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 ${\alpha }_1,\dots ,{\alpha }_n\in A$ 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 ${\beta }_1,\dots ,{\beta }_n$, 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 $V_{{\beta }_1}\cup \cdots \cup V_{{\beta }_n}=Y$. $◻$

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 $U_{{\alpha }_1},\dots ,U_{{\alpha }_n}$ 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, $V_{{\alpha }_1},\dots ,V_{{\alpha }_n}$ is an open subcover of $A$. $◻$

Theorem (Cantor's intersection theorem):

Let $X$ be a topological space, let $(S,\le )$ 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 $s\le t\Rightarrow K_s\supseteq K_t$. Then

$\bigcap _{s\in S}{K}_s\ne \mathrm{\varnothing }$.

Proof: Suppose that

$\bigcap _{s\in S}{K}_s=\mathrm{\varnothing }$.

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 $U_{n_1},\dots ,U_{n_k}$, and upon choosing $N:=max\{n_1,\dots ,n_k\}$, 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 $K_N=\mathrm{\varnothing }$, a contradiction. $◻$

Proposition (compact nonempty Kolmogorov spaces contain a closed point):

Let $X$ be a nonempty, compact T₀ space. Then $X$ contains a point $x_0$ such that $\{x_0\}$ is closed in $X$.

(On the condition of the axiom of choice.)

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 $x\ne y$. Then by the hypothesis, select an open set $U\subseteq X$ that contains one point, but not the other, eg. $x\notin U$. Then $x\in$ $◻$

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 $U_{x,y}\cap V_{x,y}=\mathrm{\varnothing }$, $x\in U_{x,y}$ and $y\in V_{x,y}$. Since $K$ is compact, select among the $U_{x,y}$ a finite subcover $U_{x_1,y},\dots ,U_{x_1,y}$; 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. $◻$

Conversely, we have:

Proposition (compact sets being closed implies T₁):

Let $X$ be a topological space where all compact sets are closed. Then $X$ is T₁.

Proof: Any finite subset of $X$ is compact, so that we may apply the characterisation of T₁ spaces. $◻$

Proposition (R₁ space is Hausdorff iff all compact sets are closed):

Let $X$ be an R₁ 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 T₁. $◻$

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. $◻$

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 $U_{x_1,y},\dots ,U_{x_{k_y},y}$ 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 $V_{y_1},\dots ,V_{y_n}$ 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. $◻$

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

$\bigcap _{\alpha \in A}{F}_{\alpha }=\mathrm{\varnothing }$,

there exists a finite set of indices $\{{\alpha }_1,\dots ,{\alpha }_n\}$ such that

$\bigcap _{j=1}^n{F}_{{\alpha }_j}=\mathrm{\varnothing }$.

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. $◻$

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.

(On the condition of the axiom of choice.)

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 $W_{y,z}\cap V_{y,z}=\mathrm{\varnothing }$. Consider the family of sets $(U,f^{-1}(W_{y,z}){)}_{z\ne y}$; it forms an open cover of $X$, so that we may extract a finite subcover $U,f^{-1}(W_{y,z_1}),\dots ,f^{-1}(W_{y,z_n})$ (note that $U$ is needed, since it's the only set of the cover that contains $x$). Then set $V:=V_{y,z_1})\cap \dots \cap V_{y,z_n}$ 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 $f(w)\in V\cap W_{y,z_j}$. $◻$

Proposition (a finite union of compact sets is compact):

Let $X$ be a topological space and let $K_1,\dots ,K_n\subseteq X$ 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 $U_{{\alpha }_{j,1},j},\dots ,U_{{\alpha }_{j,m_j},j}$. Then observe that

$U_{{\alpha }_{1,1}},\dots ,U_{{\alpha }_{1,m_1}},\dots ,U_{{\alpha }_{n,1}},\dots ,U_{{\alpha }_{n,m_n}}$

is an open cover of $K_1\cup \cdots \cup K_n$, since each $K_j$ is covered. $◻$

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 $y\in \bar{B}$, 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 $V_z\cap W_z=\mathrm{\varnothing }$. 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 $f^{-1}(V_{z_1})\cap \cdots \cap f^{-1}(V_{z_n})\cap f^{-1}(K)=f^{-1}(V_{z_1}\cap \cdots \cap V_{z_n}\cap K)=\mathrm{\varnothing }$, which contradicts the fact that $y\in \bar{B}$. $◻$

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 $X_{\mathrm{\infty }}$ of $X$ is defined by adjoining to $X$ a formal symbol $\mathrm{\infty }\notin X$, ie. $X_{\mathrm{\infty }}=X\cup \{\mathrm{\infty }\}$, and by defining the topology on $X_{\mathrm{\infty }}$ as the union of the following sets:

1. The topology of $X$
2. All $X_{\mathrm{\infty }}$-complements of closed, compact sets of $X$

Proposition (Alexandroff compactification is well-defined):

Let $X$ be a topological space, and let $X_{\mathrm{\infty }}$ be its Alexandroff compactification. Then $X_{\mathrm{\infty }}$ is a compact topological space, and if $X$ is not already compact, together with the inclusion $\iota :X\to X_{\mathrm{\infty }}$ it gives a compactification of $X$.

Proof: First, we prove that the given topology is indeed a topology. Clearly, $\mathrm{\varnothing }$ is compact and closed. Hence, $X_{\mathrm{\infty }}$ is in the topology, as is $\mathrm{\varnothing }$, 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 $X_{\mathrm{\infty }}$-complements of closed, compact sets $K,L$ of $X$, then

$U\cap V=(X_{\mathrm{\infty }}\setminus L)\cap (X_{\mathrm{\infty }}\setminus K)=X_{\mathrm{\infty }}\setminus (K\cup L)$,

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 $(V_{\beta }{)}_{\beta \in B}=(X_{\mathrm{\infty }}\setminus K_{\beta }{)}_{\beta \in B}$ of complements of compact and closed sets. Then

$\bigcup _{\alpha \in A}{U}_{\alpha }\cup \bigcup _{\beta \in B}{V}_{\beta }=X\setminus (\bigcap _{\alpha \in A}(X\setminus U_{\alpha })\cap \bigcap _{\beta \in B}(X_{\mathrm{\infty }}\setminus K_{\beta }))=X\setminus (\bigcap _{\alpha \in A}(X\setminus U_{\alpha })\cap \bigcap _{\beta \in B}(X\setminus K_{\beta }))$,

and if $B\ne \mathrm{\varnothing }$, we conclude since closed subsets of compact sets are compact; if $B=\mathrm{\varnothing }$, we conclude since $X$ is a topological space.

Note now that the inclusion $\iota :X\to X_{\mathrm{\infty }}$ is a homeomorphism onto its image by definition of the topology on $X_{\mathrm{\infty }}$; it is continuous, open and bijective. Then suppose that $X$ is not compact; we claim that $\iota (X)=X$ is dense in $X_{\mathrm{\infty }}$. Indeed, let $O$ be any open set in $X_{\mathrm{\infty }}$. Since $X$ is not compact, $O$ must intersect $X$. Hence, $X$ is dense in $X_{\mathrm{\infty }}$. $◻$

Proposition (Alexandroff compactification of locally compact Hausdorff space is Hausdorff):

Let $X$ be a locally compact Hausdorff space. Then the Alexandroff compactification $X_{\mathrm{\infty }}$ is Hausdorff.

Proof: Let $x,y\in X_{\mathrm{\infty }}$, $x\ne y$; as usual, we denote by $\mathrm{\infty }$ the point that was added to $X$ in forming $X_{\mathrm{\infty }}$. Suppose first that neither $x$ nor $y$ are $\mathrm{\infty }$. Then $x$ and $y$ are separated by disjoint neighbourhoods because $X$ is Hausdorff. Suppose now wlog. that $x=\mathrm{\infty }$, then $y\ne \mathrm{\infty }$, 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. $◻$

Exercises

1. 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.
2. 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.
3. 1. 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.
   2. 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.
4. Use Alexander's subbasis theorem to prove Tychonoff's theorem.
5. 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.
6. Let $Z_1,\dots ,Z_n$ 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 $X$.
7. 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.
8. Let $X$ be a noncompact connected topological space. Prove that its Alexandroff compactification $X_{\mathrm{\infty }}$ is connected.