Delta$, a contradiction. Therefore, $X$ is Hausdorff.

14. In the finite complement topology on $\mathbb{R}$, to what point or points does the sequence $x_n=\frac{1}{n}$ converge?

Recall that a sequence $x_n$ converges to $x$ if there is a positive integer $N$ that depends on each neighbourhood $U$ of $x$ such that $x_n \in U$ whenever $n \geqslant N$.

Denote the finite complement topology on $\mathbb{R}$ by $\mathcal{T}_f$. Each open set $V$ of $\mathcal{T}_f$ must satisfy that $\mathbb{R} \backslash V$ is finite or all of $\mathbb{R}$.

Since every open set under $\mathcal{T}_f$ contains either all, or, all but finitely many values of $x_n$ (because each non-empty open set excludes only finitely many elements of $\mathbb{R}$), we can say that the sequence $x_n$ converges to every real number in $\mathbb{R}$.

To see this, we fix some $x \in \mathbb{R}$. For each neighbourhood $U$ of $x$ (so $U$ is non-empty), as the number of elements in $\mathbb{R}$ that $U$ excludes is finite, it suffices to choose $N$ to be $N'+1$ where $N'$ is the maximum positive integer such that $\frac{1}{N'}$ is not in $U$ if $U$ does not contain some (finitely many) numbers in the form of $\frac{1}{n}$, or choose $N$ to be $1$ if $U$ contains every number in the form of $\frac{1}{n}$.

15. Show that $T_1$ axiom is equivalent to the condition that for each pair of points of $X$, each has a neighbourhood not containing the other (point).

Let $X$ be a topological space that satisfies the $T_1$ axiom, then every one-point subsets of $X$ is closed. Take $x_1, x_2 \in X$ such that $x_1 \neq x_2$, then $X \backslash \{x_1\}$ is a neighbourhood of $x_2$ that does not contain $x_1$ whereas $X \backslash \{x_2\}$ is a neighbourhood of $x_1$ that does not contain $x_2$, thus proving one direction of the equivalence.

Conversely, suppose that $x, x' \in X$ with $x \neq x'$ such that $x'$ has a neighbourhood $U$ that does not contain $x$ (and vice versa), then $U \cap \{x\} = \emptyset$. By Theorem 17.5, this implies that $x'$ is not in the closure of $\{x\}$. Since the choice of $x'$ is arbitrary, it follows that the closure of $\{x\}$ is $\{x\}$ itself, so $\{x\}$ is closed. Since the choice of $x \in X$ is also arbitrary, every one-point subset of $X$ is closed.

17. Consider the lower limit topology on $\mathbb{R}$ and the topology given by the basis $\mathcal{C}$ of Exercise 8 of §13. Determine the closures of the intervals $A = (0,\sqrt{2})$ and $B = (\sqrt{2},3)$ in these two topologies.

Using Theorem 17.6, it suffices to find all the limit points of $A$ and $B$ under the two respective topologies.

Under the lower limit topology on $\mathbb{R}$, if $x$ is a limit point of $A$, then every neighbourhood $U$ of $x$ intersects $A$ in a point other than $x$. This implies that $U$ must contain some half-open interval in the form of $[a,b)$ that contains some point of $A$. If $x \notin A$, it follows that $x = 0$ necessarily because if $x=-\varepsilon$ where $\varepsilon > 0$ is arbitrary, then we obtain a neighbourhood $\left[-\varepsilon,-\frac{\varepsilon}{2}\right)$ that does not intersect $A$. Similarly, if $x = \sqrt{2}+\delta$ where $\delta \geqslant 0$, then $[\sqrt{2}+\delta,\infty)$ is a neighbourhood that does not intersect $A$. On the other hand, if $x = 0$, then every neighbourhood must contain some points in between $0$ and $\sqrt{2}$, whether or not $0$ is the lower endpoint of a half-open interval. Therefore, the closure of $A$ under this topology is $[0,\sqrt{2})$. By a similar argument, the closure of the $B$ under this topology is $[\sqrt{2},3)$.

Under the topology given by the basis $\mathcal{C}$, if $x$ is a limit point of $A$, then every neighbourhood $U$ of $x$ must contain some half-open interval in the form of $[a,b)$ where $a$ and $b$ are rational that contains some point of $A$. If $x \notin A$, then $x=0$ or $x=\sqrt{2}$ necessarily because any neighbourhood that contains $0$ (resp. $\sqrt{2}$) must contain some points in between $0$ and $\sqrt{2}$, since $\sqrt{2}$ cannot be the lower endpoint of a half-open interval under the $\mathcal{C}$-generated topology. This implies that the closures of $A$ and $B$ under this topology are respectively $[0,\sqrt{2}]$ and $[\sqrt{2},3]$.

19. If $A \subseteq X$, we define the boundary of $A$ by the equation

$$\mathrm{Bd} \; A = \overline{A} \cap \overline{(X \backslash A)}.$$

(a) Show that $\mathrm{Int} A$ and $\mathrm{Bd} \; A$ are disjoint, and $\overline{A}=\mathrm{Int} A \cup \mathrm{Bd} \; A$.

(b) Show that $\mathrm{Bd} \; A = \emptyset \iff A \text{ is both open and closed}$.

(c) Show that $U \text{ is open } \iff \mathrm{Bd}\; U=\overline{U} \backslash U$.

(d) If $U$ is open, is it true that $U = \mathrm{Int}(\overline{U})$? Justify your answer.

21. (Kuratowski) Consider the collection of all subsets $A$ of the topological space $X$. The operations of closure $A \to \overline{A}$ and complementation $A \to X \backslash A$ are functions from this collection to itself.

(a) Show that starting with a given set $A$, one can form no more than 14 distinct sets by applying these two operations successively.

(b) Find a subset $A$ of $\mathbb{R}$ (in its usual topology) for which the maximum of 14 is obtained.

Exercise after §18

1. Prove that for functions $f : \mathbb{R} \to \mathbb{R}$, the $\varepsilon$-$\delta$ definition of continuity implies the open set definition.

note

The arguments to show Exercise 18.1 can be modified to prove the same for real multivariate functions, though this may require the notion of box topology to define the standard topology on $\mathbb{R}^n$.

Let $x_0 \in \mathbb{R}$. Suppose that for a every $\varepsilon > 0$, there exists $\delta > 0$ such that for $x \in \mathbb{R}$ such that $|x-x_0|<\delta$, we have $|f(x)-f(x_0)|<\varepsilon$.

Let $U$ be an open subset of $\mathbb{R}$ that contains $f(x_0)$, then it contains some open interval about $f(x_0)$, which we can express in the form $(f(x_0)-\varepsilon, f(x_0)+\varepsilon)$, for some $\varepsilon>0$. For that open interval, there then exists a corresponding $\delta$ such that $| x - x_0 | < \delta$ where $x \in \mathbb{R}$, and $\{x \in \mathbb{R} : |x-x_0|<\delta \} = (x_0-\delta,x_0+\delta)$ is open in $\mathbb{R}$. This implies that $f^{-1}((f(x_0)-\varepsilon,f(x_0)+\varepsilon)) = (x_0-\delta,x_0+\delta)$ is open.

Note that $U = (U \cap f(\mathbb{R})) \cup (U \backslash f(\mathbb{R}))$ (why?). For points in $U \cap f(\mathbb{R})$, we can apply the argument above to apply an open subset of the domain. For subsets of $U \backslash f(\mathbb{R})$, their inverse images are empty, which is also open.

Since arbitrary unions of open sets are open, and inverse images preserve unions, if follows that $f^{-1}(U)$ is an open subset of $\mathbb{R}$, thus proving the open set definition.

2. Suppose that $f : X \to Y$ is continuous. If $x$ is a limit point of the subset $A$ of $X$, is it necessarily true that $f(x)$ is a limit point of $f(A)$?

Yes. We prove this assertion by contraposition.

Suppose that $f(x)$ is not a limit point of $f(A)$. There then exists a neighbourhood $U \subseteq Y$ of $f(x)$ that does not intersect $f(A)$ in a point different from $f(x)$. Note that $f(X)$ must intersect $U$ since $f(x) \in U$, so since $f : X \to Y$ is continuous, $f^{-1}(U)$ must be a neighbourhood of $x$. However, since $U$ does not contain any point in $f(A)$ other than $f(x)$ itself, it follows that $f^{-1}(U)$ does not contain any point in $A$ other than $x$ itself. This implies that $x$ is not a limit point of $A$.

4. Given $x_0 \in X$ and $y_0 \in Y$, show that the maps $f : X \to X \times Y$ and $g : Y \to X \times Y$ defined by

   $$f(x) = (x, y_0) \quad \text{and} \quad g(y)=(x_0,y)$$

   are imbeddings.

We prove the desired statement through a number of steps.

1. We claim that $f$ and $g$ are injective.

   Indeed, for $x_1, x_2 \in X$ and $y_1, y_2 \in Y$, $f(x_1)=f(x_2) \implies (x_1,y_0)=(x_2,y_0) \implies x_1=x_2$ and $g(y_1)=g(y_2) \implies (x_0,y_1)=(x_0,y_2) \implies y_1=y_2$.

2. We claim that $f$ and $g$ are continuous.

   Using Theorem 18.4, it suffices to prove that the respective coordinate functions of $f$ and $g$ are continuous. If we write $f(x)=(f_1(x),f_2(x))$ and $g(y)=(g_1(y),g_2(y))$, then $f_1(x)=x$, $f_2(x)=y_0$, $g_1(y)=x_0$ and $g_2(y)=y$. Since $f_1$ and $g_2$ are identity functions, they are continuous as for every open subset $U$ of $X$ (resp. $Y$), we have $f^{-1}(U)=U$. Since $f_2$ and $g_1$ are constant functions, they are also continuous (by Theorem 18.2).

3. We claim that the functions $f' : X \to f(X)$ and $g' : Y \to g(Y)$ obtained by restricting their codomains to their respective images are homeomorphisms.

   Indeed, the inverses of $f'$ and $g'$ are respectively given by the projections $f'^{-1}(x,y)=x$ and $g'^{-1}(x,y)=y$. We know that projections are continuous, so $f'$ and $g'$ are homeomorphisms.

This concludes the proof that $f$ and $g$ are embeddings.

5. Show that the subspace $(a,b)$ of $\mathbb{R}$ is homeomorphic with $(0,1)$ and the subspace $[a,b]$ of $\mathbb{R}$ is homeomorphic with $[0,1]$.

Define the function $f : (a,b) \to (0,1)$ by $f(x)=\dfrac{x-a}{b-a}$. The inverse is then given by $f^{-1}(x)=x(b-a)+a$. Since $f$ is a rational function whereas $f^{-1}$ is a polynomial function, they are both continuous. Thus, $(a,b)$ and $(0,1)$ are homeomorphic. A very similar argument can be used to show that $[a,b]$ and $[0,1]$ are homeomorphic.

6. Find a function $f : \mathbb{R} \to \mathbb{R}$ that is continuous at precisely one point.

Define the function $f : \mathbb{R} \to \mathbb{R}$ by

$$f(x)=\begin{cases} x & \text{if } x \in \mathbb{Q}, \\ 0 & \text{if } x \notin \mathbb{Q}. \end{cases}$$

We claim that $f$ is continuous at $x=0$ and nowhere else. Due to Exercise 18.1, we use the $\varepsilon$-$\delta$ definition here.

Let $\varepsilon>0$ be arbitrary. Set $\delta=\varepsilon$. It follows that for all $x \in \mathbb{R}$ such that $0<|x-0|=|x|<\delta$, we have that $|f(x)-f(0)|=\begin{cases} |x| & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \notin \mathbb{Q} \end{cases} < \delta=\varepsilon$. Thus, $f$ is continuous at $x = 0$.

Now let $x_0 \neq 0$. Take $\varepsilon = \frac{1}{2}|x_0|$. It follows that for all $\delta>0$, due to the density of rationals and irrationals in the real numbers, there must exist some $x \notin \mathbb{Q}$ such that $0<|x-x_0|<\delta$ if $x_0 \in \mathbb{Q}$, and vice versa. This implies that $|f(x)-f(x_0)|=\begin{cases} |x_0| & \text{if } x_0 \in \mathbb{Q}, x \notin \mathbb{Q} \\ |x| & \text{if } x \in \mathbb{Q}, x_0 \notin \mathbb{Q} \end{cases} \geqslant \varepsilon$ holds by choosing $x$ such that $|x|>|x_0|$, thus $f$ is not continuous at $x_0$.

8. Let $Y$ be an ordered set in the order topology. Let $f,g : X \to Y$ be continuous.

(a) Show that the set $\{x : f(x) \leqslant g(x)\}$ is closed in $X$.

(b) Let $h : X \to Y$ be the function

$$h(x)=\min\{f(x),g(x)\}.$$

Show that $h$ is continuous. [Hint: Use the pasting lemma.]

(a) We show that its complement, given by the set $W := \{x : f(x)>g(x)\}$, is open in $X$. Take an element $x_0 \in W$, then $f(x_0)>g(x_0)$. Since $f$ is continuous, we can take a neighbourhood $U$ of $f(x_0)$ such that $U \subseteq g(A)$ and $f(x_0)>u$ for all $u \in U$; there then exists a neighbourhood $U'$ of $x_0$ with $f(U') \subseteq U$. Note that this implies that $U' \subseteq W$. Similarly, since $g$ is continuous, we can take a neighbourhood $V$ of $g(x_0)$ such that $V \subseteq f(A)$ and $v>g(x_0)$ for all $v \in V$; there then exists a neighbourhood $V'$ of $x_0$ with $f(V') \subseteq V$, so that $V' \subseteq W$.

Taking unions of such $U'$ and $V'$ over all $x_0 \in W$, we obtain the entire $W$. This implies that $W$ is a union of open sets, so it is open, thus $\{x : f(x) \leqslant g(x)\}$ is closed in $X$.

(Alternative proofs.)

(b) Note that $f : X \to Y$ and $g : X \to Y$ are continuous. Note as well that from (a), $U := \{x : f(x) \leqslant g(x)\}$ and $V := \{x : f(x) \geqslant g(x) \}$ are closed subsets of $X$ (since $f$ and $g$ as continuous functions are arbitrary).

Consider the restriction of $f$ to $U$ and the restriction of $g$ to $V$. It follows that $U \cup V = X$, as for any $x \in X$, exactly one of $f(x) \leqslant g(x)$ or $g(x) \leqslant f(x)$ holds. We also have that $f(x)=g(x)$ for any $x \in U \cap V$, since $U \cap V$ is precisely $\{x : f(x)=g(x)\}$. By Pasting Lemma, these two restrictions combine to give a continuous function $h : X \to Y$ defined by $h(x)=\begin{cases} f(x) & \text{if } x \in U, \\ g(x) & \text{if } x \in V \end{cases}$, which in this case is equivalent to $h(x)=\min\{f(x),g(x)\}$.

10. Let $f : A \to B$ and $g : C \to D$ be continuous functions. Let us define a map $f \times g : A \times C \to B \times D$ by the equation

    $$(f \times g)(a,c)=(f(a),g(c)).$$

    Show that $f \times g$ is continuous.

Note that any open subset of $B \times D$ is generated by products in the form of $U \times V$, where $U$ and $V$ are open in $B$ and $D$ respectively. Since $f : A \to B$ and $g : C \to D$ are continuous, we have that $f^{-1}(U)$ is open in $A$ and $g^{-1}(V)$ is open in $C$.

Since the inverse image $(f \times g)^{-1}(U \times V)$ equals $f^{-1}(U) \times g^{-1}(V)$, it is open in $A \times C$. Therefore, $f \times g$ is continuous.

11. Let $F : X \times Y \to Z$. We say that $F$ is continuous in each variable separately if for each $y_0$ in $Y$, the map $h : X \to Z$ defined by $h(x)=F(x, y_0)$ is continuous, and for each $x_0$ in $X$, the map $k : Y \to Z$ defined by $k(y)=F(x_0, y)$ is continuous. Show that if $F$ is continuous, then $F$ is continuous in each variable separately.

Let $U$ be an open subset of $Z$, then $h^{-1}(U) = \{ x \in X : (x,y_0) \in F^{-1}(U) \}$ where $y_0 \in Y$ is arbitrarily fixed. Since $F : X \times Y \to Z$ is continuous, the inverse image $F^{-1}(U)$ is open, so $F^{-1}(U)$ is generated by products in the form of $V \times W$ where $V$ and $W$ are open in $X$ and $Y$ respectively. This implies that $h^{-1}(U)$ is generated by some open sets in $X$, and hence open in $X$. Thus, $h$ is continuous.

Similarly, arbitrarily fixing some $x_0 \in X$, we have that $k^{-1}(U) = \{ y \in Y : (x_0,y) \in F^{-1}(U) \}$, so $k^{-1}(U)$ is generated by some open sets in $Y$, and thus open in $Y$. Therefore, $k$ is continuous.

12. Let $F : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be defined by the equation

$$F(x,y)=\begin{cases} \frac{xy}{x^2+y^2} & \text{if } (x,y) \neq (0,0), \\ 0 & \text{if } (x,y)=(0,0)\end{cases}.$$

(a) Show that $F$ is continuous in each variable separately.

(b) Compute the function $g : \mathbb{R} \to \mathbb{R}$ defined by $g(x)=F(x,x)$.

(c) Show that $F$ is not continuous.

note

The function given by Exercise 18.12 is a classic example in the study of multivariable calculus to demonstrate a multivariate function that is not differentiable at the origin. In fact, this exercise shows that it is not even continuous at the origin to begin with, so it is certainly not differentiable at the origin.

(a) Let $x_0 \in \mathbb{R}$ and $y_0 \in \mathbb{R}$ be arbitrarily fixed. Define the maps $h : \mathbb{R} \to \mathbb{R}$ and $k : \mathbb{R} \to \mathbb{R}$ in a similar manner as Exercise 18.11. We use the $\varepsilon$-$\delta$ definition to prove the continuity of $h$ and $k$ thanks to Exercise 18.1.

Let $\varepsilon>0$ be arbitrary and fix some $x' \in \mathbb{R}$. Take $\delta=\dfrac{|y_0|\varepsilon}{2}$. It follows that when for $x \in \mathbb{R}$ such that $0 < |x-x'| < \delta$, we have that $|h(x)-h(x')|=\left|\dfrac{x y_0}{x^2+y_0^2}-\dfrac{x'y_0}{x'^2+y_0^2}\right|=|y_0| \cdot \left|\dfrac{x(x'^2+y_0^2)-x'(x^2+y_0^2)}{(x^2+y_0^2)(x'^2+y_0^2)}\right| \leqslant |y_0|\left|\dfrac{y_0^2(x-x')}{y_0^4}\right|+|y_0|\left|\dfrac{xx'(x'-x)}{y_0^4}\right| < \dfrac{\delta}{|y_0|}+\dfrac{\delta}{|y_0|} = \varepsilon$. Thus, $h$ is continuous.

Using the same argument as above, by replacing $h$ with $k$, $y_0$ with $x_0$, $x'$ with $y'$ and $x$ with $y$, we then prove the continuity of $k$.

(b) The function $g : \mathbb{R} \to \mathbb{R}$ defined by $g(x)=F(x,x)$ is given by

$$g(x)=\begin{cases} \frac{1}{2} & \text{if } x \neq 0, \\ 0 & \text{if } x=0. \end{cases}$$

(c) Let $U$ be the open interval $\left(-\frac{1}{3},\frac{1}{3}\right)$ in $\mathbb{R}$. Note that it is a neighbourhood of $0$ in $\mathbb{R}$.

If $F$ were continuous, then $F^{-1}(U)$ is open, so every point $x$ of $F^{-1}(U)$ has a neighbourhood $V$ that is contained in $F^{-1}(U)$. However, this is not true because if we consider the point $(0,0) \in F^{-1}(U)$, which makes sense as $F(0,0)=0 \in U$, then its corresponding neighbourhood that is contained in $F^{-1}(U)$ must contain some point in the form of $(x_0,x_0)$ where $x_0 \neq 0$, but $F(x_0,x_0)=g(x_0)=\frac{1}{2} \notin U$, which leads to a contradiction.

Therefore, $F$ is not continuous.

13. Let $A \subseteq X$; let $f : A \to Y$ be continuous; let $Y$ be Hausdorff. Show that if $f$ may be extended to a continuous function $g : \overline{A} \to Y$, then $g$ is uniquely determined by $f$.

Exercise after §19

1. Prove Theorem 19.2.

2. Prove Theorem 19.3.

3. Prove Theorem 19.4.

4. Show that $(X_1 \times \cdots X_{n-1}) \times X_n$ is homeomorphic with $X_1 \times \cdots \times X_n$.

5. One of the implications stated in Theorem 19.6 holds for the box topology. Which one?

6. Let $\mathbf{x}_1$, $\mathbf{x}_2$ be a sequence of the points of the product space $\prod X_\alpha$. Show that this sequence converges to the points $\mathbf{x}$ if and only if the sequence $\pi_\alpha(\mathbf{x}_1)$, $\pi_\alpha(\mathbf{x}_2)$, ... converges to $\pi_\alpha(\mathbf{x})$ for each $\alpha$. Is this fact true if one uses the box topology instead of the product topology?

8. Given sequences $(a_1,a_2,\ldots)$ and $(b_1,b_2,\ldots)$ of real numbers with $a_i > 0$ for all $i$, define $h : \mathbb{R}^\omega \to \mathbb{R}^\omega$ by the equation

$$h((x_1,x_2,\ldots)) = (a_1 x_1+b_1, a_2 x_2+b_2, \ldots).$$

Show that if $\mathbb{R}^\omega$ is given the product topology, $h$ is a homeomorphism of $\mathbb{R}^\omega$ with itself. What happens if $\mathbb{R}^\omega$ is given the box topology?

9. Show that the choice axiom is equivalent to the statement that for any indexed family $\{A_\alpha\}_{\alpha \in J}$ of nonempty sets, with $J \neq 0$, the cartesian product

$$\prod_{\alpha \in J}A_\alpha$$

is not empty.

Exercise after §20

1. (a) In $\mathbb{R}^n$, define

$$d'(\mathbf{x},\mathbf{y})=|x_1-y_1|+\cdots+|x_n-y_n|.$$

Show that $d'$ is a metric that induces the usual topology of $\mathbb{R}^n$. Sketch the basis elements under $d'$ when $n=2$.

(b) More generally, given $p \geqslant 1$, define

$$d'(\mathbf{x},\mathbf{y})=\left[\sum_{i=1}^{n}|x_i-y_i|^p\right]^{\frac{1}{p}}$$

for $\mathbf{x} , \mathbf{y} \in \mathbb{R}^n$. Assume that $d'$ is a metric. Show that it induces the usual topology on $\mathbb{R}^n$.

2. Show that $\mathbb{R} \times \mathbb{R}$ in the dictionary order topology is metrizable.

3. Let $X$ be a metric space with metric $d$.

(a) Show that $d : X \times X \to \mathbb{R}$ is continuous.

(b) Let $X'$ denote a space having the same underlying set as $X$. Show that if $d' : X \times X \to \mathbb{R}$ is continuous, then the topology of $X'$ is finer than the topology of $X$.

note

One can summarise the result of this exercise as follows: if $X$ has a metric $d$, then the topology induced by $d$ is the coarsest topology relative to which the function $d$ is continuous.

9. Show that the Euclidean metric $d$ on $\mathbb{R}^n$ is a metric, as follows: if $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$ and $c \in \mathbb{R}$, define

$$\mathbf{x}+\mathbf{y}=(x_1+y_1,\ldots,x_n+y_n),$$$$c\mathbf{x}=(cx_1,\ldots,cx_n),$$$$\mathbf{x} \cdot \mathbf{y} = x_1 y_1 +\cdots + x_n y_n.$$

(a) Show that $\mathbf{x} \cdot (\mathbf{y}+\mathbf{z})=(\mathbf{x}\cdot\mathbf{y})+(\mathbf{x}\cdot\mathbf{z})$.

(b) (Cauchy-Schwarz inequality.) Show that $| \mathbf{x} \cdot \mathbf{y} | \leqslant \Vert\mathbf{x}\Vert\Vert\mathbf{y}\Vert$. [Hint: If $\mathbf{x}, \mathbf{y} \neq 0$, let $a = \frac{1}{\Vert\mathbf{x}\Vert}$ and $b = \frac{1}{\Vert\mathbf{y}\Vert}$, and use the fact that $\Vert a\mathbf{x}+b\mathbf{y}\Vert \geqslant 0$.]

(c) Show that $\Vert \mathbf{x}+\mathbf{y} \Vert \leqslant \Vert\mathbf{x}\Vert + \Vert\mathbf{y}\Vert$. [Hint: Compute $(\mathbf{x}+\mathbf{y}) \cdot (\mathbf{x}+\mathbf{y})$ and apply (b).]

(d) Verify that $d$ is a metric.

11. Show that if $d$ is a metric for $X$, then

$$d'(x,y)=\frac{d(x,y)}{1+d(x,y)}$$

is a bounded metric that gives the topology of $X$.

[Hint: If $f(x) = \dfrac{x}{1+x}$ for $x > 0$, use the mean-value theorem to show that $f(a+b)-f(b) \leqslant f(a)$.]

Following the hint, define $f(x)=\dfrac{x}{1+x}$ with $x>0$. It follows that $f'(x)=\dfrac{1}{(1+x)^2}$. Let $a,b>0$ such that $a>b$ without loss of generality. By mean-value theorem, since $b < a < a+b$, we have that $f'(a)=\dfrac{f(a+b)-f(b)}{a} \implies af'(a)=f(a+b)-f(b)$. Meanwhile, $\dfrac{a}{(1+a)^2}-\dfrac{a}{1+a}=-\dfrac{a^2}{(1+a)^2} \leqslant 0 \implies af'(a) \leqslant f(a)$, so $f(a+b) - f(b) \leqslant f(a) \implies f(a+b) \leqslant f(a)+f(b)$.

Note as well that $f(x)$ is increasing within $x>0$, as $f(x)=1-\dfrac{1}{1+x}$, whence $\dfrac{1}{1+x}$ decreases as $x$ increases and is positive.

Let $x,y,z \in \mathbb{R}$ be arbitrary. Since $d(x,y) \geqslant 0$ and the equality holds if and only if $x=y$, it follows that $1+d(x,y) \geqslant 0$ as well, so $d'(x,y) \geqslant 0$ and the equality holds if and only if $x=y$ as the numerator in $d'$ is determined solely by $d$. By definition, we can also see that $d'(x,y)=d'(y,x)$. To show the triangle inequality property of $d'$, we can apply the fact proven in the previous paragraph and let $a=d(x,y)$ and $b=d(y,z)$; note as well that $d' = f \circ d$. We then have that $d'(x,z) = f(d(x,z)) \leqslant f(d(x,y)+d(y,z)) \leqslant f(d(x,y))+f(d(y,z))=d'(x,y)+d'(y,z)$, with the first $\leqslant$ relation being a consequence of the triangle inequality of $d$ and $f$ increasing when $x>0$.

Since for any $x,y \in \mathbb{R}$, $\dfrac{d(x,y)}{1+d(x,y)} \leqslant \dfrac{1+d(x,y)}{1+d(x,y)}=1$, we conclude that $d'$ is bounded by $M=1$.

Now, we show that $d'$ gives the topology of $X$ that is equal to that induced by $d$. [TODO]

Exercise after §21

1. Let $A \subseteq X$. If $d$ is a metric for the topology of $X$, show that $d|_{A \times A}$ is a metric for the subspace topology on $A$.

2. Let $X$ and $Y$ be metric spaces with metrics $d_X$ and $d_Y$, respectively. Let $f : X \to Y$ have the property that for every pair of points $x_1, x_2$ of $X$,

$$d_Y(f(x_1),f(x_2))=d_X(x_1,x_2).$$

Show that $f$ is an embedding. It is caleld an isometric embedding of $X$ in $Y$.

3. Let $X_n$ be a metric space with metric $d_n$, for $n \in \mathbb{Z}_+$.

(a) Show that

$$\rho(x,y)=\max\{d_1(x_1,y_1), \ldots, d_n(x_n,y_n)\}$$

is a metric for the product space $X_1 \times \cdots \times X_n$.

(b) Let $\overline{d_i}=\min\{d_i,1\}$. Show that

$$D(x,y)=\sup\left\{\frac{\overline{d_i}(x_i,y_i)}{i}\right\}$$

is a metric for the product space $\prod X_i$.

5. Theorem (i.e. very useful result). Let $x_n \to x$ and $y_n \to y$ in the space $\mathbb{R}$. Then

$$x_n+y_n \to x+y,$$$$x_n-y_n \to x-y,$$$$x_n y_n \to xy,$$

and provided that each $y_n \neq 0$ and $y \neq 0$,

$$x_n/y_n \to x/y.$$

[Hint: Apply Lemma 21.4; recall from the exercises of §19 that if $x_n \to x$ and $y_n \to y$, then $(x_n,y_n) \to (x,y)$.]

6. Define $f_n : [0,1] \to \mathbb{R}$ by the equation $f_n(x)=x^n$. Show that the sequence $(f_n(x))$ converges for each $x \in [0,1]$, but that the sequence $(f_n)$ does not converge uniformly.

8. Let $X$ be a topological space and let $Y$ be a metric space. Let $f_n : X \to Y$ be a sequence of continuous functions. Let $x_n$ be a sequence of points of $X$ converging to $x$. Show that if the sequence $(f_n)$ converges uniformly to $f$, then $(f_n(x_n))$ converges to $f(x)$.

11. Prove the following standard facts about infinite series:

(a) (Monotone convergence theorem). Show that if $(s_n)$ is a bounded sequence of real numbers and $s_n \leqslant s_{n+1}$ for each $n$, then $(s_n)$ converges.

(b) Let $(a_n)$ be a sequence of real numbers; define

$$s_n=\sum_{i=1}^{n}a_i.$$

If $s_n \to s$, we say that the infinite series

$$\sum_{i=1}^{\infty}a_i$$

converges to $s$ also. Show that if $\sum a_i$ converges to $s$ and $\sum b_i$ converges to $t$, then $\sum (ca_i+b_i)$ converges to $cs+t$.

(c) Prove the comparison test for infinite series: If $|a_i| \leqslant b_i$ for each $i$, and if the series $\sum b_i$ converges, then the series $\sum a_i$ converges. [Hint: show that the series $\sum |a_i|$ and $\sum c_i$ converge, where $c_i=|a_i|+a_i$.]

(d) Given a sequence of functions $f_n : X \to \mathbb{R}$, let

$$s_n(x)=\sum_{i=1}^{n}f_i(x).$$

Prove the Weierstrass M-test for uniform convergence: If $|f_i(x)| \leqslant M_i$ for all $x \in X$ and for all $i$, and if the series $\sum M_i$ converges, then the sequence $(s_n)$ converges uniformly to a function $s$. [Hint: Let $r_n=\sum_{i=n+1}^{\infty}M_i$. Show that if $k<n$, then $|s_k(x)-s_n(x)| \leqslant r_n$; conclude that $|s(x)-s_n(x)| \leqslant r_n$.]

12. Prove continuity of the algebraic operations on $\mathbb{R}$, as follows: Use the metric $d(a,b)=|a-b|$ on $\mathbb{R}$ and the metric on $\mathbb{R}^2$ given by the equation

$$\rho((x,y),(x_0,y_0))=\max\{|x-x_0|,|y-y_0|\}.$$

(a) Show that addition is continuous. [Hint: Given $\varepsilon$, let $\delta=\varepsilon/2$ and note that $d(x+y,x_0+y_0) \leqslant |x-x_0|+|y-y_0|$.]

(b) Show that multiplication is continuous. [Hint: Given $(x_0,y_0)$ and $0 < \varepsilon < 1$, let $3\delta=\varepsilon/(|x_0|+|y_0|+1)$ and note that $d(xy,x_0y_0) \leqslant |x_0||y-y_0|+|y_0||x-x_0|+|x-x_0||y-y_0|$.]

(c) Show that the operation of taking reciprocals is a continuous map from $\mathbb{R} \backslash \{0\}$ to $\mathbb{R}$. [Hint: show the inverse image of the interval $(a,b)$ is open. Consider five cases, according as $a$ and $b$ are positive, negative, or zero.]

(d) Show that the subtraction and quotient operations are continuous.

Exercise after §22

1. Check the details of Example 3.

   i.e.

   Let $p$ be the map of the real line $\mathbb{R}$ onto the three-point set $A = \{a,b,c\}$ defined by

   $$p(x)=\begin{cases} a & \text{if } x>0, \\ b & \text{if } x<0, \\ c & \text{if } x=0. \end{cases}$$

   Check that the quotient topology on $A$ induced by $p$ is given by $\mathcal{T} := \{\emptyset, A, \{a\},\{b\}, \{a,b\}\}$.

2. (a) Let $p : X \to Y$ be a continuous map. Show that if there is a continuous map $f : Y \to X$ such that $p \circ f$ equals the identity map of $Y$ (i.e. $f$ is the right inverse [section] of $p$), then $p$ is a quotient map.

   (b) If $A \subseteq X$, a retraction of $X$ onto $A$ is a continuous map $r : X \to A$ such that $r(a)=a$ for each $a \in A$. Show that a retraction is a quotient map.

3. Let $\pi_1 : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be projection on the first coordinate. Let $A$ be the subspace of $\mathbb{R} \times \mathbb{R}$ consisting of all points $(x,y)$ for which either $x \geqslant 0$ or $y=0$ (or both); let $q : A \to \mathbb{R}$ be obtained by restricting $\pi_1$.

   Show that $q$ is a quotient map that is neither open nor closed.

5. Let $p : X \to Y$ be an open map. Show that if $A$ is open in $X$, then the map $q : A \to p(A)$ obtained by restricting $p$ is an open map.

Footnotes

1. Note that finitude is a special case of countability. ↩