Notes

The Countability Axioms

In Chapter 2, it is discussed that every metrizable space satisfies the following first countability axiom:

Definition. A space $X$ is said to have a countable basis at $x$ if there is a countable collection $\mathcal{B}$ of neighbourhoods of $x$ such that each neighbourhood of $x$ contains at least one of the elements of $\mathcal{B}$. A space that has a countable basis at each of its points is said to satisfy the first countability axiom, or to be first-countable.

It turns out that metrizable spaces are not the only family of spaces that satisfy the first countability axiom, and this is further discussed in this chapter.

Theorem 30.1 shows that it suffices to determine limit points of sets and check continuity of functions via convergent sequences if the space is first-countable.

[Theorem 30.1] Let $X$ be a topological space.

(a) Let $A$ be a subset of $X$. If there is a sequence of points of $A$ converging to $x$, then $x \in \overline{A}$; the converse holds if $X$ is first-countable.

(b) Let $f : X \to Y$. If $f$ is continuous, then for every convergent sequence $x_n \to x$ in $X$, the sequence $f(x_n)$ converges to $f(x)$. The converse holds if $X$ is first-countable.

The proof of Theorem 30.1 is basically the same as the proofs of Lemma 21.2 and Theorem 21.3 presented in Chapter 2. Even though metrizability (which is stronger than first countability) is assumed there, in the proof itself actually only first countability is applied instead of the full strength of metrizability.

There is also a more important countability axiom: the second countability axiom.

Definition. If a space $X$ has a countable basis for its topology, then $X$ is said to satisfy the second-countability axiom, or to be second-countable.

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Note that the second countability axiom is stronger than the first countability axiom: the former implies the latter. In fact, it is so strong that not even every metric space satisfies it; examples include $\mathbb{R}^\omega$ in the uniform topology (see Chapter 2, Metric Topology).

Both countability axioms are preserved under the operations of taking subspaces or countable products.

[Theorem 30.2] A subspace of a first-countable space is first-countable, and a countable product of first-countable spaces is first-countable. A subspace of a second-countable space is second-countable, and a countable product of second-countable spaces is second-countable.

Definition. A subset $A$ of a space $X$ is said to be dense in $X$ if $\overline{A}=X$.

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An example of a dense subset of a space is the rational numbers $\mathbb{Q}$ as a subset of the real numbers $\mathbb{R}$, because every real number is the limit point of a sequence in $\mathbb{Q}$, applying Theorem 30.1.

[Theorem 30.3] Suppose that $X$ has a countable basis, then:

(a) Every open covering of $X$ contains a countable subcollection covering $X$. (This property is also called the Lindelöf condition. A space having this property is called a Lindelöf space.)

(b) There exists a countable subset of $X$ that is dense in $X$. (A space with this property is called separable.)

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• The two properties above are weaker in general than the second countability axiom, but are equivalent to it when the space is metrizable (see Exercise 30.5).
• If a space is metrizable, it follows that $X$ is also second-countable.

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To show that a space $X$ is Lindelöf, it in fact suffices to check that every open covering of $X$ by its basis elements contains a countable subcollection covering $X$.

Why? Suppose that $X$ is a space such that every open covering of $X$ by its basis elements contains a countable subcollection covering $X$. Note that an open set is a union

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Do not confuse separable spaces with a separation of a space which is used to define the connectedness of a space!

Separable spaces ≠ Disconnected spaces!

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• The product of two Lindelöf spaces need not be Lindelöf. For example, $\mathbb{R}_l$ is Lindelöf but the product space $\mathbb{R}_l \times \mathbb{R}_l$ is not.

  The space $\mathbb{R}_l^2$ is also called the Sorgenfrey plane.

• A subspace of a Lindelöf space need not be Lindelöf. For example, the ordered square $I_o^2$ is Lindelöf but the subspace $A = I \times (0,1)$ is not.

The Separation Axioms

Note that the Hausdorff axiom mentioned in Chapter 2 is also a separation axiom.

Definition. Suppose that one-point sets are closed in $X$, then $X$ is said to be regular if for each pair consisting of a point $x$ and a closed set $B$ disjoint from $x$, there exists disjoint open sets containing $x$ and $B$, respectively. The space $X$ is said to be normal if for each pair $A,B$ of disjoint closed sets of $X$, there exists disjoint open sets containing $A$ and $B$, respectively.

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Note that normality is stronger than regularity, and regularity is stronger than the Hausdorff condition.

The $T_1$ axiom is necessary in the definition of regularity of normality in order for this to be the case. A two-point space in the indiscrete topology satisfies the other part of the definitions of regularity and normality despite not being Hausdorff.

[Lemma 31.1] (Equivalent formulation of regularity and normality) Let $X$ be a topological space. Let one-point sets in $X$ be closed.

(a) $X$ is regular if and only if given a point $x$ of $X$ and a neighbourhood $U$ of $x$, there is a neighbourhood $V$ of $x$ such that $\overline{V} \subseteq U$.

(b) $X$ is normal if and only if given a closed set $A$ and an open set $U$ containing $A$, there is an open $V$ containing $A$ such that $\overline{V} \subseteq U$.

[Theorem 31.2] (a) A subspace of a Hausdorff space is Hausdorff; a product of Hausdorff spaces is Hausdorff.

(b) A subspace of a regular space is regular; a product of regular spaces is regular.

(a) Let $X$ be Hausdorff. Let $x$ and $y$ be two (distinct) points of the subspace $Y$ of $X$. If $U$ and $V$ are neighbourhoods in $X$ of $x$ and $y$, respectively, such that $U \cap V = \emptyset$, then $U \cap Y$ and $V \cap Y$ are also disjoint neighbourhoods of $x$ and $y$ respectively in $Y$.

Let $\{X_\alpha\}$ be a family of Hausdorff spaces. Let $\mathbf{x}=(x_\alpha)$ and $\mathbf{y}=(y_\alpha)$ be distinct points of the product space $\prod X_\alpha$, so that there exists some index $\beta$ such that $x_\beta \neq y_\beta$.

Using the Hausdorff condition, take disjoint open sets $U$ and $V$ in $X_\beta$ containing $x_\beta$ and $y_\beta$, respectively, then the sets $\pi_\beta^{-1}(U)$ and $\pi_\beta^{-1}(V)$ are disjoint open sets in $\prod X_\alpha$ containing $\mathbf{x}$ and $\mathbf{y}$ (projections are continuous, and inverse images preserve intersections), respectively.

(b) Let $Y$ be a subspace of the regular space $X$. Take a point $y$ in $Y$ (which is also in $X$) and a closed subset $B$ of $Y$ that does not contain $x$.

Note that $\overline{B} \cap Y = B$, where $\overline{B}$ denotes the closure of $B$ in $X$ (see Theorem 17.4 in Chapter 2). This implies that $x \notin \overline{B}$, so, since $X$ is regular, we can obtain disjoint open sets $U$ and $V$ of $X$ containing $x$ and $\overline{B}$, respectively. It follows that $U \cap Y$ and $V \cap Y$ are disjoint open sets in $Y$ containing $x$ and $B$, respectively.

Let $\{X_\alpha\}$ be a family of regular spaces and $X = \prod X_\alpha$. By (a), $X$ is Hausdorff, so that one-point sets are closed in $X$.

Let $\mathbf{x}=(x_\alpha)$ be a point of $X$ and $U$ be a neighbourhood of $\mathbf{x}$ in $X$. Choose a basis element $\prod U_\alpha$ about $\mathbf{x}$ that lies in in $U$. For each $\alpha$, choose a neighbourhood $V_\alpha$ of $x_\alpha$ in $X_\alpha$ such that $\overline{V_\alpha} \in U_\alpha$; if it happens that $U_\alpha = X_\alpha$, choose $V_\alpha = X_\alpha$. It follows that $V = \prod V_\alpha$ is a neighbourhood of $\mathbf{x}$ in $X$. By Theorem 19.5, we have that $\overline{V}=\prod \overline{V_\alpha}$, which implies that $\overline{V} \subseteq \prod U_\alpha \subseteq U$, and thus by Lemma 31.1, $X$ is regular.

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• The space $\mathbb{R}_K$ is Hausdorff but not regular.
• The space $\mathbb{R}_l$ is normal.
• The Sorgenfrey plane $\mathbb{R}_l^2$ is not normal.

Normal Spaces

Recall that a space $X$ (with one-point sets being closed) is said to be normal if for each pair $A,B$ of disjoint closed sets of $X$, there exists disjoint open sets containing $A$ and $B$, respectively.

[Theorem 32.1] Every regular space with a countable basis is normal.

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Recall that $X$ is said to be regular if for each pair consisting of a point $x$ and a closed set $B$ disjoint from $x$, there exists disjoint open sets containing $x$ and $B$, respectively.

Let $X$ be a regular space with a countable basis $\mathcal{B}$. Let $A$ and $B$ be disjoint closed subsets of $X$.

Each point $x$ of $A$ has a neighbourhood $U$ not intersecting $B$. Using Lemma 31.1, choose a neighbourhood $V$ of $x$ such that $\overline{V} \subseteq U$. Finally, choose an element of $\mathcal{B}$ that contains $x$ and is contained in $V$. Doing this for each $x$ in $A$, we obtain a countable covering of $A$ by open sets whose closures do not intersect $B$, which we can then index with positive integers and denote it by $\{U_n\}$.

We then repeat the steps for $B$ and obtain a countable covering $\{V_n\}$ for $B$ such that the closure of each set in the covering does not intersect $A$.

The sets $U=\bigcup U_n$ and $V=\bigcup V_n$ are open sets containing $A$ and $B$ respectively, but they need not be disjoint. However, we can perform a trick to construct two disjoint open sets $U'$ and $V'$ that contain $A$ and $B$ respectively from them.

Define $U'=\bigcup_{n \in \mathbb{Z}_+}U_n'$ and $V'=\bigcup_{n \in \mathbb{Z}_+}V_n'$, where $U_n'=U_n \backslash \bigcup_{i=1}^{n} \overline{V_i}$ and $V_n'=V_n \backslash \bigcup_{i=1}^{n} \overline{U_i}$.

Note that each set $U_n'$, as well as $V_n'$ is open since it is the difference of an open set and a closed set. We see that $U'$ contains $A$, because each $x$ in $A$ belongs to $U_n$ for some $n$, and $x$ belongs to none of the sets $\overline{V_i}$. Similarly, $V'$ contains $B$.

Finally, $U'$ and $V'$ are disjoint. If not, then there exists some $x \in U' \cap V'$, which implies that $x \in U_j' \cap V_k'$ for some $j$ and $k$. If $j \leqslant k$, then by definition of $U_j'$, we have that $x \in U_j$ but from the definition of $V_k'$, we also have that $x \notin U_j$, causing a contradiction. The situation is similar for $j \geqslant k$.

[Theorem 32.2] Every metrizable space is normal.

Let $X$ be a metrizable space with metric $d$. Let $A$ and $B$ be disjoint closed subsets of $X$.

For each $a \in A$, choose $\varepsilon_a$ such that the ball $B(a,\varepsilon_a)$ does not intersect $B$. Similarly, for each $b \in B$, choose $\varepsilon_b$ such that $B(b,\varepsilon_b)$ does not intersect $A$.

Define

$$U=\bigcup_{a \in A}B(a,\varepsilon_a/2), \quad V=\bigcup_{b \in B}B(b,\varepsilon_b/2).$$

Note that $U$ and $V$ are open sets containing $A$ and $B$, respectively; we claim that they are disjoint.

If $U \cap V \neq \emptyset$, then there exists $z$ such that

$$z \in B(a,\varepsilon_a/2) \cap B(b,\varepsilon_b/2)$$

for some $a \in A$ and some $b \in B$. By triangle inequality, we have that $d(a,b) \leqslant d(a,z)+d(z,b) < (\varepsilon_a+\varepsilon_b)/2$. If $\varepsilon_a \leqslant \varepsilon_b$, then $d(a,b)<\varepsilon_b$, so that the ball $B(b,\varepsilon_b)$ contains the point $a$. If $\varepsilon_b \leqslant \varepsilon_a$, then $d(a,b)<\varepsilon_a$, so that the ball $B(a,\varepsilon_a)$ contains the point $b$. Both of the situations are impossible.

The bolded statement above is the key that motivates the definitions of $U$ and $V$.

[Theorem 32.3] Every compact Hausdorff space is normal.

Recall that Lemma 26.4 in Chapter 3 states that: if $Y$ is a compact subspace of the Hausdorff space $X$ and $x_0$ is not in $Y$, then there exists disjoint open sets $U$ and $V$ of $X$ containing $x_0$ and $Y$, respectively.

Let $X$ be a compact Hausdorff space. We have already essentially proved that $X$ is regular due to Lemma 26.4.

Essentially the same argument as given in Lemma 26.4 can be applied to show that $X$ is normal: Given disjoint closed sets $A$ and $B$ in $X$, thanks to the regularity of $X$, we can choose, for each point $a$ of $A$, disjoint open sets $U_a$ and $V_a$ containing $a$ and $B$, respectively.

The collection $\{U_a\}$ covers $A$. Since $A$ is compact, $A$ may be covered by finitely many sets $U_{a_1},\ldots,U_{a_m}$, then

$$U=\bigcup_{i=1}^{m} U_{a_i} \quad \text{and} \quad V=\bigcap_{i=1}^{m}V_{a_i}$$

are disjoint open sets containing $A$ and $B$, respectively.

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Why not define $V=\bigcup_{i=1}^{m}V_{a_i}$ ?

This is because while each $V_{a_i}$ does not contain some $a_i \in A$, it does not necessarily mean that it does not intersect other elements in $A$ as well.

[Theorem 32.4] Every well-ordered set $X$ is normal in the order topology.

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In fact, the stronger result is true: every order topology is normal. See here for proofs.

Let $X$ be a well-ordered set. We assert that every interval of the form $(x,y]$ is open in $X$: If $X$ has a largest element and $y$ is that element, $(x,y]$ is just a basis element about $y$. If $y$ is not the largest element of $X$, then $(x,y]$ equals the open set $(x,y')$, where $y'$ is the immediate successor of $y$.

Now, let $A$ and $B$ be disjoint closed sets in $X$; assume first that neither $A$ nor $B$ contains the smallest element $a_0$ of $X$. For each $a \in A$, there exists a basis element about $a$ disjoint from $B$; it contains some interval of the form $(x,a]$, since $a$ is not the smallest element of $X$.

For each $a \in A$, choose such an interval $(x_a,a]$ disjoint from $B$. Similarly, for each $b \in B$, choose an interval $(y_b,b]$ disjoint from $A$. The sets

$$U=\bigcup_{a \in A}(x_a,a] \quad \text{and} \quad V=\bigcup_{b \in B}(y_b,b]$$

are open sets containing $A$ and $B$, respectively; we claim that they are disjoint.

If there exists some $z \in U \cap V$, then $z \in (x_a.a] \cap (y_b,b]$ for some $a \in A$ and $b \in B$. Assume that $a < b$, then if $a \leqslant y_b$, the two intervals are disjoint, while if $a > y_b$, we have $a \in (y_b,b]$, contrary to the fact that $(y_b,b]$ is disjoint from $A$. A similar contradiction occurs if $b < a$.

Finally, assume that $A$ and $B$ are disjoint closed sets in $X$, and $A$ contains the smallest element $a_0$ of $X$. The set $\{a_0\}$ is both open and closed in $X$. (This is because $\{a_0\} = [a_0, a')$, where $a'$ is the immediate successor of $a_0$, whilst $X \backslash \{a_0\} = (a_0,+\infty)$, both of which are open.) By the result of the preceding paragraph, there exists disjoint open sets $U$ and $V$ containing the closed sets $A \backslash \{a_0\}$ and $B$ respectively, then $U \cup \{a_0\}$ and $V$ are disjoint open sets containing $A$ and $B$ respectively.

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1. If $J$ is uncountable, the product space $\mathbb{R}^J$ is not normal.
2. The product space $S_\Omega \times \overline{S_\Omega}$ is not normal. ($S_\Omega$ denotes the minimal uncountable well-ordered set whereas $\overline{S_\Omega}$ denotes the set $S_\Omega \cup \{\Omega\}$, where $\Omega$ is the first uncountable ordinal.)

The Urysohn Lemma

[Theorem 33.1] (Urysohn lemma). Let $X$ be a normal space; let $A$ and $B$ be disjoint closed subsets of $X$. Let $[a,b]$ be a closed interval in the real line. There then exists a continuous map

$$f : X \to [a,b]$$

such that $f(x)=a$ for every $x$ in $A$, and $f(x)=b$ for every $x$ in $B$.

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Such continuous function is also called a Urysohn function.

In fact, the converse of Urysohn lemma also holds. See here for a proof.

We need only consider the case where the interval in question is the interval $[0,1]$; the general case follows from that one (through a continuous bijection from $[0,1]$ to $[a,b]$ defined by $x \mapsto (b-a)x+a$).

Step 1. Using normality, we construct a certain family $U_p$ of open sets of $X$, indexed by the rational numbers (or just any countable dense subset of $[0,1]$ that contains the points $0$ and $1$), then use these sets to define the continuous function $f$.

Let $P$ be the set of all rational numbers in the interval $[0,1]$. We shall define, for each $p$ in $P$, an open set $U_p$ of $X$, such that whenever $p<q$, we have $\overline{U_p} \subseteq U_q$. Thus, the sets $U_p$ will be simply ordered by inclusion ($\subseteq$) in the same way their subscripts are ordered by the usual ordering in the real line, i.e.

$$p < q \implies \overline{U_p} \subseteq U_q. \tag{*}$$

Since $P$ is countable, we can use induction (principle of recursive definition) to define the sets $U_p$.

Arrange the elements of $P$ in an infinite sequence in some way; for simplicity, suppose that the numbers $1$ and $0$ are the first two elements of the sequence. Define the sets $U_p$ as follows: [Base case] First, let $U_1=X \backslash B$. Next, since $A$ is a closed set contained in the open set $U_1$, by normality of $X$ (Theorem 31.1), we may choose an open set $U_0$ such that

$$A \subseteq U_0 \quad \text{and} \quad \overline{U_0} \subseteq U_1.$$

In general, let $P_n$ denote the set consisting of the first $n$ rational numbers in the sequence. Suppose that $U_p$ is defined for all rational numbers $p$ belonging to the set $P_n$, satisfying the condition (*) [Induction hypothesis].

Let $r$ denote the next rational number in the sequence; we want to define $U_r$.

[Inductive case] Consider the set $P_{n+1}=P_n \cup \{r\}$. It is a finite subset of the interval $[0,1]$, thus it has a simple ordering derived from the usual order relation $<$ on the real line. In a finite simply ordered set, every element (except the smallest and the largest) has an immediate predecessor and an immediate successor. The number $0$ is the smallest element, and $1$ is the largest element, of the simply ordered set $P_{n+1}$, and $r$ is neither $0$ nor $1$. This implies that $r$ has an immediate predecessor $p$ and an immediate successor $q$ in $P_{n+1}$ (so that $p < r < q$). The sets $U_p$ and $U_q$ are already defined, and $\overline{U_p} \subseteq U_q$ by induction hypothesis. By normality of $X$ (Theorem 31.1), we can find an open set $U_r$ of $X$ such that

$$\overline{U_p} \subseteq U_r \quad \text{and} \quad \overline{U_r} \subseteq U_q.$$

We claim that (*) now holds for every pair of elements of $P_{n+1}$. If both elements lie in $P_n$, (*) holds by induction hypothesis. If one of them is $r$ and the other is a point of $P_n$, denoted here by $s$, then either $s \leqslant p$, in which case $\overline{U_s} \subseteq \overline{U_p} \subseteq U_r$, or $s \geqslant q$, in which case $\overline{U_r} \subseteq U_q \subseteq U_s$. This proves the claim.

Hence, we manage to define $U_p$ for all $p \in P$ by induction.

Step 2. We now extend the definition of $U_p$ from all rational numbers in the interval $[0,1]$ to all rational numbers $p$ in $\mathbb{R}$ by defining

$$\begin{cases} U_p = \emptyset & \text{if } p<0, \\ U_q = X & \text{if } p>1. \end{cases}$$

Indeed, if $r<0$, then $r<q$ for all $q \in [0,1]$, whilst $\overline{U_r}=\emptyset \subseteq U_q$. Similarly, if $r>1$, then $r>q$ for all $q \in [0,1]$; meanwhile, $\overline{U_q} \subseteq X = U_r$. Therefore, it is still true that for any pair of rational numbers $p$ and $q$, we have $p<q \implies \overline{U_p} \subseteq U_q$.

Step 3. Given a point $x$ of $X$, let us define $\mathbb{Q}(x)$ to be the set of those rational numbers $p$ such that the corresponding open sets $U_p$ contain $x$, i.e.

$$\mathbb{Q}(x)=\{p : x \in U_p\}$$

This set contains no number less than $0$, since no $x$ is in $U_p$ for $p<0$, and it contains every number greater than $1$, since every $x$ is in $U_p$ for $p > 1$. Therefore, $\mathbb{Q}(x)$ is bounded below, and its greatest lower bound (infimum) is a point of the interval $[0,1]$. Define

$$f(x)=\inf\mathbb{Q}(x)=\{ p : x \in U_p\}.$$

Step 4. We show that $f$ is the desired continuous function.

If $x \in A$, then $x \in U_p$ for every $p \geqslant 0$, so that $\mathbb{Q}(x)$ equals the set of all nonnegative rationals, and $f(x)=\inf\mathbb{Q}(x)=0$. Similarly, if $x \in B$, then $x \in U_p$ for no $p \leqslant 1$, so that $\mathbb{Q}(x)$ consists of all rational numbers greater than $1$, and $f(x)=1$.

Now here comes the hard part: we wish to show that $f$ is continuous. For this purpose, we first prove the following facts:

1. $x \in \overline{U_r} \implies f(x) \leqslant r$.
2. $x \notin U_r \implies f(x) \geqslant r .$

To prove 1., note that if $x \in \overline{U_r}$, then $x \in U_s$ for every $s<r$ (by definition of $U_p$ in Step 1). Therefore, $\mathbb{Q}(x)$ contains all rational numbers greater than $r$, so that by definition we have $f(x)=\inf\mathbb{Q}(x) \leqslant r$.

To prove 2., note that if $x \notin U_r$, then $x \notin U_s$ for any $s < r$. Therefore, $\mathbb{Q}(x)$ contains no rational numbers less than $r$, so that $f(x)=\inf\mathbb{Q}(x) \geqslant r$.

Now we prove continuity of $f$. By Theorem 18.1 (see Chapter 2), it suffices to show that given a point $x_0$ of $X$ and an open interval $(c,d)$ in $\mathbb{R}$ containing the point $f(x_0)$, we can find a neighbourhood $U$ of $x_0$ such that $f(U) \subseteq (c,d)$.

Choose rational numbers $p$ and $q$ such that $c < p < f(x_0) < q < d$. (This is possible since rational numbers are dense in $\mathbb{R}$.) We assert that the open set $U = U_q \backslash \overline{U_p}$ is the desired neighbourhood of $x_0$.

First, we note that $x_0 \in U$, since the fact that $f(x_0)<q$ implies by the contraposition of 2. that $x_0 \in U_q$, whereas the fact that $f(x_0)>p$ implies by the contraposition of 1. that $x_0 \notin \overline{U_p}$.

Finally, we show that $f(U) \subseteq (c,d)$. Let $x \in U$, then $x \in U_q \subseteq \overline{U_q}$, so that by 1., $f(x) \leqslant q$. Meanwhile, $x \notin U_p$ so by 2., $f(x)>p$. Hence, $f(x) \in [p,q] \subseteq (c,d)$, as desired.

Definition. If $A$ and $B$ are two (not necessarily closed) subsets of the topological space $X$, and if there is a continuous function $f : X \to [0,1]$ such that $f(A)=\{0\}$ and $f(B)=\{1\}$, we say that $A$ and $B$ can be separated by a continuous function.

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Is there a version of Urysohn lemma for regular spaces?

Not quite. Regularity is not enough to separate a point from a closed set by a continuous function, as one runs into difficulty when trying to find an open set $U_p$ such that $\overline{U_0} \subseteq U_p$ and $\overline{U_p} \subseteq U_1$, following the proof of Urysohn lemma given above.

In fact, being able to separate a point from a closed set by a continuous function is a stronger condition, which gives rise to a new separation axiom, as defined below.

Definition. A space $X$ is completely regular if one-point sets are closed in $X$ and if for each point $x_0$ and each closed set $A$ not containing $x_0$, there is a continuous function $f : X \to [0,1]$ such that $f(x_0)=1$ and $f(A)=\{0\}$.

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• A normal space is completely regular.
• A completely regular space is regular (to see this, consider the sets $f^{-1}([0,\frac{1}{2}))$ and $f^{-1}((\frac{1}{2},1])$).

Complete regularity behaves nicely with regards to subspaces and products:

[Theorem 33.2] A subspace of a completely regular space is completely regular. A product of completely regular spaces is completely regular.

Let $X$ be completely regular; let $Y$ be a subspace of $X$. Let $x_0$ be a point of $Y$, and let $A$ be a closed set of $Y$ not containing $x_0$. Note that $A = \overline{A} \cap Y$ (see Theorem 17.4 in Chapter 2 or here for why), where $\overline{A}$ is the closure of $A$ in $X$. This implies that $x_0 \notin \overline{A}$. Since $X$ is completely regular, we can choose a continuous function $f: X \to [0,1]$ such that $f(x_0)=1$ and $f(\overline{A})=\{0\}$. The restriction of $f$ to $Y$ is the desired continuous function on $Y$.

Let $X = \prod X_\alpha$ be a product of completely regular spaces. Let $\mathbf{b}=(b_\alpha)$ be a point of $X$ and $A$ be a closed set of $X$ disjoint from $\mathbf{b}$. Choose a basis element $\prod U_\alpha$ containing $\mathbf{b}$ that does not intersect $A$, then $U_\alpha = X_\alpha$ except for finitely many $\alpha$, say, $\alpha=\alpha_1,\ldots,\alpha_n$ (see Theorem 19.1 in Chapter 2). Given $i=1,\ldots,n$, choose a continuous function

$$f_i : X_{\alpha_i} \to [0,1]$$

such that $f_i(b_{\alpha_i})=1$ and $f_i(X \backslash U_{\alpha_i})=\{0\}$. Let $\phi_i(\mathbf{x})=f_i(\pi_{\alpha_i}(\mathbf{x}))$, then $\phi_i$ maps $X$ continuously into $\mathbb{R}$ and vanishes outside $\pi_{\alpha_i}^{-1} (U_{\alpha_i})$. The product

$$f(\mathbf{x})=\phi_1(\mathbf{x}) \cdot \phi_2(\mathbf{x}) \cdot \cdots \cdot \phi_n(\mathbf{x})$$

is the desired continuous function on $X$, because it equals $1$ at $\mathbf{b}$ and vanishes outside $\prod U_\alpha$.

The Urysohn Metrization Theorem

[Theorem 34.1] (Urysohn metrization theorem). Every regular space $X$ with a countable basis is metrizable.

We shall prove that $X$ is metrizable by embedding $X$ in a metrizable space $Y$, i.e. by showing $X$ is homeomorphic with a subspace of $Y$.

Two versions of the proof are presented in this book. The first version defines $Y$ as the space $\mathbb{R}^\omega$ in the product topology, which is shown in Theorem 20.5 (see Chapter 2) to be metrizable. In the second version, the space $Y$ is also $\mathbb{R}^\omega$, but this time in the topology induced by the uniform metric $\bar{\rho}$. In each case, it turns out that the construction actually embeds $X$ in the subspace $[0,1]^\omega$ of $\mathbb{R}^\omega$.

Step 1. We prove the following:

There exists a countable collection of continuous functions $f_n : X \to [0,1]$ having the property that given any point $x_0$ of $X$ and any neighbourhood $U$ of $x_0$, there exists an index $n$ such that $f_n(x_0)>0$ and $f_n(x)=0$ for points $x$ that are outside $U$.

Due to Theorem 32.1 and the Urysohn lemma, given $x_0$ and $U$, there exists such a function. However, if we choose one such function for each pair $(x_0,U)$, the resulting collection will not in general be countable. Our task is to cut the collection down to size. Here is one way to proceed:

Let $\{B_n\}$ be a countable basis for $X$. For each pair $n,m$ of indices for which $\overline{B_n} \subseteq B_m$, apply the Urysohn lemma to choose a continuous function $g_{n,m} : X \to [0,1]$ such that $g_{n,m}(\overline{B_n})=\{1\}$ and $g_{n,m}(X \backslash B_m)=\{0\}$. The collection $\{g_{n,m}\}$ then satisfies our requirement: Given $x_0$ and a neighbourhood $U$ of $x_0$, one can choose a basis element $B_m$ containing $x_0$ that is contained in $U$. Using regularity, one can then choose $B_n$ so that $x_0 \in B_n$ and $\overline{B_n} \subseteq B_m$, then $n,m$ is a pair of indices for which the function $g_{n,m}$ is defined, and it is positive at $x_0$ and vanishes outside $U$. Since the collection $\{g_{n,m}\}$ is indexed with a subset of $\mathbb{Z}_+ \times \mathbb{Z}_+$, it is countable; hence it can be reindexed with the positive integers, giving us the desired collection $\{f_n\}$.

Step 2 (first version). Now that we have the functions $f_n$ from Step 1, take $\mathbb{R}^\omega$ in the product topology and define a map $F : X \to \mathbb{R}^\omega$ as such:

$$F(x)=(f_1(x),f_2(x),\ldots).$$

We claim that $F$ is an embedding. First, $F$ is continuous because $\mathbb{R}^\omega$ has the product topology and each $f_n$ is continuous (see Theorem 19.6 in Chapter 2). Next, $F$ is injective because given $x \neq y$, we know that there is an index $n$ such that $f_n(x)>0$ and $f_n(y)=0$ (note that the fact in Step 1 applies for every point in $X$); thus, $F(x) \neq F(y)$. Finally, we prove that $F$ is a homeomorphism of $X$ onto its image, the subspace $Z=F(X)$ of $\mathbb{R}^\omega$. We know that $F$ defines a continuous bijection of $X$ with $Z$, so we need only show that for each open set $U$ in $X$, the set $F(U)$ is open in $Z$ (see the 'Homeomorphisms' section of Chapter 2 notes).

Let $z_0$ be a point of $F(U)$. We shall find an open set $W$ of $Z$ such that $z_0 \in W \subseteq F(U)$ (see Exercise 13.1). Let $x_0$ be the point of $U$ such that $F(x_0)=z_0$. Choose an index $N$ for which $f_N(x_0)>0$ and $f_N(X \backslash U)=\{0\}$. Take the open ray $(0,+\infty)$ in $\mathbb{R}$, and let $V$ be the open set $V=\pi_N^{-1}((0,+\infty))$ of $\mathbb{R}^\omega$. Let $W$ = $V \cap Z$. Note that $W$ is open in $Z$, by definition of subspace topology. We assert that $z_0 \in W \subseteq F(U)$. First, $z_0 \in W$ because $\pi_N(z_0)=\pi_N(F(x_0))=f_N(x_0)>0$. Next, $W \subseteq F(U)$, because if $z \in W$, then $z = F(x)$ for some $x \in X$, and $\pi_N(Z) \in (0,+\infty)$. Since $\pi_N(z)=\pi_N(F(x))=f_N(x)$, and $f_N$ vanishes outside $U$, the point $x$ must be in $U$, then $z = F(x)$ is in $F(U)$, as desired. Thus, $F$ is an embedding of $X$ in $\mathbb{R}^\omega$.

Step 3 (second version). In this version, we embed $X$ in the metric space $(\mathbb{R}^\omega,\bar{\rho})$. Actually, we embed $X$ in the subspace $[0,1]^\omega$, on which $\bar{\rho}$ equals the metric $\rho(\mathbf{x},\mathbf{y})=\sup\{|x_i-y_i|\}$ (since the interval $[0,1]$ is bounded).

We use the countable collection of collections $f_n : X \to [0,1]$ constructed in Step 1, but now we impose the additional condition that $f_n(x) \leqslant \frac{1}{n}$ for all $x$. (This condition can be satisfied by just dividing each function $f_n$ by $n$.)

Define $F : X \to [0,1]^\omega$ by the equation $F(x)=(f_1(x),f_2(x),\ldots)$ as in Step 2. We assert that $F$ is now an embedding relative to the metric $\rho$ on $[0,1]^\omega$.

We know from Step 2 that $F$ is injetive. Furthermore, we know that if we use the product topology on $[0,1]^\omega$, the map $F$ carries open sets of $X$ onto open sets of the subspace $Z = F(X)$. This statement still holds if one passes to the finer topology (see Theorem 20.4) on $[0,1]^\omega$ induced by the metric $\rho$.

It remains to prove that $F$ is continuous. This does not follow from the fact that each component function is continuous, as we are not using the product topology on $\mathbb{R}^\omega$ now. Here is where the assumption that $f_n(x) \leqslant \frac{1}{n}$ comes in.(

Let $x_0$ be a point of $X$, and let $\varepsilon>0$ (be arbitrary). To prove continuity, we need to find a neighbourhood $U$ of $x_0$ such that $x \in U \implies \rho(F(x),F(x_0))<\varepsilon$.

First choose $N$ large enough so that $1/N \leqslant \varepsilon/2$, then for each $n=1,\ldots,N$ use the continuity of $f_n$ to choose a neighbourhood $U_n$ of $x_0$ such that $|f_n(x)-f_n(x_0)| \leqslant \varepsilon/2$ for $x \in U_n$.

Let $U = U_1 \cap \cdots \cap U_N$; we show that $U$ is the desired neighbourhood of $x_0$. Let $x \in U$. If $n \leqslant N$, then $|f_n(x)-f_n(x_0)| \leqslant \varepsilon/2$ by choice of $U$, and if $n > N$, then $|f_n(x)-f_n(x_0)|<1/N \leqslant \varepsilon/2$ because $f_n$ maps $X$ into $[0,1/n]$ (and $1/n < 1/N$). Therefore, for all $x \in U$, $\rho(F(x),F(x_0)) \leqslant \varepsilon/2 < \varepsilon$, as desired.

[Theorem 34.2] (Embedding theorem). Let $X$ be a space in which one-point sets are closed. Suppose that $\{f_\alpha\}_{\alpha \in J}$ is an indexed family of continuous functions $f_\alpha : X \to \mathbb{R}$ satisfying the requirement that for each point $x_0$ of $X$ and each neighbourhood $U$ of $x_0$, there is an index $\alpha$ such that $f_\alpha$ is positive at $x_0$ and vanishes outside $U$, then the function $F : X \to \mathbb{R}^J$ defined by

$$F(x)=(f_\alpha(x))_{\alpha \in J}$$

is an embedding of $X$ in $\mathbb{R}^J$. If $f_\alpha$ maps $X$ into $[0,1]$ for each $\alpha$, then $F$ embeds $X$ in $[0,1]^J$.

note

One needs one-point sets in $X$ to be closed to ensure that given $x \neq y$, there is an index $\alpha$ such that $f_\alpha(x) \neq f_\alpha(y)$, due to the fact given in Step 1 of the proof of Theorem 34.1 given above.

The proof is almost identical to Step 2 of the proof for Theorem 34.1 above, with $n$ replaced by $\alpha$, and $\mathbb{R}^\omega$ by $\mathbb{R}^J$, throughout.

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A family of continuous functions that satisfies the hypotheses of Theorem 34.2 is said to separate points from closed sets in $X$.

[Theorem 34.3] (Corollary from Theorem 34.2) A space $X$ is completely regular if and only if it is homeomorphic to a subspace of $[0,1]^J$ for some $J$.

The Tietze Extension Theorem

[Theorem 35.1] (Tietze extension theorem). Let $X$ be a normal space; let $A$ be a closed subspace of $X$.

(a) Any continuous map of $A$ into the closed interval $[a,b]$ of $\mathbb{R}$ may be extended to a continuous map of all of $X$ into $[a,b]$.

(b) Any continuous map of $A$ into $\mathbb{R}$ may be extended to a continuous map of all of $X$ into $\mathbb{R}$.

The idea of the proof is to construct a sequence of continuous functions $s_n$ defined on the entire space $X$, where the sequence $s_n$ converges uniformly, and the restriction of $s_n$ to $A$ approximates $f$ more and more closely as $n$ increases. The limit function (called $g$) will then be continuous (see 'Uniform limit theorem' in Chapter 2), and its restriction to $A$ will be equal to $f$.

Step 1. We first construct a particular function $g$ defined on all of $X$ such that $g$ is 'not too large', and such that $g$ approximates $f$ on the set $A$ 'fairly accurately'. More precisely, take a continuous map $f : A \to [-r,r]$. We assert that there exists a continuous function $g : X \to \mathbb{R}$ such that

$$|g(x)| \leqslant \frac{1}{3}r \quad \text{for all } x \in X, \tag{1}$$$$|g(a)-f(a)| \leqslant \frac{2}{3}r \quad \text{for all } a \in A. \tag{2}$$

The claim can be proven by dividing the interval $[r,r]$ into three equal closed intervals of length $\frac{2}{3}r$, taking the inverse images of the first and third invervals, applying the Urysohn lemma (thanks to normality) to obtain a continuous function from $X$ to the second interval having the desired property (1), and considering three cases (whether $a$ is in the first or third interval, or neither) to prove property (2).

Step 2. We now prove part (a) of the theorem. Without loss of generality, we can replace the arbitrary closed interval $[a,b]$ of $\mathbb{R}$ by the interval $[-1,1]$.

Let $f : X \to [-1,1]$ be a continuous map, then $f$ satisfies the hypotheses in Step 1, with $r = 1$, so there exists a continuous function $g_1 : X \to \mathbb{R}$ such that

$$|g_1(x)| \leqslant \frac{1}{3} \quad \text{for all } x \in X,$$$$|f(a)-g_1(a)| \leqslant \frac{2}{3} \quad \text{for all } a \in A.$$

By induction (by considering $f-g_1-\cdots-g_k$ and applying Step 1 repeatedly), we can define a sequence of functions $g_n$ such that

$$|g_{n}(x)| \leqslant \frac{1}{3}\left(\frac{2}{3}\right)^{n-1} \quad \text{for all } x \in X, \tag{1'}$$$$|f(a)-g_1(a)-\cdots-g_n(a)| \leqslant \left(\frac{2}{3}\right)^{n} \quad \text{for all } a \in A. \tag{2'}$$

Next, we define $g(x)=\sum\limits_{n=1}^{\infty}g_n(x)$. By comparison test with the geometric series $\frac{1}{3}\sum\limits_{n=1}^{\infty}\left(\frac{2}{3}\right)^{n-1}$ (and triangle inequality), we know that $g$ converges.

To show that $g$ is continuous, we must show that the partial sum sequence $s_n$ converges to $g$ uniformly. This follows from the 'Weierstrass $M$-test' in analysis.

We also see that $g(a)=f(a)$ for all $a \in A$, as $g$ is the limit of the infinite sequence $s_n(x)$ of partial sums, and the property (2') implies that $s_n(a) \to f(a)$ for all $a \in A$.

Finally, we show that $g$ maps $X$ into the interval $[-1,1]$. Pretending that we do not see that $(1/3)\sum(2/3)^n$ converges to $1$ and all we knew was that $g$ mapped $X$ into $\mathbb{R}$, we can still compose $g$ with the 'normalisation map' to 'scale down' values with norms exceeding $1$ to within $1$ to obtain the desired outcome.

Step 3. We now prove part (b) of the theorem. Take a continuous function $f$ that maps $A$ into $\mathbb{R}$. We can replace $\mathbb{R}$ by the open interval $(-1,1)$ because it is homeomorphic to $\mathbb{R}$.

Part (a) of the theorem gives us a continuous map $g : X \to [-1,1]$. From here, we define a subset $D$ of $X$ like so: $D = g^{-1}(\{-1\}) \cup g^{-1}(\{1\})$. Note that $D$ is closed in $X$ due to the continuity of $g$, and the set $A$ is disjoint from $D$ because $g(A)=f(A) \subseteq (-1,1)$. By the Urysohn lemma, there is a continuous function $\phi : X \to [0,1]$ such that $\phi(D)=\{0\}$ and $\phi(A)=\{1\}$.

Now, define $h(x)=\phi(x)g(x)$. Note that $h$ is the product of two continuous functions, so $h$ is continuous. It is also an extension of $f$, since $\phi(a)=1$ for all $a \in A$. Finally, $h$ maps all of $X$ into the open interval $(-1,1)$, because $\phi(x)=0$ if $x \in D$; if $x \notin D$, then $|g(x)| < 1 \implies |h(x)| \leqslant 1 \cdot |g(x)| < 1$.

Embeddings of Manifolds

What conditions are required so that a space $X$ can be embedded in some finite-dimensional Euclidean space $\mathbb{R}^N$?

Definition. An $m$-manifold is a Hausdorff space $X$ with a countable basis such that each point $x$ of $X$ has a neighbourhood that is homeomorphic with an open subset of $\mathbb{R}^m$.

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• A 1-manifold is often called a curve, and a 2-manifold is called a surface.
• If $\phi : X \to \mathbb{R}$, then the (closed) support of $\phi$ is defined to be the closure of the set $\phi^{-1}(\mathbb{R} \backslash \{0\})$, that is, if $x$ lies outside the support of $\phi$, then there is some neighbourhood $U$ of $x$ on which $\phi(u)=0$ for all $u \in U$ (note that this is the negation of Theorem 17.5, see Chapter 2).

Definition. Let $\{U_1,\ldots,U_n\}$ be a finite indexed open covering of the space $X$. An indexed family of continuous functions

$$\phi_i : X \to [0,1] \quad \text{for } i =1,\ldots,n,$$

is said to be a partition of unity dominated by $\{U_i\}$ if:

1. $\mathrm{supp}(\phi_i) \subseteq U_i$ for each $i$.
2. $\sum\limits_{i=1}^{n}\phi_i(x)=1$ for each $x$.

[Theorem 36.1] (Existence of finite partitions of unity). Let $\{U_1,\ldots,U_n\}$ be a finite open covering of the normal space $X$, then there exists a partition of unity dominated by $\{U_i\}$.

Step 1. We first prove by induction that one can 'shrink' the covering $\{U_i\}$ to an open covering $\{V_1,\ldots,V_n\}$ of $X$ such that $\overline{V_i} \subseteq U_i$ for each $i$.

In general, given open sets $V_1, \ldots, V_{k-1}$ such that the collection $\{V_1,\ldots,V_{k-1},U_k,U_{k+1},U_n\}$ covers $X$, let $A$ be the relative complement of the union of such covering (excluding $U_k$) with respect to $X$, then $A$ is a closed subset of $X$ which is contained in the open set $U_k$. Choose $V_k$ to be an open set containing $A$ such that $\overline{V_k} \subseteq U_k$ by normality, then $\{V_1,\ldots,V_{k-1},V_k,U_{k+1},\ldots,U_n\}$ covers $X$. Repeating until the $n$-th step yields the required result.

Step 2. Given the finite open covering $\{U_i\}$ of $X$, choose a finite open covering $\{V_i\}$ of $X$ such that $\overline{V_i} \subseteq U_i$ for each $i$. Again, choose an open covering $\{W_i\}$ of $X$ such that $\overline{W_i} \subseteq V_i$ for each $i$. Using the Urysohn lemma, for each $i$, take a continuous function $\psi_i : X \to [0,1]$ such that $\psi_i(\overline{W_i})=\{1\}$ and $\psi_i(X \backslash V_i)=\{0\}$.

Since $\psi_i^{-1}(R \backslash \{0\})$ is contained in $V_i$, we have $\mathrm{supp}(\psi_i) \subseteq \overline{V_i} \subseteq U_i$. Since the collection $\{W_i\}$ covers $X$, the sum $\Psi(x)=\sum_{i=1}^{n}\psi_i(x)$ is positive for each $x$ (as $\psi_i(x)=1$ for some $i$). Thus, we may define, for each $j$,

$$\phi_j(x)=\frac{\psi_j(x)}{\Psi(x)}.$$

We see that the functions $\phi_i,\ldots,\phi_n$ form the desired partition of unity, as each $\phi_j$ maps $X$ to $[0,1]$ (as $\psi_j$ also maps $X$ to $[0,1]$ and $\Psi$ is positive for each $x$), we have $\mathrm{supp}(\phi_j) \subseteq U_j$ for each $j$ (since $\mathrm{supp}(\phi_j)$ is determined by $\mathrm{supp}(\psi_j)$), and $\sum_{j=1}^{n}\phi_j(x)=1$ for each $x$ (as the numerator in the definition of $\phi_j$ adds up to $\Psi(x)$ when we sum over the indices $j$).

[Theorem 36.2] If $X$ is a compact $m$-manifold, then $X$ can be embedded in $\mathbb{R}^N$ for some positive integer $N$.

Let $\{U_1,\ldots, U_n\}$ be finitely many open sets that cover $X$, each of which may be embedded in $\mathbb{R}^M$. Take embeddings $g_i : U_i \to \mathbb{R}^m$ for each $i$. Since $X$ is compact and Hausdorff, it is also normal. Let $\phi_1,\ldots,\phi_n$ be a partition of unity dominated by $\{U_i\}$ and $A_i$ be $\mathrm{supp}(\phi_i)$. Define, for each $i=1,\ldots,n$, a function $h_i : X \to \mathbb{R}^m$ as follows:

$$h_i(x)=\begin{cases} \phi(x) \cdot g_i(x) & \text{for } x \in U_i, \\ \mathbf{0}=(0,\ldots,0) & \text{for } x \in X \backslash A_i. \end{cases}$$

The function $h_i$ is well-defined as the two definitions agree on the intersection of their domains. It is also continuous as its restrictions to the open sets $U_i$ and $X \backslash A_i$ are continuous (see point 6. of Theorem 18.2 in Chapter 2).

Now define $F : X \to (\underbrace{\mathbb{R} \times \cdots \times \mathbb{R}}_{n \text{ times}} \times \underbrace{\mathbb{R}^m \times \cdots \times \mathbb{R}^m}_{n \text{ times}})$ as such:

$$F(x)=(\phi_1(x),\ldots,\phi_n(x),h_1(x),\ldots,h_n(x)).$$

Due to Theorem 18.4, $F$ is continuous. Since $X$ is compact, to prove that $F$ is an embedding, it suffices to show that $F$ is injective (see Theorem 26.6 in Chapter 3). Suppose that $F(x)=F(y)$, then $\phi_i(x)=\phi_i(y)$ and $h_i(x)=h_i(y)$ for all $i$. Since $\sum\phi_i(x)=1$, for some $i$ we have that $\phi_i(x)>0$, so $\phi_i(y)>0$ as well, and thus $x, y \in U_i$, then

$$\phi_i(x) \cdot g_i(x)=h_i(x)=h_i(y)=\phi_i(y) \cdot g_i(y).$$

Since $\phi_i(x)=\phi_i(y)>0$, it follows that $g_i(x)=g_i(y)$, and $g_i$ is injective, so $x=y$, as desired.