Notes

Introduction

This chapter generalises the concepts of connectedness, which is depended on by a basic theorem regarding continuous functions called the 'intermediate value theorem, as well as compactness, which sees application in establishing important results such as extreme value theorem and uniform continuity, from real numbers which is central in mathematical analysis to arbitrary topological spaces.

Connectedness

Connected Spaces

Definition. Let $X$ be a topological space. A separation of $X$ is pair $U, V$ of disjoint nonempty open subsets of $X$ whose union is $X$. The space is $X$ is said to be connected if there does not exist a separation of $X$.

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• If $X$ is connected, then so is any space homeomorphic to $X$, as $X$ is a topological property.

• Another way of phrasing the definition of connectedness is the following:

  A space $X$ is connected if and only if the only subsets of $X$ that are both open and closed in $X$ are the empty set and $X$ itself.

  This is because if $A$ is a nonempty proper subset of $X$ that is both open and closed in $X$, then the sets $U=A$ and $V=X \backslash A$ form a separation of $X$, since they are open, disjoint, nonempty, and have $X$ as their union. Conversely, if $U$ and $V$ form a separation of $X$, then $U$ is nonempty and different from $X$, and it is both open and closed in $X$.

[Lemma 23.1] (Subspace connectedness). If $Y$ is a subspace of $X$, a separation of $Y$ is a pair of disjoint nonempty sets $A$ and $B$ whose union is $Y$, neither of which contains a limit point of the other. The space $Y$ is connected if there exists no separation of $Y$.

Constructing new connected spaces from old.

[Lemma 23.2] If the sets $C$ and $D$ form a separation of $X$, and if $Y$ is a connected subspace of $X$, then $Y$ lies entirely within either $C$ or $D$.

[Theorem 23.3] The union of a collection of connected subspaces of $X$ that have a point in common (i.e. the intersection of the collection is nonempty) is connected.

[Theorem 23.4] Let $A$ be a connected subspace of $X$. If $A \subseteq B \subseteq \overline{A}$, then $B$ is also connected.

Equivalently: if $B$ is formed by adjoining to the connected subspace $A$ some or all of tis limit points, then $B$ is connected.

[Theorem 23.5] (Continuous functions preserve connectedness). The image of a connected space under a continuous map is connected.

Let $f : X \to Y$ be a continuous map where $X$ is connected. We want to prove that $f(X)$ is connected.

Since the map obtained from $f$ by restricting its range to the space $f(X)$ is also continuous (see Theorem 18.2 in Chapter 2), it suffices to consider the case of a continuous surjective map $g : X \to f(X)$.

Assume that there exists two disjoint nonempty open subsets $A$ and $B$ of $Y$ such that $f(X)=A \cup B$, then $g^{-1}(A)$ and $g^{-1}(B)$ are disjoint sets whose union is $X$. Since $g$ is continuous, $g^{-1}(A)$ and $g^{-1}(B)$ are open in $X$; they are also nonempty since $g$ is surjective. Thus, they form a separation of $X$, contradicting the connectedness of $X$.

[Theorem 23.6] A finite Cartesian product of connected spaces is connected.

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Does Theorem 23.6 extend to arbitrary products of connected space? The answer is: it depends on the topology used for the product.

In fact, an arbitrary product of connected space is connected in the product topology. A proof of this result is a solution to Exercise 23.10.

Connected Subspaces of the Real Line

Definition. A simply ordered (a.k.a. strictly totally ordered) set $L$ having more than one element is called a linear continuum if the following hold:

1. $L$ has the least upper bound property.
2. If $x < y$, there exists $z$ such that $x < z < y$ (i.e. the order on $L$ is dense).

[Theorem 24.1] If $L$ is a linear continuum in the order topology, then $L$ is connected, and so are intervals and rays in $L$.

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Since $\mathbb{R}$ is a linear continuum in order topology, we immediately see that:

[Corollary 24.2] The real line $\mathbb{R}$ is connected and so are intervals and rays in $\mathbb{R}$.

One may also ask: does the converse of Theorem 24.1 hold? The answer is in fact yes, and a proof of this is a solution to Exercise 24.4.

[Theorem 24.3] (Intermediate Value Theorem). Let $f : X \to Y$ be a continuous map, where $X$ is a connected space and $Y$ is an ordered set in the order topology. If $a$ and $b$ are two points of $X$ and if $r$ is a point of $Y$ lying between $f(a)$ and $f(b)$, then there exists a point $c$ of $X$ such that $f(c)=r$.

Definition. (Path-connectedness). Given points $x$ and $y$ of the space $X$, a path in $X$ from $x$ to $y$ is a continuous map $f : [a,b] \to X$ of some closed interval in the real line into $X$, such that $f(a)=x$ and $f(b)=y$. A space $X$ is said to be path connected if every pair of points of $X$ can be joined by a path in $X$.

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The converse of the definition above does not hold; a connected space need not be path connected.

[Example 6]. The orderred square $I_o^2$ (recall that it is defined as $I \times I \subset \mathbb{R}^2$, where $I := [0,1]$, endowed with the dictionary order) is connected but not path connected.

Since it is a linear continuum (as a subset of $\mathbb{R}^2$), it is connected. We show that $I_o^2$ is not path connected as follows:

Let $p = (0,0)$ and $q = (1,1)$ (note that they are respectively the smallest and largest element of $I_o^2$). Assume for the sake of contradiction that there is a path $f : [a,b] \to I_o^2$ joining $p$ and $q$. It follows that the image set $f([a,b])$ must contain every point $(x,y)$ of $I_o^2$ by the intermediate value theorem (recall that a path is a continuous map).

Therefore, for each $x \in I$, the set

$$U_x = f^{-1}(\{x\} \times (0,1))$$

is a nonempty subset of $[a,b]$; by continuity, it is open in $[a,b]$. Choose, for each $x \in I$, a rational number $q_x$ belonging to $U_x$. Since the sets $U_x$ are disjoint, the map $x \mapsto q_x$ is an injective mapping of $I$ into $\mathbb{Q}$. This contradicts the fact that the inverval $I$ is uncountable.

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The uncountability of $I$ comes from the fact that if there were a bijective mapping from the open interval $(0,1)$ (note that this is just $I$ but 0 and 1 are excluded) to the natural numbers (which is the definition of being countable) so that each element of it can be indexed as $x_i=0.d_{i1}d_{i2}d_{i3}\cdots$ where $i \in \mathbb{N}$, then one can come up with an element in $I$ given by $x=0.a_1 a_2 a_3 \cdots$ where $a_i=d_{ii}+1 \pmod{10}$, which is not equal to any of the $x_i$'s, contradicting the existence of said bijective mapping.

There is also a topological proof for the uncountability of $I$, which is discussed in Section 27 (Compact Subspaces of the Real Line), specifically Theorem 27.7.

Components and Local Connectedness

Breaking up into smaller pieces.

The notion of local connectedness refers to a natural process of breaking up an arbitrary (topological) space into connected components.

Definition. ([Connected] components). Given $X$, define an equivalence relation on $X$ by setting $x \sim y$ if there is a connected subspace of $X$ containing both $x$ and $y$. The equivalence classes are then called the components (or the 'connected components') of $X$.

[Theorem 25.1] The components of $X$ are connected disjoint subspaces of $X$ whose union is $X$, such that each nonempty connected subspace of $X$ intersects only one of them.

This comes from the fact that the components form equivalence classes in $X$, which induces a partition of $X$.

Definition. (Path components). We define another equivalence relation on the space $X$ by defining $x \sim y$ if there is a path in $X$ from $x$ to $y$. The equivalence classes are called the path components of $X$.

[Theorem 25.2] The path components of $X$ are path-connected disjoint subspaces of $X$ whose union is $X$, such that each nonempty path-connected subspace of $X$ intersects only one of them.

Definition. (Local connectedness). A space $X$ is said to be locally connected at $x$ if for every neighbourhood $U$ of $x$, there is a connected neighbourhood $V$ of $x$ contained in $U$. If $X$ is a locally connected at each of its points, it is said simply to be locally connected.

Similarly, a space $X$ is said to be locally path connected at $x$ if for every neighbourhood $U$ of $x$, there is a path-connected neighbourhood $V$ of $x$ contained in $U$. If $X$ is locally path connected at each of its points, then it is said to be locally path connected.

[Theorem 25.3] A space $X$ is locally connected if and only if for every open set $U$ of $X$, each component of $U$ is open in $X$.

[Theorem 25.4] A space $X$ is locally path connected if and only if for every open set $U$ of $X$, each path component of $U$ is open in $X$.

[Theorem 25.5] (Relation between path components and components). If $X$ is a topological space, each path component of $X$ lies in a component of $X$. If $X$ is locally path connected, then the components and the path components of $X$ are the same.

Compactness

According to the Wikipedia article on compact spaces, the notion of compactness seeks to generalise the notion of closed and bounded subsets of Euclidean space.

Compactness is also key in extending local properties to global properties (i.e. properties that apply to the whole space).

Definition. A collection $\mathcal{A}$ of subsets of a space $X$ is said to cover $X$, or to be a covering of $X$, if the union of the elements of $A$ is equal to $X$. It is called an open covering of $X$ if its elements are open subsets of $X$.

Definition. A space $X$ is said to be compact if every open covering $\mathcal{A}$ of $X$ contains a finite subcover (i.e. a finite subcollection that also covers $X$).

[Lemma 26.1] Let $Y$ be a subspace of $X$, then $Y$ is compact if and only if every covering of $Y$ by sets open in $X$ contains a finite subcollection covering $Y$.

[Theorem 26.2] Every closed subspace of a compact space is compact.

[Theorem 26.3] Every compact subspace of a Hausdorff space is closed.

[Lemma 26.4] 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.

[Theorem 26.5] The image of a compact space under a continuous map is compact.

Let $f : X \to Y$ be continuous and $X$ be compact. Let $\mathcal{A}$ be a covering of the set $f(X)$ by sets open in $Y$. It follows that

$$\{f^{-1}(A) : A \in \mathcal{A}\}$$

is a collection of sets covering $X$; since $f$ is continuous, these sets are open. Since $X$ is compact, there are finitely many of them, say,

$$f^{-1}(A_1),\ldots, f^{-1}(A_n),$$

that cover $X$, so the sets $A_1,\ldots,A_n$ cover $f(X)$.

[Theorem 26.6] Let $f : X \to Y$ be a bijective continuous function. If $X$ is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism.

[Theorem 26.7] The product of finitely many compact spaces is compact.

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A natural question arises from Theorem 26.7: Is the product of infinitely many compact spaces compact?. The answer is in fact 'yes.' It is called the Tychonoff's theorem, which is further discussed in Chapter 5 of this book.

[Lemma 26.8] (The tube lemma). Consider the product space $X \times Y$, where $Y$ is compact. If $N$ is an open set of $X \times Y$ containing the slice $x_0 \times Y$ of $X \times Y$, then $N$ contains some tube $W \times Y$ about $x_0 \times Y$, where $W$ is a neighbourhood of $x_0$ in $X$.

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Definition. A tube about $x_0 \times Y$ is the set $W \times Y$, where $W$ is a neighbourhood $W$ of $x_0$ in $X$ such that $N$ (as defined above) contains the entire set $W \times Y$.

Definition. A collection $\mathcal{C}$ of subsets of $X$ is said to have the finite intersection property if for every finite subcollection $\{C_1,\ldots,C_n\}$ of $\mathcal{C}$, the intersection $C_1 \cap \cdots \cap C_n$ is nonempty.

[Theorem 26.9] Let $X$ be a topological space. Then $X$ is compact if and only if for every collection $\mathcal{C}$ of closed sets in $X$ having the finite intersection property, the intersection $\bigcap_{C \in \mathcal{C}}C$ of all the elements of $\mathcal{C}$ is nonempty.

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A special case of Theorem 26.9 is the Cantor's intersection theorem: given a nested sequence $C_1 \supseteq C_2 \supseteq \cdots \supseteq C_n \supseteq C_{n+1} \supseteq \cdots$ of closed sets in a compact space $X$, if each of the sets $C_n$ is nonempty, then the collection $\mathcal{C} = \{C_n\}_{n \in \mathbb{Z}_+}$ automatically has the finite intersection property, so the intersection $\bigcap_{n \in \mathbb{Z}_+} C_n$ is nonempty.

Compact Subspaces of the Real Line

[Theorem 27.1] Let $X$ be a simply ordered set having the least upper bound property. In the order topology, each closed interval in $X$ is compact.

[Corollary 27.2] Every closed interval in $\mathbb{R}$ is compact.

[Theorem 27.3] A subspace $A$ of $\mathbb{R}^n$ is compact if and only if it is closed and is bounded in the Euclidean metric $d$ or the square metric $\rho$.

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Theorem 27.3 is also known as the Heine-Borel theorem in real analysis.

[Theorem 27.4] (Extreme value theorem, generalised). Let $f : X \to Y$ be continuous, where $Y$ is an ordered set in the order topology. If $X$ is compact, then there exists points $c$ and $d$ in $X$ such that $f(c) \leqslant f(x) \leqslant f(d)$ for every $x \in X$.

Definition. Let $(X,d)$ be a metric space; let $A$ be a nonempty subset of $X$. For each $x \in X$, we define the distance from $x$ to $A$ by the equation

$$d(x,A)=\inf\{d(x,a) : a \in A\}$$

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For fixed $A$, the function $d(x,A)$ is a continuous function of $x$. Indeed, given $x, y \in X$, we have the inequalities

$$d(x,A) \leqslant d(x,a) \leqslant d(x,y)+d(y,a)$$

for each $a \in A$, which implies that

$$d(x,A)-d(x,y) \leqslant \inf_{a \in A}d(y,a)=d(y,A) \implies d(x,A)-d(y,A) \leqslant d(x,y).$$

The same inequality holds with $x$ and $y$ interchanged. The continuity of the function $d(x,A)$ follows from the continuity of $d(x,y)$ and Theorem 21.1 (by setting $\delta=\varepsilon$).

[Lemma 27.5] (The Lebesgue number lemma). Let $\mathcal{A}$ be an open covering of the metric space $(X,d)$. If $X$ is compact, there is a $\delta>0$ such that for each subset of $X$ having diameter less than $\delta$, there exists an element of $\mathcal{A}$ containing it.

The number $\delta$ is called a Lebesgue number for the covering $\mathcal{A}$.

Definition. A function $f$ from the metric space $(X,d_X)$ to the metric space $(Y,d_Y)$ is said to be uniformly continuous if given $\varepsilon>0$, there is a $\delta>0$ such that for every pair of points $x_0,x_1$ of $X$, we have $d_X(x_0,x_1)<\delta \implies d_Y(f(x_0),f(x_1)) < \varepsilon$.

[Theorem 27.6] (Uniform continuity theorem). Let $f : X \to Y$ be a continuous map of the compact metric space $(X,d_X)$ to the metric space $(Y,d_Y)$, then $f$ is uniformly continuous.

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Theorem 27.6 is closely related to the continuous extension theorem in real analysis.

Definition. If $X$ is a space, a point $x$ of $X$ is called an isolated point of $X$ if the one-point set $\{x\}$ is open in $X$.

[Theorem 27.7] Let $X$ be a nonempty compact Hausdorff space. If $X$ has no isolated points, then $X$ is uncountable.

Step 1. We show first that for any nonempty open set $U$ of $X$ and any point $x$ in $X$, there exists a nonempty open set $V$ contained in $U$ such that $x \in \overline{V}$.

Choose a point $y \in U$ such that $y \neq x$; this is possible if $x \in U$ because $x$ is not an isolated point of $X$ and it is possible if $x \notin U$ since $U \neq \emptyset$. Now choose disjoint open sets $W_1$ and $W_2$ about $x$ and $y$, respectively (i.e. $x \in W_1$ and $y \in W_2$; this is possible since $X$ is Hausdorff), then the set $V=W_2 \cap U$ is the desired open set; it is contained in $U$, it is nonempty because it contains $y$, and its closure does not contain $x$.

Step 2. We show that given $f : \mathbb{Z}_+ \to X$, the function $f$ is not surjective, which implies that $X$ is uncountable.

Let $x_n = f(n)$. Apply Step 1 to the nonempty open set $U = X$ to choose a nonempty open set $V_1 \subseteq X$ such that $x_1 \notin \overline{V_1}$. In general, given $V_{n-1}$ open and nonempty, choose $V_n$ to be a nonempty open set such that $V_n \subseteq V_{n-1}$ and $x_n \notin \overline{V_n}$.

Consider the nested sequence $\overline{V_1} \supseteq \overline{V_2} \supseteq \cdots$ of nonempty closed sets of $X$. Since $X$ is compact, by Theorem 26.9, there is a point $x \in \bigcap \overline{V_n}$. Now $x$ cannot equal $x_n$ for any $n$, since $x$ belongs to $\overline{V_n}$ and $x_n$ does not.

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[Corollary 27.8] (from Theorem 27.7) Every closed interval in $\mathbb{R}$ is uncountable.

Every closed interval $I$ in $\mathbb{R}$ is nonempty, compact and Hausdorff. It also has no isolated points, since if we consider $I$ as a subspace of $\mathbb{R}$, the intersection of $I$ with every nonempty open set of $\mathbb{R}$ must contain more than one point.

Limit Point Compactness

Definition. A space $X$ is said to be limit point compact if every infinite subset of $X$ has a limit point.

[Theorem 28.1] Compactness implies limit point compactness, but the converse is not true.

Definition. (Subsequence and sequential compactness). Let $X$ be a topological space. If $(x_n)$ is a sequence of points of $X$, and if $0 < n_1 < n_2 < \cdots < n_i < \cdots$ is an increasing sequence of positive integers, then the sequence $(y_i)$ defined by setting $y_i = x_{n_i}$ is called a subsequence of the sequence $(x_n)$. The space $X$ is said to be sequentially compact if every sequence of poitns of $X$ has a convergent subsequence.

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The notion of sequential compactness is closely related to the Bolzano-Weierstrass theorem.

[Theorem 28.2] Let $X$ be a metrizable space, then $X$ is compact iff $X$ is limit point compact iff $X$ is sequentially compact.

Local Compactness

Definition. A space $X$ is said to be locally compact at $x$ if there is some compact subspace $C$ of $X$ that contains a neighbourhood of $x$. If $X$ is locally compact at each of its points, $X$ is said simply to be locally compact.

Note that a compact space is automatically locally compact.

[Theorem 29.1] Let $X$ be a space, then $X$ is locally compact Hausdorff if and only if there exists a space $Y$ satisfying the following conditions:

1. $X$ is a subspace of $Y$.
2. The set $Y \backslash X$ consists of a single point.
3. $Y$ is a compact Hausdorff space.

In fact, such $Y$ can be constructed by adjoining a point not in $X$ to $X$, and defining its topology to consist of all sets $U$ that are open in $X$ and all sets of the form $Y \backslash C$ where $C$ is a compact subspace of $X$.

If $Y$ and $Y'$ are two spaces satisfying the conditions above, then there is a homeomorphism of $Y$ with $Y'$ which equals the identity map on $X$. (Another way of phrasing this is that the existence of such space is unique up to a homeomorphism that maps $X$ identically.)

Definition. If $Y$ is a compact Hausdorff space and $X$ is a proper subspace of $Y$ whose closure equals $Y$, then $Y$ is said to be a compactification of $X$. If $Y \backslash X$ is a single point, then $Y$ is called the one-point compactification of $X$ (a.k.a. Alexandroff compactification).

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From Theorem 29.1, we see that $X$ has a one-point compactification $Y$ if and only if $X$ is a locally compact Hausdorff space that is not itself compact, since in this case the point of $Y \backslash X$ is a limit point of $X$, so that $\overline{X}=Y$.

If $X$ itself happens to be compact, then the space $Y$ obtained from Theorem 29.1 is obtained from $X$ by adjoining a single isolated point.

[Theorem 29.2] (Equivalent formulation of local compactness for Hausdorff spaces) Let $X$ be a Hausdorff space, then $X$ is locally compact if and only if given $x$ in $X$, and given a neighbourhood $U$ of $x$, there is a neighbourhood $V$ of $x$ such that $\overline{V}$ is compact and $\overline{V} \subseteq U$.

[Corollary 29.3] Let $X$ be locally compact Hausdorff and $A$ be a subspace of $X$. If $A$ is closed in $X$ or open in $X$, then $A$ is locally compact.

[Corollary 29.4] A space $X$ is homeomorphic to an open subspace of a compact Hausdorff space if and only if $X$ is locally compact Hausdorff.

Suppose that $X$ is locally compact Hausdorff. If $X$ itself is compact, then $X$ itself is an open subspace of a compact Hausdorff space that is homeomorphic to $X$, as desired. If $X$ is not compact, then $X$ has a one-point compactification $Y$ given by Theorem 29.1, such that $X$ is an open subspace of $Y$, which is as desired because again, $X$ is indeed homeomorphic to itself.

Suppose that $X$ is homeomorphic to an open subspace of a compact Hausdorff space. If $X$ is compact, then $X$ must be locally compact and the assumed homeomorphism implies that $X$ is also Hausdorff due to Exercise 17.12, so we are done. If $X$ is not compact, then the one-point compactification $Y$ given by Theorem 29.1 has $X$, which is Hausdorff, as its open subspace. It is also locally compact by Corollary 29.3.