# General Topology/Compact spaces

Definition (compact space):

Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is called compact if and only if for every open cover ${\displaystyle (U_{\alpha })_{\alpha \in A}}$ of ${\displaystyle X}$ there exists a finite subcover, that is, indices ${\displaystyle \alpha _{1},\ldots ,\alpha _{n}\in A}$ so that ${\displaystyle X=U_{\alpha _{1}}\cup \cdots \cup U_{\alpha _{n}}}$.

Definition (compact subset):

Let ${\displaystyle X}$ be a topological space and ${\displaystyle S\subseteq X}$ be a subset. ${\displaystyle S}$ is called compact iff it is compact with respect to the subspace topology induced on ${\displaystyle S}$ by the topology of ${\displaystyle X}$.

Proposition (the image of a compact set via a continuous map is compact):

Let ${\displaystyle X,Y}$ be topological spaces, ${\displaystyle S\subseteq X}$ a compact subset and ${\displaystyle f:X\to Y}$ a continuous function. Then ${\displaystyle f(S)}$ is a compact subset of ${\displaystyle Y}$.

Proof: Define ${\displaystyle T:=f(S)}$ to ease notation. Let ${\displaystyle (V_{\beta })_{\beta \in B}}$ be an open cover of ${\displaystyle T}$, that is, by definition of the subspace topology, ${\displaystyle V_{\beta }=W_{\beta }\cap T}$ for suitable ${\displaystyle W_{\beta }}$. Since

${\displaystyle f^{-1}(W_{\beta }\cap T)=f^{-1}(T)\cap f^{-1}(W_{\beta })}$,

we obtain upon noting that ${\displaystyle S\subseteq f^{-1}(T)}$, that the sets

${\displaystyle U_{\beta }:=S\cap f^{-1}(W_{\beta })}$

form an open cover of ${\displaystyle S}$ with respect to its subspace topology; indeed, the continuity of ${\displaystyle f}$ insures that each set ${\displaystyle f^{-1}(W_{\beta })}$ is open. Since ${\displaystyle S}$ is compact, a finite subcover, indexed by ${\displaystyle \beta _{1},\ldots ,\beta _{n}}$, may be chosen. Let ${\displaystyle y\in T}$ be arbitrary. By the definition of ${\displaystyle T}$, pick ${\displaystyle x\in S}$ such that ${\displaystyle f(x)=y}$, and then ${\displaystyle j\in [n]}$ such that ${\displaystyle x\in U_{\beta _{j}}}$. Then ${\displaystyle f(x)\in S\cap W_{\beta _{n}}=V_{\beta _{n}}}$, so that ${\displaystyle V_{\beta _{1}}\cup \cdots \cup V_{\beta _{n}}=Y}$. ${\displaystyle \Box }$

Definition (proper map):

A function ${\displaystyle f:X\to Y}$ between topological spaces is called proper if and only if for each compact subset ${\displaystyle T\subseteq Y}$, the preimage ${\displaystyle f^{-1}(T)}$ is a compact subset of ${\displaystyle X}$.

Note that the composition of proper maps is proper.

Proposition (closed subsets of a compact space are compact):

Let ${\displaystyle X}$ be a compact space, and let ${\displaystyle A\subseteq X}$ be closed. Then ${\displaystyle A}$ is compact.

Proof: Let ${\displaystyle (V_{\alpha })_{\alpha \in A}}$ be an open cover of ${\displaystyle A}$. By definition of the subspace topology of ${\displaystyle A}$, we take ${\displaystyle V_{\alpha }=A\cap U_{\alpha }}$ where ${\displaystyle U_{\alpha }}$ is open in ${\displaystyle X}$; in order to avoid the axiom of choice, we may replace ${\displaystyle U_{\alpha }}$ by the union of all ${\displaystyle U_{\alpha }}$ for which ${\displaystyle V_{\alpha }=A\cap U_{\alpha }}$. Then an open cover of ${\displaystyle X}$ is given by

${\displaystyle \{X\setminus A\}\cup (U_{\alpha })_{\alpha \in A}}$,

and by compactness of ${\displaystyle X}$, we may extract a finite subcover. Suppose that ${\displaystyle U_{\alpha _{1}},\ldots ,U_{\alpha _{n}}}$ are the sets of ${\displaystyle (U_{\alpha })_{\alpha \in A}}$ that made it into the subcover. Then

${\displaystyle A\subseteq U_{\alpha _{1}}\cup \cdots \cup U_{\alpha _{n}}}$,

since ${\displaystyle A}$ is contained in the whole subcover, but the only additional set in this subcover may be ${\displaystyle X\setminus A}$, which doesn't change whether or not ${\displaystyle A}$ is covered. Thus, ${\displaystyle V_{\alpha _{1}},\ldots ,V_{\alpha _{n}}}$ is an open subcover of ${\displaystyle A}$. ${\displaystyle \Box }$

Theorem (Cantor's intersection theorem):

Let ${\displaystyle X}$ be a topological space, let ${\displaystyle (S,\leq )}$ be a directed set and let ${\displaystyle (K_{s})_{s\in S}}$ be a family of nonempty sets ${\displaystyle K_{s}}$ which are simultaneously compact and closed, such that ${\displaystyle s\leq t\Rightarrow K_{s}\supseteq K_{t}}$. Then

${\displaystyle \bigcap _{s\in S}K_{s}\neq \emptyset }$.

Proof: Suppose that

${\displaystyle \bigcap _{s\in S}K_{s}=\emptyset }$.

Note that ${\displaystyle K_{1}}$ is compact, and since each ${\displaystyle K_{j}}$ is closed, its complement ${\displaystyle U_{j}:=X\setminus K_{j}}$ is open. Further, by definition of the subspace topology, the sets ${\displaystyle V_{j}:=U_{j}\cap K_{1}}$ are open in ${\displaystyle K_{1}}$, and by de Morgan (${\displaystyle X=\bigcup _{n\in \mathbb {N} }U_{n}}$) and distributivity of intersection over union, we get that the ${\displaystyle V_{j}}$ form an open cover of ${\displaystyle K_{1}}$. By compactness of ${\displaystyle K_{1}}$, we may extract a finite subcover ${\displaystyle U_{n_{1}},\ldots ,U_{n_{k}}}$, and upon choosing ${\displaystyle N:=\max\{n_{1},\ldots ,n_{k}\}}$, we get that ${\displaystyle V_{N}=K_{1}}$ since ${\displaystyle U_{1}\subseteq U_{2}\subseteq \cdots \subseteq U_{N}}$ and thus, since ${\displaystyle K_{N}\subseteq K_{1}}$, also ${\displaystyle K_{N}=\emptyset }$, a contradiction. ${\displaystyle \Box }$

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

Let ${\displaystyle X}$ be a nonempty, compact T0 space. Then ${\displaystyle X}$ contains a point ${\displaystyle x_{0}}$ such that ${\displaystyle \{x_{0}\}}$ is closed in ${\displaystyle X}$.

(On the condition of the axiom of choice.)

Proof: The set of nonempty, closed subsets of ${\displaystyle X}$, 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 ${\displaystyle A}$. Suppose that ${\displaystyle A}$ contains two distinct points ${\displaystyle x\neq y}$. Then by the hypothesis, select an open set ${\displaystyle U\subseteq X}$ that contains one point, but not the other, eg. ${\displaystyle x\notin U}$. Then ${\displaystyle x\in }$ ${\displaystyle \Box }$

Proposition (compact subsets of Hausdorff spaces are closed):

Let ${\displaystyle X}$ be a Hausdorff space, and let ${\displaystyle K\subseteq X}$ be compact. Then ${\displaystyle K}$ is closed.

Proof: Let ${\displaystyle y\in X\setminus K}$ be given. For each ${\displaystyle x\in K}$, there exist open sets ${\displaystyle U_{x,y}}$ and ${\displaystyle V_{x,y}}$ such that ${\displaystyle U_{x,y}\cap V_{x,y}=\emptyset }$, ${\displaystyle x\in U_{x,y}}$ and ${\displaystyle y\in V_{x,y}}$. Since ${\displaystyle K}$ is compact, select among the ${\displaystyle U_{x,y}}$ a finite subcover ${\displaystyle U_{x_{1},y},\ldots ,U_{x_{1},y}}$; note that this step does not use the axiom of choice, since the ${\displaystyle U_{x,y}}$ cover ${\displaystyle X}$ in their totality; that is, we include in the cover not only one specific ${\displaystyle U_{x,y}}$ for each ${\displaystyle x}$, but all sets of this form. Then set ${\displaystyle V_{y}:=V_{x_{1},y}\cap \cdots \cap V_{x_{n},y}}$ and obtain that ${\displaystyle V_{y}}$ is disjoint from ${\displaystyle K}$; indeed, it can't contain ${\displaystyle x\in U_{x_{j},y}}$ for any ${\displaystyle j\in [n]}$. Therefore,

${\displaystyle X\setminus K\subseteq \bigcup _{y\in X\setminus K}V_{y}\subseteq X\setminus K}$, ie. ${\displaystyle X\setminus K=\bigcup _{y\in X\setminus K}V_{y}}$ open. ${\displaystyle \Box }$

Conversely, we have:

Proposition (compact sets being closed implies T1):

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

Proof: Any finite subset of ${\displaystyle X}$ is compact, so that we may apply the characterisation of T1 spaces. ${\displaystyle \Box }$

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

Let ${\displaystyle X}$ be an R1 space. Then ${\displaystyle X}$ 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 ${\displaystyle X}$ is T1. ${\displaystyle \Box }$

Proposition (intersection of compact sets in Hausdorff spaces is compact):

Let ${\displaystyle (K_{\alpha })_{\alpha \in A}}$ be compact subsets of a Hausdorff space ${\displaystyle X}$. Then

${\displaystyle \bigcap _{\alpha \in A}K_{\alpha }}$

is compact.

Proof: Since ${\displaystyle X}$ is Hausdorff, all the ${\displaystyle K_{\alpha }}$s are closed. Hence, the given set is a closed subset of the compact set ${\displaystyle K_{\alpha _{0}}}$, where ${\displaystyle \alpha _{0}\in A}$ is arbitrary. ${\displaystyle \Box }$

Proposition (compact Hausdorff spaces are normal):

Let ${\displaystyle X}$ be a compact Hausdorff space. Then ${\displaystyle X}$ is normal.

Proof: Let ${\displaystyle A,B\subseteq X}$ be two closed subsets of ${\displaystyle X}$ which are disjoint. First, we note that ${\displaystyle A,B}$ are compact, since closed subsets of a compact space are compact. Then let ${\displaystyle x\in A,y\in B}$ be arbitrary. Since ${\displaystyle X}$ is Hausdorff, we may choose ${\displaystyle U_{x,y}}$ and ${\displaystyle V_{x,y}}$ disjoint open so that ${\displaystyle x\in U_{x,y}}$ and ${\displaystyle y\in V_{x,y}}$. Since ${\displaystyle A}$ is compact, choose a finite subcover ${\displaystyle U_{x_{1},y},\ldots ,U_{x_{k_{y}},y}}$ of ${\displaystyle A}$. Then set ${\displaystyle V_{y}:=V_{x_{1},y}\cap \cdots \cap V_{x_{k_{y}},y}}$ and observe (as in the proof of the last proposition) that ${\displaystyle V_{y}}$ is an open subset of ${\displaystyle X}$ disjoint from ${\displaystyle A}$. Then note that since ${\displaystyle B}$ is compact, we may choose a finite subcover ${\displaystyle V_{y_{1}},\ldots ,V_{y_{n}}}$ of ${\displaystyle B}$. Then define

${\displaystyle V:=V_{y_{1}}\cup \cdots \cup V_{y_{n}}}$, ${\displaystyle U:=\bigcap _{m=1}^{n}\bigcup _{j=1}^{k_{y_{m}}}U_{x_{j},y_{m}}}$

and observe that ${\displaystyle A\subseteq U}$, ${\displaystyle B\subseteq V}$ and that ${\displaystyle U,V}$ are open and disjoint. ${\displaystyle \Box }$

Definition (finite intersection property):

Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is said to possess the finite intersection property if and only if for all families ${\displaystyle (F_{\alpha })_{\alpha \in A}}$ of closed subsets of ${\displaystyle X}$ such that

${\displaystyle \bigcap _{\alpha \in A}F_{\alpha }=\emptyset }$,

there exists a finite set of indices ${\displaystyle \{\alpha _{1},\ldots ,\alpha _{n}\}}$ such that

${\displaystyle \bigcap _{j=1}^{n}F_{\alpha _{j}}=\emptyset }$.

Proposition (compactness is equivalent to the finite intersection property):

Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is compact if and only if it satisfies the finite intersection property.

Proof: ${\displaystyle X}$ being compact is equivalent to the assertion that for all families ${\displaystyle (U_{\alpha })_{\alpha \in A}}$ that cover ${\displaystyle X}$, there exists a finite subcover. Such covers are in bijective correspondence to families ${\displaystyle (F_{\alpha })_{\alpha \in A}}$ with empty intersection via

${\displaystyle (U_{\alpha })_{\alpha \in A}\mapsto (X\setminus U_{\alpha })_{\alpha \in A}}$, with inverse ${\displaystyle (F_{\alpha })_{\alpha \in A}\mapsto (X\setminus F_{\alpha })_{\alpha \in A}}$

Note further that under this correspondence, families of open sets cover ${\displaystyle X}$ 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 ${\displaystyle X}$, and if we have compactness, an analogous argument works. ${\displaystyle \Box }$

Proposition (continuous bijection from compact to Hausdorff is homeomorphism):

Let ${\displaystyle X}$ be a compact space and ${\displaystyle Y}$ be a Hausdorff space. Suppose that ${\displaystyle f:X\to Y}$ is a bijective function which is continuous. Then ${\displaystyle f}$ is a homeomorphism.

(On the condition of the axiom of choice.)

Proof: Let ${\displaystyle y\in Y}$ be given; we prove that ${\displaystyle f}$ is continuous at ${\displaystyle y}$. Set ${\displaystyle x:=f^{-1}(y)}$ and suppose that ${\displaystyle x\in U}$ where ${\displaystyle U}$ is open. Note that for each ${\displaystyle z\in Y\setminus \{y\}}$, we may choose open neighbourhoods ${\displaystyle V_{y,z}}$ and ${\displaystyle W_{y,z}}$ such that ${\displaystyle y\in V_{y,z}}$, ${\displaystyle z\in W_{y,z}}$ and ${\displaystyle W_{y,z}\cap V_{y,z}=\emptyset }$. Consider the family of sets ${\displaystyle (U,f^{-1}(W_{y,z}))_{z\neq y}}$; it forms an open cover of ${\displaystyle X}$, so that we may extract a finite subcover ${\displaystyle U,f^{-1}(W_{y,z_{1}}),\ldots ,f^{-1}(W_{y,z_{n}})}$ (note that ${\displaystyle U}$ is needed, since it's the only set of the cover that contains ${\displaystyle x}$). Then set ${\displaystyle V:=V_{y,z_{1}})\cap \ldots \cap V_{y,z_{n}}}$ so that ${\displaystyle V}$ is an open neighbourhood of ${\displaystyle y}$, and observe that ${\displaystyle f^{-1}(V)\subseteq U}$, because if ${\displaystyle w\in f^{-1}(V)\setminus U}$, then ${\displaystyle w\in f^{-1}(W_{y,z_{j}})}$ for a suitable ${\displaystyle j\in [n]}$, a contradiction since then ${\displaystyle f(w)\in V\cap W_{y,z_{j}}}$. ${\displaystyle \Box }$

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

Let ${\displaystyle X}$ be a topological space and let ${\displaystyle K_{1},\ldots ,K_{n}\subseteq X}$ be compact subsets of ${\displaystyle X}$. Then ${\displaystyle K_{1}\cup \cdots \cup K_{n}}$ is a compact subset of ${\displaystyle X}$.

Proof: Let ${\displaystyle (U_{\alpha })_{\alpha \in A}}$ be an open cover of ${\displaystyle K_{1}\cup \cdots \cup K_{n}}$. By definition of the subspace topology, this means that ${\displaystyle U_{\alpha }=(K_{1}\cup \cdots \cup K_{n})\cap V_{\alpha }}$, where ${\displaystyle V_{\alpha }}$ is open in ${\displaystyle X}$, for all ${\displaystyle \alpha }$. Note that upon defining ${\displaystyle U_{\alpha ,j}:=K_{j}\cap V_{\alpha }}$ for ${\displaystyle j\in [n]}$, we obtain that ${\displaystyle (U_{\alpha ,k})_{\alpha \in A}}$ forms an open cover of ${\displaystyle K_{j}}$, so that we may extract a finite subcover ${\displaystyle U_{\alpha _{j,1},j},\ldots ,U_{\alpha _{j,m_{j}},j}}$. Then observe that

${\displaystyle U_{\alpha _{1,1}},\ldots ,U_{\alpha _{1,m_{1}}},\ldots ,U_{\alpha _{n,1}},\ldots ,U_{\alpha _{n,m_{n}}}}$

is an open cover of ${\displaystyle K_{1}\cup \cdots \cup K_{n}}$, since each ${\displaystyle K_{j}}$ is covered. ${\displaystyle \Box }$

Definition (locally compact):

Let ${\displaystyle X}$ be a topological space. Then ${\displaystyle X}$ is said to be locally compact if and only if for each ${\displaystyle x\in X}$ and each open neighbourhood ${\displaystyle U}$ of ${\displaystyle x}$, there exists a compact neighbourhood ${\displaystyle K}$ of ${\displaystyle x}$ so that ${\displaystyle K\subseteq X}$.

Proposition (proper continuous maps to a locally compact Hausdorff space are closed):

Let ${\displaystyle X,Y}$ be topological spaces, where ${\displaystyle Y}$ is locally compact. Let ${\displaystyle f:X\to Y}$ be a continuous and proper function. Then ${\displaystyle f}$ is in fact closed.

Proof: Suppose that ${\displaystyle A\subseteq X}$ is closed, and set ${\displaystyle B:=f(A)\subseteq Y}$. Let ${\displaystyle y\in {\overline {B}}}$, we are then to show that ${\displaystyle y\in B}$. Since ${\displaystyle Y}$ is locally compact, pick a compact neighbourhood ${\displaystyle K}$ of ${\displaystyle y}$. Since ${\displaystyle f}$ is proper and continuous, ${\displaystyle f^{-1}(K)}$ will be a compact set. Suppose that ${\displaystyle K}$ does not contain a point which is mapped via ${\displaystyle f}$ to ${\displaystyle y}$. Since ${\displaystyle Y}$ is Hausdorff, whenever ${\displaystyle z\in f^{-1}(K)}$, we find open neighbourhoods ${\displaystyle V_{z}}$ of ${\displaystyle y}$ and ${\displaystyle W_{z}}$ of ${\displaystyle f(z)}$ such that ${\displaystyle V_{z}\cap W_{z}=\emptyset }$. Therefore, the sets ${\displaystyle f^{-1}(V_{z})}$ and ${\displaystyle f^{-1}(W_{z})}$ are disjoint. Now the ${\displaystyle f^{-1}(W_{z})}$ cover ${\displaystyle f^{-1}(K)}$, where ${\displaystyle z}$ runs through all points of ${\displaystyle f^{-1}(K)}$. Therefore, by compactness of ${\displaystyle f^{-1}(K)}$, we may pick a finite subcover ${\displaystyle f^{-1}(W_{z_{1}}),\cdots ,f^{-1}(W_{z_{n}})}$, and then ${\displaystyle f^{-1}(V_{z_{1}})\cap \cdots \cap f^{-1}(V_{z_{n}})\cap f^{-1}(K)=f^{-1}(V_{z_{1}}\cap \cdots \cap V_{z_{n}}\cap K)=\emptyset }$, which contradicts the fact that ${\displaystyle y\in {\overline {B}}}$. ${\displaystyle \Box }$

Definition (compactification):

Let ${\displaystyle X}$ be a topological space. A compactification of ${\displaystyle X}$ is a pair ${\displaystyle (f,Y)}$, where ${\displaystyle Y}$ is a compact topological space and ${\displaystyle f:X\to Y}$ is continuous, such that ${\displaystyle f}$ is an embedding and ${\displaystyle f(X)}$ is dense in ${\displaystyle Y}$.

Often, ${\displaystyle X\subset Y}$ and ${\displaystyle f}$ is the inclusion.

Definition (Alexandroff compactification):

Let ${\displaystyle X}$ be a topological space. The Alexandroff compactification ${\displaystyle X_{\infty }}$ of ${\displaystyle X}$ is defined by adjoining to ${\displaystyle X}$ a formal symbol ${\displaystyle \infty \notin X}$, ie. ${\displaystyle X_{\infty }=X\cup \{\infty \}}$, and by defining the topology on ${\displaystyle X_{\infty }}$ as the union of the following sets:

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

Proposition (Alexandroff compactification is well-defined):

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

Proof: First, we prove that the given topology is indeed a topology. Clearly, ${\displaystyle \emptyset }$ is compact and closed. Hence, ${\displaystyle X_{\infty }}$ is in the topology, as is ${\displaystyle \emptyset }$, since ${\displaystyle X}$ is a topological space. Then, let ${\displaystyle U,V}$ be open. If either ${\displaystyle U}$ or ${\displaystyle V}$ are open subsets of ${\displaystyle X}$, then so is ${\displaystyle U\cap V=(U\cap X)\cap (V\cap X)}$. If both are ${\displaystyle X_{\infty }}$-complements of closed, compact sets ${\displaystyle K,L}$ of ${\displaystyle X}$, then

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

and again ${\displaystyle U\cap V}$ 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 ${\displaystyle (U_{\alpha })_{\alpha \in A}}$ of open sets of ${\displaystyle X}$, and a family ${\displaystyle (V_{\beta })_{\beta \in B}=(X_{\infty }\setminus K_{\beta })_{\beta \in B}}$ of complements of compact and closed sets. Then

${\displaystyle \bigcup _{\alpha \in A}U_{\alpha }\cup \bigcup _{\beta \in B}V_{\beta }=X\setminus \left(\bigcap _{\alpha \in A}(X\setminus U_{\alpha })\cap \bigcap _{\beta \in B}(X_{\infty }\setminus K_{\beta })\right)=X\setminus \left(\bigcap _{\alpha \in A}(X\setminus U_{\alpha })\cap \bigcap _{\beta \in B}(X\setminus K_{\beta })\right)}$,

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

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

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

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

Proof: Let ${\displaystyle x,y\in X_{\infty }}$, ${\displaystyle x\neq y}$; as usual, we denote by ${\displaystyle \infty }$ the point that was added to ${\displaystyle X}$ in forming ${\displaystyle X_{\infty }}$. Suppose first that neither ${\displaystyle x}$ nor ${\displaystyle y}$ are ${\displaystyle \infty }$. Then ${\displaystyle x}$ and ${\displaystyle y}$ are separated by disjoint neighbourhoods because ${\displaystyle X}$ is Hausdorff. Suppose now wlog. that ${\displaystyle x=\infty }$, then ${\displaystyle y\neq \infty }$, so ${\displaystyle y\in X}$. Since ${\displaystyle X}$ is locally compact, pick a compact neighbourhood ${\displaystyle K}$ of ${\displaystyle y}$. Since ${\displaystyle K}$ is a neighbourhood of ${\displaystyle x}$, pick an open neighbourhood ${\displaystyle U\subseteq K}$ of ${\displaystyle x}$. Set ${\displaystyle V:=X\setminus K}$. Since ${\displaystyle X}$ is Hausdorff, ${\displaystyle K}$ is closed, so that ${\displaystyle U,V}$ are open neighbourhoods of ${\displaystyle y,x}$ that satisfy the requirements of the definition of a Hausdorff space. ${\displaystyle \Box }$

## Exercises

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