Point-Set Topology

If $A$ is a subset of $X$ that is both open and closed, then so is $A^c$, and then $X=\mathit{𝐴\coprod }{A}^c$. Conversely, if $X=U\mathit{\coprod 𝑉}$ with $U$ and $V$ nonempty and open in $X$, then $U$ and $V$ are also closed. □

We have that $A\cap U$ and $A\cap V$ are open and closed in $A$, so $A$ is the disjoint union of these intersections. If $A$ is connected, this forces one of the $A\cap U$ and $A\cap V$ to be empty, and then the other is $A$. □

Let $X$ be a topological space and $A$ a connected subset. Suppose that $\overline{A}=U\mathit{\coprod 𝑉}$ for disjoint subspaces $U$ and $V$ in $X$. By Lemma 3.1.6, we have that $A$ is contained in either $U$ or $V$: without loss of generality, we suppose $A\subseteq U$. As $U$ is closed, we have that $\overline{A}\subseteq U$ as well, and therefore $V=\varnothing$. □

Suppose $X$ is connected and $f:X\to Y$ is continuous. If $Y$ is the disjoint union of (open) subspaces $U$ and $V$, then $X$ is the union of the disjoint open subsets $f^{-1}(U)$ and $f^{-1}(V)$, hence equal to their disjoint union as topological spaces. □

Let $a\in {\bigcap }_{i\in I}{A}_i$. If $X=U\mathit{\coprod 𝑉}$ for subspaces $U$ and $V$, then without loss of generality we may suppose $a\in U$. Since $A_i$ is connected and contains $a$, we must then have $A_i\subseteq U$ for all $i\in I$, forcing $V=\varnothing$. □

Clearly $\sim$ is reflexive. For symmetry, it’s enough show that for $x,y\in X$, the connected components $A$ of $x$ and $B$ of $y$ are either equal or disjoint. Suppose that $z\in A\cap B$. Then $A\cup B$ is connected by Proposition 3.1.9, but in that $A$ (resp., $B$) is the largest connected subset of $X$ containing $x$ (resp., $y$), we must have $A=A\cap B=B$. Finally, if $x\sim y$ and $y\sim z$ for $x,y,z\in X$, then we’ve just seen that $x$ and $y$ have the same connected component, as do $y$ and $z$, so $x$ and $z$ do as well. □

We know that $X$ is the union of its connected components, which are disjoint by Proposition 3.1.11. It follows from Proposition 3.1.7 that connected components are closed, being the largest connected subsets containing a given point. The second statement follows from the definition of a disjoint union of topological spaces. If $X$ has finitely many connected components, then any union of all but one of them is closed, and then they are all open as complements of these unions. □

Let $X$ be a path connected space. Suppose that $X$ is a disjoint union of nonempty open subspaces $U$ and $V$, and let $x\in X$ and $y\in Y$. Let $\gamma$ be a path from $x$ to $y$, and let $A=\gamma ([0,1])$. Then $A$ is the disjoint union of the nonempty sets $A\cap U$ and $A\cap V$, so $A$ is disconnected, but it is the image of a connected space under a continuous function. □

As all closed intervals of finite length are homeomorphic, we can and will consider the interval $[0,1]$. Let $𝒰$ be an open cover of $[0,1]$, and let $A$ be the subset of $[0,1]$ consisting of those $x$ such that $[0,x]$ has a finite subcover by elements in $𝒰$. Let $b$ be the supremum of of the elements of $A$. If $b<1$, then let $U$ be an element $𝒰$ containing $b$. Since $U$ contains an interval $(b-𝜖,b+𝜖)$ for some $𝜖>0$ with $b+𝜖\le 1$, we can find a finite subcover $𝒱$ of $[0,b-\frac{𝜖}{2}]$ inside $𝒰$. Ten $𝒱\cup \{U\}$ is a finite subcover of $[0,b+\frac{𝜖}{2}]$ in $𝒰$. This means that $b+\frac{𝜖}{2}\in A$, contradicting the fact that $b$ is the supremum of all elements of $A$. Thus $b=1$, and therefore $𝒰$ has a finite subcover. □

Given an open cover $𝒱$ of $A$ by open sets in $A$, we can find a set $𝒰$ of open sets in $X$ covering $A$ such that $𝒱=\{A\cap U\mid U\in 𝒰\}$. Conversely, given a collection $𝒰$ of open sets in $X$ covering $A$, we may define a cover $𝒱$ of $A$ by open sets in $A$ by taking intersections with $A$ as in the latter formula. Any (finite) subset $𝒞$ of $𝒰$ covers $A$ if and only if the (finite) subset $\{A\cap U\mid A\in 𝒞\}$ of $𝒱$ covers $A$. □

Let $X$ be compact, and let $A\subseteq X$ be closed. Let $𝒰$ be an open cover of $A$ in $X$. Then $𝒰\cup \{A^c\}$ is an open cover of $X$, so it has a finite subcover $𝒱$. If $A^c\in 𝒱$, then $𝒱-\{A^c\}$ is an open cover of $A$ in $X$, and otherwise, $𝒱$ is an open cover of $A$ in $X$. □

For each $a\in A$, choose open disjoint neighborhoods $U_a$ of $a$ and $V_a$ of $x$ in $X$. The collection $\{U_a\mid a\in A\}$ is an open cover of $A$, and it has a finite subcover, say by $U_{a_1},\dots ,U_{a_n}$. Then $U={\bigcup }_{i=1}^n{U}_{a_i}$ and $V={\bigcap }_{i=1}^n{V}_{a_i}$ are the desired open subsets of $X$. □

Let $X$ be Hausdorff, and let $A\subseteq X$ be compact. By Lemma 3.2.7, for each $x\in A^c$, there exists an open neighborhood $V_x$ of $x$ in $A^c$. The union of the $V_x$ is $A^c$, so $A^c$ is open, and thus $A$ is closed. □

Let $𝒱$ be an open cover of $f(X)$ in $Y$. Then $𝒰=\{f^{-1}(V)\mid V\in 𝒱\}$ is an open cover of $X$. Since $X$ is compact, it has a finite subcover $\{f^{-1}(V_1),\dots ,f^{-1}(V_n)\}$ with each $V_i\in 𝒱$, and then $\{V_1,\dots ,V_n\}$ is an open cover of $f(X)$. □

To see that $f$ is a quotient map, we may show that $B\subseteq Y$ is closed if $f^{-1}(B)$ is closed. Since $f$ is continuous and surjective, we may write $B=f(f^{-1}(B))$, so it suffices to show that $f$ is a closed map.

Let $A$ be a closed set in $X$, which is necessarily compact. As $f$ is continuous, its image $f(A)$ is compact as well. As $Y$ is Hausdorff, we then have that $f(A)$ is closed. □

For each $x\in X$, let $U_x$ be an open neighborhood of $x$ in $X$ and $V_x$ be an open neighborhood of $y$ in $Y$ such that $U_x\times V_x\subseteq W$, which exist since $(x,y)\in W$ and $W$ is open. The collection $\{U_x\mid x\in X\}$ covers $X$ so has a finite subcover, say consisting of $U_{x_1},\dots ,U_{x_n}$. If we set $V={\bigcap }_{i=1}^n{V}_{x_i}$, then $V$ is an open neighborhood of $y$ in $Y$, and since each $U_{x_i}\times V$ is contained in $W$, so is their union $X\times V$. □

It suffices by recursion to consider the case $n=2$, so the product of compact spaces $X$ and $Y$. Let $𝒲$ be an open covering of $X\times Y$. For any $y\in Y$, the set $X\times \{y\}$ is compact, being homeomorphic to $X$ via the projection map, and so there exist $Z_1,\dots ,Z_m\in 𝒲$ that together cover $X\times \{y\}$. Set $W_y={\bigcup }_{i=1}^m{Z}_i$. We then have by Lemma 3.2.12 that there exists an open neighborhood of $V_y$ of $y$ such that $X\times V_y\subseteq W_y$. The collection $\{V_y\mid y\in Y\}$ then covers $Y$, and it has a finite subcover as $Y$ is compact. But then $X\times Y$ is covered by the finitely many $W_y$, and each of these is in turn a union of finitely many sets in $𝒲$. Thus, $𝒲$ has a finite subcover. □

Fix a subset $A$ of ${ℝ}^n$. We may consider the cover of $A$ by all open balls in ${ℝ}^n$ of radius $1$. If $A$ is unbounded, then it cannot have a finite subcover, since the union of any finite number of balls of radius $1$ is bounded. Therefore, only bounded subsets can be compact.

Now suppose that $A$ is bounded. It is then contained in a direct product of closed intervals, which is compact as a finite product of compact sets. If $A$ is also closed, then it is compact by Proposition 3.2.6. On the other hand, if $A$ is not closed, then it cannot be compact by Proposition 3.2.8. □

Let $𝒜$ be a collection of closed subsets of $X$ with the finite intersection property. Then $𝒰=\{A^c\mid A\in 𝒜\}$ is a collection of open subsets of $X$ with the property that no finite subset of $𝒰$ covers $X$. If $X$ is compact, this forces $𝒰$ not to cover $X$, which is exactly to say that the intersection of elements of $𝒜$ is nonempty.

Conversely, if $X$ is not compact, there exists an open cover $𝒰$ of $X$ which has no finite subcover, which is to say that the collection $𝒜=\{U^c\mid U\in 𝒰\}$ of closed sets has the finite intersection property but has empty intersection. □

If $X$ is finite Hausdorff, then it is discrete, so has singleton open sets. So, suppose by way of contradiction that $X$ is countably infinite. Label its points $x_i$ for $i\ge 1$. Let $X=U_0$, which is an open neighborhood of $x_1$. By Lemma 1.5.13, we may recursively let $U_i$ be an open neighborhood of $x_{i+1}$ contained in $U_{i-1}$ with $x_i\notin \overline{U_i}$ for each $i\ge 1$.

The intersection of any finite subcollection of $\{U_i\mid i\ge 1\}$ is some $U_n$ for $n\ge 1$ since these $U_i$ are nested. Since $x_{n+1}\in U_n$, this collection satisfies the FIP. In particular, the collection $\{\overline{U_i}\mid i\ge 1\}$ of closed sets does as well. Now, $X$ being compact, we then have that ${\bigcap }_{i=1}^{\infty }\overline{U_i}$ is nonempty. At the same time, since $x_i\notin \overline{U_i}$, this intersection does not contain any of the points of $X=\{x_i\mid i\ge 1\}$, providing the desired contradiction. □

By Theorem 3.2.17, the interval $[0,1]$ is uncountable, and therefore so is $ℝ$. □

Let $X$ be a compact space, and let $A$ be a subset without a limit point. Then $A$ is necessarily closed. Moreover, for each $a\in A$, there exists an open set $U_a$ containing $a$ and no other point of $A$. Then $X$ has an open cover by $\{U_a\mid a\in A\}\cup \{A^c\}$. Since $X$ has a finite subcover, $A$ is contained in a finite union of sets $U_{a_1},\dots ,U_{a_n}$, which implies that $A=\{a_1,\dots ,a_n\}$, so $A$ is finite. □

Let $A$ be an infinite set in $X$, and let ${(a_n)}_{n\ge 1}$ be a sequence of distinct elements in $A$. It has a convergent subsequence, say with limit $a$. Then every neighborhood of $a$ contains some point of the subsequence not equal to $a$. Thus, $a$ is a limit point of $A$. □

Suppose that some open cover $𝒰$ of $X$ has infinite Lebesgue number. For each $n\ge 1$, there exists a subset $A_n$ of $X$ of diameter less than $\frac{1}{n}$ not contained in any element of $𝒰$. For each such $A_n$, choose $x_n\in A_n$. The sequence ${(x_n)}_{n\ge 1}$ has a convergent subsequence ${(x_{n_k})}_{k\ge 1}$, say with limit $x$. Let $U\in 𝒰$ be an open neighborhood of $x$, and let $B(x,\delta )$ be an open ball inside of it. Let $k$ be sufficiently large such that $\frac{1}{n_k}<\frac{\delta }{2}$ and $d(x,x_{n_k})<\frac{\delta }{2}$. By the triangle inequality, we then have $A_{n_k}\subseteq B(x,\delta )\subseteq U$, providing the desired contradiction. □

Let $X$ be a metric space, and suppose that $𝜖>0$ is such that $X$ cannot be covered by finitely many balls of radius $𝜖$. Choose $x_1\in X$ and then recursively choose $x_n$ in the complement of ${\bigcup }_{i=1}^{n-1}B(x_i,𝜖)$. The sequence ${(x_n)}_{n\ge 1}$ satisfies $d(x_m,x_n)\ge 𝜖$ for all $m>n$, and consequently it cannot have a convergent subsequence (since every convergent subsequence is necessarily Cauchy). Thus, $X$ not sequentially compact. □

Let $X$ be a limit point compact space. Let ${(x_n)}_{n\ge 1}$ be a sequence in $X$. If the set $A=\{x_n\mid n\ge 1\}$ of values of the sequence is finite, then ${(x_n)}_{n\ge 1}$ has a constant, hence convergent, subsequence. Otherwise, $A$ is infinite, so has a limit point $x\in X$. Inductively, we have that every ball among the $B(x,\frac{1}{k})$ for $k\ge 1$ contains some $x_{n_k}$ with $n_k\ge n_{k-1}$ if $k\ge 2$. The sequence ${(x_{n_k})}_{k\ge 1}$ then converges to $x$. Thus, $X$ is sequentially compact.

Now suppose that $X$ is sequentially compact, and let $𝒰$ be an open cover of $X$. Let $𝜖>0$ be such that every subset of diameter less than $𝜖$ is contained in an element of $𝒰$. Choose a finite open cover of $X$ by open balls of radius $\frac{𝜖}{3}$, and note that they have diameter at most $\frac{2𝜖}{3}$, hence are each contained in some element of $𝒰$. Then $𝒰$ has a finite subcover by these elements, and therefore $X$ is compact. □

Suppose that $X$ is not compact, and let $𝒮$ be a subbase of $X$. We need to show that $𝒮$ contains an open cover $𝒰$ that does not have a finite subcover.

Let $Q$ be the set of all open covers of $X$ that have no finite subcover, and note that $Q\ne \varnothing$ by the noncompactness of $X$. Let $𝒞$ be a chain in $Q$, and let $𝒲={\bigcup }_{𝒱\in 𝒞}𝒱$ be the union of all covers in $𝒞$. Any finite collection of elements of $𝒲$, being each contained in some element of $𝒞$, are all contained in the largest such element $𝒱$ under inclusion. Being that $𝒱$ has no finite subcover, such a finite collection cannot cover $X$, so $𝒲$ has no finite subcover, which is to say that $𝒲\in Q$. Thus, every chain in $Q$ has an upper bound, and so by Zorn’s lemma, $Q$ has a maximal element $M$.

Now consider the subset $𝒰=𝒮\cap M$ of $𝒮$. Being a subset of $M$, no finite subset of $𝒰$ covers $M$. We claim that $𝒰$ covers $X$, which will finish the proof. Let $x\in X$, let $U\in M$ with $x\in U$. Let $V_1,\dots ,V_n\in 𝒮$ be such that $x\in V_1\cap \cdots \cap V_n\subseteq U$, which exist as $𝒮$ is a subbase. If $V_i\notin M$ for all $1\le i\le n$, we can find by the maximality of $M$ finite subsets ${𝒩}_i$ of $M$ such that the collections ${𝒩}_i\cap \{V_i\}$ cover $X$. Then $𝒩={\bigcup }_{i=1}^n{𝒩}_i$ and $V_1\cap \dots \cap V_n$ together cover $X$, so $𝒩\cup \{U\}\subset M$ covers $X$ as well, contradicting the fact that $M$ has no finite subcover. Thus, there exists $i$ such that $V_i\in 𝒰$, and $V_i$ contains $x$ by definition. Thus $𝒰$ is a cover of $X$. □

Let $\{X_i\mid i\in I\}$ be a collection of compact spaces, and let $X={\prod }_{i\in I}{X}_i$. For each $i\in I$, let ${𝒯}_i$ be the topology on $X_i$, and let

where ${\pi }_i:X\to X_i$ is the $i$th projection map. Then $𝒮$ is a subbase for the topology on $X$. We claim that every open cover of $X$ by elements in $𝒮$ has a finite subcover, which by the Alexander subbase theorem will finish the proof.

Let $𝒰\subseteq 𝒮$ be an open cover of $X$. Then

for some collections ${𝒰}_i$ of open sets in $X_i$ for each $i\in I$. If no ${𝒰}_i$ covers $X_i$, then for each $i\in I$, we may by the axiom of choice find $x_i\in X_i$ such that $x_i\notin {\bigcup }_{U_i\in {𝒰}_i}{U}_i$. Set $x={(x_i)}_{i\in I}$. Then $x\notin U$ for any $U\in 𝒰$, for any such $U$ has the form $U={\pi }_i^{-1}(U_i)$ for some $U_i\in {𝒰}_i$ for some $i\in I$, and $x_i={\pi }_i(x)\notin U_i$. Thus there exists $i\in I$ such that ${𝒰}_i$ covers $X_i$. It has a finite subcover $𝒱$, and the finite set $\{{\pi }_i^{-1}(V)\mid V\in 𝒱\}$ then covers $X$. □

Suppose $X$ is locally connected. Let $U$ be open in $X$, and let $A$ be a connected component of $U$. If $x\in A$, then there is a connected open neighborhood $V$ of $x$ that is contained in $U$, and $V$ is contained in $A$ by its connectedness. Since $x$ was arbitrary, $A$ is open.

Conversely, suppose that all connected components of open sets in $X$ are open in $X$. Let $x\in X$, and let $V$ be an open neighborhood of $x$. Let $A$ be the connected component of $x$ in $V$. Note that $A$ is open by assumption, so it is a neighborhood of $x$. Thus, $X$ is locally connected. □

Note that arbitrary intersections of closed, compact subsets $A$ of $X$ are closed, and then compact, and the union of an open subset $U$ of $X$ and the complement $Y-A$ of a closed, compact subset $A$ of $X$ is the complement in $Y$ of a smaller closed, and then compact, subset of $X$. It follows that the collection $𝒯$ of open sets defined in the theorem is closed under arbitrary unions. Similarly, finite unions of closed, compact subsets of $X$ are closed and compact, and the intersection of $U$ and $Y-A$ as above is an open subset of $X$. Hence, $𝒯$ is closed under arbitrary intersections, so $𝒯$ is a topology.

Given an open cover $𝒱$ of $Y$, we may choose an element in it of the form $Y-A$ with $A$ closed, compact in $X$, as such a set is needed to cover $\infty$. Then $𝒱-\{Y-A\}$ is an open cover of $A$, which has a finite subcover as $A$ is compact, so $𝒱$ has a finite subcover as well. Thus, $Y$ is compact.

Finally, for any other space $Y^{\prime }$ as in the theorem, with ${\infty }^{\prime }$ its additional point, is in canonical bijection with $Y$ via the map $f:Y\to Y^{\prime }$ such that $f(x)=x$ for $x\in X$ and $f(\infty )={\infty }^{\prime }$. If $V$ is open in $Y$ and ${\infty }^{\prime }\notin Y^{\prime }$, then $f(V)=V$ is open in $X$, hence in $Y^{\prime }$. If $V$ is an open neighborhood of $\infty$ in $Y$, then its complement $A$ is closed, hence compact in $Y$, and of course also contained in $X$. Then $f(V)=Y^{\prime }-f(A)$ is the complement of a compact, closed subset of $X$, so is open in $Y^{\prime }$ as well. Thus, $f$ is continuous, and then $f^{-1}$ is continuous too, as we did not distinguish $Y$ and $Y^{\prime }$. □

Let $Y$ be the one-point compactification of $X$. Suppose that $X$ is locally compact and Hausdorff. To show that $Y$ is Hausdorff, it suffices to consider some $x\in X$ and $\infty$. Since $X$ is locally compact, we can find a compact neighborhood $A$ of $x$, which then contains some open neighborhood $U$. Then $U$ is disjoint from the open neighborhood $Y-A$ of $\infty$.

If $X$ is not Hausdorff, then there exist points $u,v\in X$ not contained in disjoint open sets in $X$. No two complements of compact sets in $X$ are disjoint, since they contain the added point $\infty$. Moreover, if $u\notin A$ and $v\in V$ for a compact subset $A$ of $X$ and an open set $V$ in $X$, then the open complement $X-A$ contains $u$ and is not disjoint from $V$, so $Y-A$ is not disjoint from $v$ either. Thus, $u$ and $v$ are not contained in disjoint open subsets of $Y$, and $Y$ is not Hausdorff.

If $X$ is not locally compact, then it is not locally compact at $\infty$. Given an open neighborhood $Y-A$ of $\infty$, where $A$ is closed in $X$ and compact, and an open neighborhood $U$ of $x$ in $X$, we cannot have $U\subseteq A$, since if this were the case, then $\overline{U}\subseteq A$ would be a compact neighborhood of $x$. Thus, $Y$ is again not Hausdorff. □

As the other direction is immediate, we may suppose that $X$ is locally compact. Let $U$ be open in $X$, and let $x\in U$. Let $Y$ be the one-point compactification of $X$, which is compact Hausdorff since $X$ is locally compact Hausdorff. Let $A=Y-U$, which is closed in $Y$, hence compact. By Lemma 3.2.7, we may find an open neighborhood $V$ of $x$ and an open set $W$ containing $A$ that are disjoint. The closure of $V$ in $Y$ is compact and disjoint from $A$, so is contained in $U$. □

Let $X$ be locally compact Hausdorff. If $A$ is closed in $X$ and $a\in A$, then there is a compact neighborhood $C$ of $a$ in $X$, and $C\cap A$ is then a compact neighborhood of $a$ in $A$. If $U$ is open in $X$ and $x\in U$, then by Proposition 3.5.14, we can find a compact neighborhood $B$ of $x$ in $U$. □

If $X$ is locally compact but not compact, take its one-point compactification, in which $X$ is open. If $X$ is open in a compact Hausdorff space, then it is locally compact by Corollary 3.5.15. □