# General Topology/Compact spaces

Definition (compact space):
Let $X$ be a topological space. $X$ is called compact if and only if for every open cover $(U_{\alpha })_{\alpha \in A}$ of $X$ there exists a finite subcover, that is, indices $\alpha _{1},\ldots ,\alpha _{n}\in A$ so that $X=U_{\alpha _{1}}\cup \cdots \cup U_{\alpha _{n}}$.
Definition (compact subset):
Let $X$ be a topological space and $S\subseteq X$ be a subset. $S$ is called compact iff it is compact with respect to the subspace topology induced on $S$ by the topology of $X$.
Proposition (the image of a compact set via a continuous map is compact):
Let $X,Y$ be topological spaces, $S\subseteq X$ a compact subset and $f:X\to Y$ a continuous function. Then $f(S)$ is a compact subset of $Y$.
Proof: Define $T:=f(S)$ to ease notation. Let $(V_{\beta })_{\beta \in B}$ be an open cover of $T$, that is, by definition of the subspace topology, $V_{\beta }=W_{\beta }\cap T$ for suitable $W_{\beta }$. Since
    $f^{-1}(W_{\beta }\cap T)=f^{-1}(T)\cap f^{-1}(W_{\beta })$,
we obtain upon noting that $S\subseteq f^{-1}(T)$, that the sets
    $U_{\beta }:=S\cap f^{-1}(W_{\beta })$
form an open cover of $S$ with respect to its subspace topology; indeed, the continuity of $f$ insures that each set $f^{-1}(W_{\beta })$ is open. Since $S$ is compact, a finite subcover, indexed by $\beta _{1},\ldots ,\beta _{n}$, may be chosen. Let $y\in T$ be arbitrary. By the definition of $T$, pick $x\in S$ such that $f(x)=y$, and then $j\in [n]$ such that $x\in U_{\beta _{j}}$. Then $f(x)\in S\cap W_{\beta _{n}}=V_{\beta _{n}}$, so that $V_{\beta _{1}}\cup \cdots \cup V_{\beta _{n}}=Y$. $\Box$
Definition (proper map):
A function $f:X\to Y$ between topological spaces is called proper if and only if for each compact subset $T\subseteq Y$, the preimage $f^{-1}(T)$ is a compact subset of $X$.
Note that the composition of proper maps is proper.
Proposition (closed subsets of a compact space are compact):
Let $X$ be a compact space, and let $A\subseteq X$ be closed. Then $A$ is compact.
Proof: Let $(V_{\alpha })_{\alpha \in A}$ be an open cover of $A$. By definition of the subspace topology of $A$, we take $V_{\alpha }=A\cap U_{\alpha }$ where $U_{\alpha }$ is open in $X$; in order to avoid the axiom of choice, we may replace $U_{\alpha }$ by the union of all $U_{\alpha }$ for which $V_{\alpha }=A\cap U_{\alpha }$. Then an open cover of $X$ is given by
    $\{X\setminus A\}\cup (U_{\alpha })_{\alpha \in A}$,
and by compactness of $X$, we may extract a finite subcover. Suppose that $U_{\alpha _{1}},\ldots ,U_{\alpha _{n}}$ are the sets of $(U_{\alpha })_{\alpha \in A}$ that made it into the subcover. Then
    $A\subseteq U_{\alpha _{1}}\cup \cdots \cup U_{\alpha _{n}}$,
since $A$ is contained in the whole subcover, but the only additional set in this subcover may be $X\setminus A$, which doesn't change whether or not $A$ is covered. Thus, $V_{\alpha _{1}},\ldots ,V_{\alpha _{n}}$ is an open subcover of $A$. $\Box$
Theorem (Cantor's intersection theorem):
Let $X$ be a topological space, let $(S,\leq )$ be a directed set and let $(K_{s})_{s\in S}$ be a family of nonempty sets $K_{s}$ which are simultaneously compact and closed, such that $s\leq t\Rightarrow K_{s}\supseteq K_{t}$. Then
    $\bigcap _{s\in S}K_{s}\neq \emptyset$.
Proof: Suppose that
    $\bigcap _{s\in S}K_{s}=\emptyset$.
Note that $K_{1}$ is compact, and since each $K_{j}$ is closed, its complement $U_{j}:=X\setminus K_{j}$ is open. Further, by definition of the subspace topology, the sets $V_{j}:=U_{j}\cap K_{1}$ are open in $K_{1}$, and by de Morgan ($X=\bigcup _{n\in \mathbb {N} }U_{n}$) and distributivity of intersection over union, we get that the $V_{j}$ form an open cover of $K_{1}$. By compactness of $K_{1}$, we may extract a finite subcover $U_{n_{1}},\ldots ,U_{n_{k}}$, and upon choosing $N:=\max\{n_{1},\ldots ,n_{k}\}$, we get that $V_{N}=K_{1}$ since $U_{1}\subseteq U_{2}\subseteq \cdots \subseteq U_{N}$ and thus, since $K_{N}\subseteq K_{1}$, also $K_{N}=\emptyset$, a contradiction. $\Box$
Proposition (compact nonempty Kolmogorov spaces contain a closed point):
Let $X$ be a nonempty, compact T$_0$ space. Then $X$ contains a point $x_{0}$ such that $\{x_{0}\}$ is closed in $X$.
Proof: The set of nonempty, closed subsets of $X$, ordered by inverse inclusion, satisfies the hypotheses of Zorn's lemma, since the arbitrary intersection of closed sets is closed and also nonempty. Hence, there exists a minimal closed set $A$. Suppose that $A$ contains two distinct points $x\neq y$. Then by the hypothesis, select an open set $U\subseteq X$ that contains one point, but not the other, eg. $x\notin U$. Then $x\in$ $\Box$
Proposition (compact subsets of Hausdorff spaces are closed):
Let $X$ be a Hausdorff space, and let $K\subseteq X$ be compact. Then $K$ is closed.
Proof: Let $y\in X\setminus K$ be given. For each $x\in K$, there exist open sets $U_{x,y}$ and $V_{x,y}$ such that $U_{x,y}\cap V_{x,y}=\emptyset$, $x\in U_{x,y}$ and $y\in V_{x,y}$. Since $K$ is compact, select among the $U_{x,y}$ a finite subcover $U_{x_{1},y},\ldots ,U_{x_{1},y}$; note that this step does not use the axiom of choice, since the $U_{x,y}$ cover $X$ in their totality; that is, we include in the cover not only one specific $U_{x,y}$ for each $x$, but all sets of this form. Then set $V_{y}:=V_{x_{1},y}\cap \cdots \cap V_{x_{n},y}$ and obtain that $V_{y}$ is disjoint from $K$; indeed, it can't contain $x\in U_{x_{j},y}$ for any $j\in [n]$. Therefore,
    $X\setminus K\subseteq \bigcup _{y\in X\setminus K}V_{y}\subseteq X\setminus K$, ie. $X\setminus K=\bigcup _{y\in X\setminus K}V_{y}$ open. $\Box$
Conversely, we have:
Proposition (compact sets being closed implies T$_1$):
Let $X$ be a topological space where all compact sets are closed. Then $X$ is T$_1$.
Proof: Any finite subset of $X$ is compact, so that we may apply the characterisation of T$_1$ spaces. $\Box$
Proposition (R$_1$ space is Hausdorff iff all compact sets are closed):
Let $X$ be an R$_1$ space. Then $X$ is Hausdorff iff all compact sets are closed.
Proof: One direction is clear since compact subsets of Hausdorff spaces are closed. For the other direction, we may apply the R-axiom characterisation of Hausdorff spaces, using the fact that $X$ is T$_1$. $\Box$
Proposition (intersection of compact sets in Hausdorff spaces is compact):
Let $(K_{\alpha })_{\alpha \in A}$ be compact subsets of a Hausdorff space $X$. Then
    $\bigcap _{\alpha \in A}K_{\alpha }$
is compact.
Proof: Since $X$ is Hausdorff, all the $K_{\alpha }$s are closed. Hence, the given set is a closed subset of the compact set $K_{\alpha _{0}}$, where $\alpha _{0}\in A$ is arbitrary. $\Box$
Proposition (compact Hausdorff spaces are normal):
Let $X$ be a compact Hausdorff space. Then $X$ is normal.
Proof: Let $A,B\subseteq X$ be two closed subsets of $X$ which are disjoint. First, we note that $A,B$ are compact, since closed subsets of a compact space are compact. Then let $x\in A,y\in B$ be arbitrary. Since $X$ is Hausdorff, we may choose $U_{x,y}$ and $V_{x,y}$ disjoint open so that $x\in U_{x,y}$ and $y\in V_{x,y}$. Since $A$ is compact, choose a finite subcover $U_{x_{1},y},\ldots ,U_{x_{k_{y}},y}$ of $A$. Then set $V_{y}:=V_{x_{1},y}\cap \cdots \cap V_{x_{k_{y}},y}$ and observe (as in the proof of the last proposition) that $V_{y}$ is an open subset of $X$ disjoint from $A$. Then note that since $B$ is compact, we may choose a finite subcover $V_{y_{1}},\ldots ,V_{y_{n}}$ of $B$. Then define
    $V:=V_{y_{1}}\cup \cdots \cup V_{y_{n}}$, $U:=\bigcap _{m=1}^{n}\bigcup _{j=1}^{k_{y_{m}}}U_{x_{j},y_{m}}$
and observe that $A\subseteq U$, $B\subseteq V$ and that $U,V$ are open and disjoint. $\Box$
Definition (finite intersection property):
Let $X$ be a topological space. $X$ is said to possess the finite intersection property if and only if for all families $(F_{\alpha })_{\alpha \in A}$ of closed subsets of $X$ such that
    $\bigcap _{\alpha \in A}F_{\alpha }=\emptyset$,
there exists a finite set of indices $\{\alpha _{1},\ldots ,\alpha _{n}\}$ such that
    $\bigcap _{j=1}^{n}F_{\alpha _{j}}=\emptyset$.
Proposition (compactness is equivalent to the finite intersection property):
Let $X$ be a topological space. $X$ is compact if and only if it satisfies the finite intersection property.
Proof: $X$ being compact is equivalent to the assertion that for all families $(U_{\alpha })_{\alpha \in A}$ that cover $X$, there exists a finite subcover. Such covers are in bijective correspondence to families $(F_{\alpha })_{\alpha \in A}$ with empty intersection via
    $(U_{\alpha })_{\alpha \in A}\mapsto (X\setminus U_{\alpha })_{\alpha \in A}$, with inverse $(F_{\alpha })_{\alpha \in A}\mapsto (X\setminus F_{\alpha })_{\alpha \in A}$
Note further that under this correspondence, families of open sets cover $X$ if and only if the corresponding family of closed sets has empty intersection; this is a consequence of de Morgan. Hence, whenever we have the finite intersection property, we may translate an open cover into a family of closed sets with empty intersection, extract a finite subfamily with empty intersection, and revert back to see that the resulting open cover (which is a subcover of the original family) is finite and covers $X$, and if we have compactness, an analogous argument works. $\Box$
Proposition (continuous bijection from compact to Hausdorff is homeomorphism):
Let $X$ be a compact space and $Y$ be a Hausdorff space. Suppose that $f:X\to Y$ is a bijective function which is continuous. Then $f$ is a homeomorphism.
Proof: Let $y\in Y$ be given; we prove that $f$ is continuous at $y$. Set $x:=f^{-1}(y)$ and suppose that $x\in U$ where $U$ is open. Note that for each $z\in Y\setminus \{y\}$, we may choose open neighbourhoods $V_{y,z}$ and $W_{y,z}$ such that $y\in V_{y,z}$, $z\in W_{y,z}$ and $W_{y,z}\cap V_{y,z}=\emptyset$. Consider the family of sets $(U,f^{-1}(W_{y,z}))_{z\neq y}$; it forms an open cover of $X$, so that we may extract a finite subcover $U,f^{-1}(W_{y,z_{1}}),\ldots ,f^{-1}(W_{y,z_{n}})$ (note that $U$ is needed, since it's the only set of the cover that contains $x$). Then set $V:=V_{y,z_{1}})\cap \ldots \cap V_{y,z_{n}}$ so that $V$ is an open neighbourhood of $y$, and observe that $f^{-1}(V)\subseteq U$, because if $w\in f^{-1}(V)\setminus U$, then $w\in f^{-1}(W_{y,z_{j}})$ for a suitable $j\in [n]$, a contradiction since then $f(w)\in V\cap W_{y,z_{j}}$. $\Box$
Proposition (a finite union of compact sets is compact):
Let $X$ be a topological space and let $K_{1},\ldots ,K_{n}\subseteq X$ be compact subsets of $X$. Then $K_{1}\cup \cdots \cup K_{n}$ is a compact subset of $X$.
Proof: Let $(U_{\alpha })_{\alpha \in A}$ be an open cover of $K_{1}\cup \cdots \cup K_{n}$. By definition of the subspace topology, this means that $U_{\alpha }=(K_{1}\cup \cdots \cup K_{n})\cap V_{\alpha }$, where $V_{\alpha }$ is open in $X$, for all $\alpha$. Note that upon defining $U_{\alpha ,j}:=K_{j}\cap V_{\alpha }$ for $j\in [n]$, we obtain that $(U_{\alpha ,k})_{\alpha \in A}$ forms an open cover of $K_{j}$, so that we may extract a finite subcover $U_{\alpha _{j,1},j},\ldots ,U_{\alpha _{j,m_{j}},j}$. Then observe that
    $U_{\alpha _{1,1}},\ldots ,U_{\alpha _{1,m_{1}}},\ldots ,U_{\alpha _{n,1}},\ldots ,U_{\alpha _{n,m_{n}}}$
is an open cover of $K_{1}\cup \cdots \cup K_{n}$, since each $K_{j}$ is covered. $\Box$
Definition (locally compact):
Let $X$ be a topological space. Then $X$ is said to be locally compact if and only if for each $x\in X$ and each open neighbourhood $U$ of $x$, there exists a compact neighbourhood $K$ of $x$ so that $K\subseteq X$.
Proposition (proper continuous maps to a locally compact Hausdorff space are closed):
Let $X,Y$ be topological spaces, where $Y$ is locally compact. Let $f:X\to Y$ be a continuous and proper function. Then $f$ is in fact closed.
Proof: Suppose that $A\subseteq X$ is closed, and set $B:=f(A)\subseteq Y$. Let $y\in {\overline {B}}$, we are then to show that $y\in B$. Since $Y$ is locally compact, pick a compact neighbourhood $K$ of $y$. Since $f$ is proper and continuous, $f^{-1}(K)$ will be a compact set. Suppose that $K$ does not contain a point which is mapped via $f$ to $y$. Since $Y$ is Hausdorff, whenever $z\in f^{-1}(K)$, we find open neighbourhoods $V_{z}$ of $y$ and $W_{z}$ of $f(z)$ such that $V_{z}\cap W_{z}=\emptyset$. Therefore, the sets $f^{-1}(V_{z})$ and $f^{-1}(W_{z})$ are disjoint. Now the $f^{-1}(W_{z})$ cover $f^{-1}(K)$, where $z$ runs through all points of $f^{-1}(K)$. Therefore, by compactness of $f^{-1}(K)$, we may pick a finite subcover $f^{-1}(W_{z_{1}}),\cdots ,f^{-1}(W_{z_{n}})$, and then $f^{-1}(V_{z_{1}})\cap \cdots \cap f^{-1}(V_{z_{n}})\cap f^{-1}(K)=f^{-1}(V_{z_{1}}\cap \cdots \cap V_{z_{n}}\cap K)=\emptyset$, which contradicts the fact that $y\in {\overline {B}}$. $\Box$
Definition (compactification):
Let $X$ be a topological space. A compactification of $X$ is a pair $(f,Y)$, where $Y$ is a compact topological space and $f:X\to Y$ is continuous, such that $f$ is an embedding and $f(X)$ is dense in $Y$.
Often, $X\subset Y$ and $f$ is the inclusion.
Definition (Alexandroff compactification):
Let $X$ be a topological space. The Alexandroff compactification $X_{\infty }$ of $X$ is defined by adjoining to $X$ a formal symbol $\infty \notin X$, ie. $X_{\infty }=X\cup \{\infty \}$, and by defining the topology on $X_{\infty }$ as the union of the following sets:
* The topology of $X$
* All $X_{\infty }$-complements of closed, compact sets of $X$
Proposition (Alexandroff compactification is well-defined):
Let $X$ be a topological space, and let $X_{\infty }$ be its Alexandroff compactification. Then $X_{\infty }$ is a compact topological space, and if $X$ is not already compact, together with the inclusion $\iota :X\to X_{\infty }$ it gives a compactification of $X$.
Proof: First, we prove that the given topology is indeed a topology. Clearly, $\emptyset$ is compact and closed. Hence, $X_{\infty }$ is in the topology, as is $\emptyset$, since $X$ is a topological space. Then, let $U,V$ be open. If either $U$ or $V$ are open subsets of $X$, then so is $U\cap V=(U\cap X)\cap (V\cap X)$. If both are $X_{\infty }$-complements of closed, compact sets $K,L$ of $X$, then
    $U\cap V=(X_{\infty }\setminus L)\cap (X_{\infty }\setminus K)=X_{\infty }\setminus (K\cup L)$,
and again $U\cap V$ is open because the union of two closed sets is closed, and the union of two compact subsets is compact.
Now suppose we are given a family of open sets $(U_{\alpha })_{\alpha \in A}$ of open sets of $X$, and a family $(V_{\beta })_{\beta \in B}=(X_{\infty }\setminus K_{\beta })_{\beta \in B}$ of complements of compact and closed sets. Then
    $\bigcup _{\alpha \in A}U_{\alpha }\cup \bigcup _{\beta \in B}V_{\beta }=X\setminus \left(\bigcap _{\alpha \in A}(X\setminus U_{\alpha })\cap \bigcap _{\beta \in B}(X_{\infty }\setminus K_{\beta })\right)=X\setminus \left(\bigcap _{\alpha \in A}(X\setminus U_{\alpha })\cap \bigcap _{\beta \in B}(X\setminus K_{\beta })\right)$,
and if $B\neq \emptyset$, we conclude since closed subsets of compact sets are compact; if $B=\emptyset$, we conclude since $X$ is a topological space.
Note now that the inclusion $\iota :X\to X_{\infty }$ is a homeomorphism onto its image by definition of the topology on $X_{\infty }$; it is continuous, open and bijective. Then suppose that $X$ is not compact; we claim that $\iota (X)=X$ is dense in $X_{\infty }$. Indeed, let $O$ be any open set in $X_{\infty }$. Since $X$ is not compact, $O$ must intersect $X$. Hence, $X$ is dense in $X_{\infty }$. $\Box$
Proposition (Alexandroff compactification of locally compact Hausdorff space is Hausdorff):
Let $X$ be a locally compact Hausdorff space. Then the Alexandroff compactification $X_{\infty }$ is Hausdorff.
Proof: Let $x,y\in X_{\infty }$, $x\neq y$; as usual, we denote by $\infty$ the point that was added to $X$ in forming $X_{\infty }$. Suppose first that neither $x$ nor $y$ are $\infty$. Then $x$ and $y$ are separated by disjoint neighbourhoods because $X$ is Hausdorff. Suppose now wlog. that $x=\infty$, then $y\neq \infty$, so $y\in X$. Since $X$ is locally compact, pick a compact neighbourhood $K$ of $y$. Since $K$ is a neighbourhood of $x$, pick an open neighbourhood $U\subseteq K$ of $x$. Set $V:=X\setminus K$. Since $X$ is Hausdorff, $K$ is closed, so that $U,V$ are open neighbourhoods of $y,x$ that satisfy the requirements of the definition of a Hausdorff space. $\Box$

## Exercises

* Let $S$ be a set, $X$ a topological space, and $f:S\to X$ a function. Then $f(S)$ is a compact subset of $X$ if and only if there exists a topology on $S$ which makes $S$ into a compact topological space.
* Let $X$ be a set with two topologies $\tau _{1}$ and $\tau _{2}$, with respect to which $X$ is compact. Prove that also, $X$ is compact with respect to the topologies $\tau _{1}\cap \tau _{2}$ and $\tau _{1}\vee \tau _{2}$, where the latter shall denote the least upper bound topology of $\tau _{1}$ and $\tau _{2}$, borrowing notation from lattice theory.
** Let $X,Y$ be topological spaces and let $K\subseteq X$ and $L\subseteq Y$ be compact sets. Prove that $K\times L$ is a compact subset of $X\times Y$, where the latter is given the product topology.
** Let $X,Y$ be Hausdorff spaces and suppose that $f:X\to Y$ and $g:Y\to Y$ are proper, continuous functions. On the condition of the axiom of choice, prove that $f\times g:X\times Y\to Y\times Y$ is proper. Hint: Prove first that it suffices to show that preimages of products of compact sets of $Y$ are compact.
* Use Alexander's subbasis theorem to prove Tychonoff's theorem.
* Prove that if $X$ is a compact space and $A\subseteq X$ is discrete with respect to the subspace topology, then $A$ is a finite set.
* Let $Z_{1},\ldots ,Z_{n}$ be compact spaces, $X$ a set and for $k\in [n]$, let $f_{k}:Z_{k}\to X$ be a function. Suppose that $X$ carries the final topology by the $f_{k}$ ($k\in [n]$). Prove that $X$ is compact if and only if $\bigcup _{k=1}^{n}f_{k}(Z_{k})$ is cofinite in $X$.
* Let $X,Y$ be topological spaces, where $X$ is compact and $Y$ is Hausdorff, and let $f:X\to Y$ be a continuous bijection. On the condition of the axiom of choice, prove that $X$ is Hausdorff and $Y$ is compact.
* Let $X$ be a noncompact connected topological space. Prove that its Alexandroff compactification $X_{\infty }$ is connected.