3.10 Compactness

Proof. (1)$\implies$(2): For each $J\suf I$, let

then $U_{J}\suf X$ is open. Suppose for contradiction that $\caps{}{i\in I}E_{i}=\emp$ then

isan open cover for $X$. By assumption, $U_{J}\sub X$ for all $J\suf I$ finite hence $\textbf{U}$ admits no finite subcover, a contradiciton.

(2)$\implies$(3): Let $\scr{F}\suf \cali{P}(X)$ be a filter, then $\curl{\cl{E}:E\in\scr{F}}$ satisfies the hypothesis of (2).

(3)$\iff$(4): The cluster points and limit points of an ultrafilter coincides. (3)$\implies$(1): For each $J\suf I$, let

then for each $J,J'\suf I,E_{J}\cap E_{J'}=E_{J\cup J'}$. Assume for contradiction that $\curl{U_i}_{i\in I}$ admits no finite subcover. Let

then $\scr{B}$ is a filter base consisting of closed sets. By assumption, there exists $x\in \caps{}{i\in I}U_{i}^{c}$, so $\curl{U_i}_{i\in I}$ is not an open cover, contradiction.$\square$

Proof.

1. (1)

   Let $\curl{U_i}_{i\in I}$ be an open cover of $E\cup F$ then there exists $J,J'\suf I$ finite such that $\cups{}{j\in J}U_{j}\supf E$ and $\cups{}{j\in J'}U_{j}\supf F$ hence $\cups{}{j\in J\cup J'}U_{j}\supf E\cup F$.

2. (2)

   Let $\curl{U_i}_{i\in I}$ be an open cover of $G$. Since $G$ is closed, $\curl{U_i}_{i\in I}\cup \curl{G^c}$ is an open cover of $E$. In which case, there exists $J\suf I$ finite such that $\cups{}{j\in J}U_{j}\cup G^{c}\supf E$, so $\cups{}{j\in J}U_{j}\supf G$.

3. (3)

   Let $\curl{U_i}_{i\in I}$ be an open cover of $f(E)$, then $\curl{f\inv (U_i)}_{i\in I}$ is an open cover of $E$. Thus there exists $J\suf I$ finite such that $\cups{}{j\in J}f\inv(U_{j})\supf E$, so $\cups{}{j\in J}U_{j}\supf f(E)$.

$\square$

Proof. Let $x\in E^{c}$. For each $y\in E$, there exists $U_{y}\in\cali{N}(y)$ and $V_{y}\in\cali{N}(x)$ such that $U_{y}\cap V_{y}=\emp$. By compactness, there exists $E_{0}\suf E$ finite such that $\cups{}{y\in E_0}U_{y}\supf E$. In which case $E\cap \caps{}{y\in E_0}V_{y}=\emp$ and $\caps{}{y\in E_0}V_{y}\in \cali{N}(x)$. Thus $E^{c}$ is open.$\square$

Proof. For each $K\suf X$ closed, $K$ and $f(K)$ are compact by Proposition 3.10.3. By Proposition 3.10.4 $f(K)$ is closed. By injectivity $f\inv (f(K))=K$ thus $f\inv$ maps closed sets to closed sets, and hence open sets to open sets.$\square$

Proof. For each $x\in A,y\in B$, there exists $U_{x,y}\in \cali{N}(x),V_{x,y}\in\cali{N}(y)$ such that $U_{x,y}\cap V_{x,y}=\emp$. By compactness of $B$ there exists $B_{x}\suf B$ finite such that $\cups{}{y\in B_x}V_{x,y}\in\cali{N}(B)$. Let $U_{x}\define \caps{}{y\in B_x}U_{x,y}\in\cali{N}(x)$ and $V_{x}\define \cups{}{y\in B_x}V_{x,y}\in\cali{N}(B)$, then $U_{x}\cap V_{x}=\emp$. By compactness of $A$, there exists $A_{0}\suf A$ finite such that $\cups{}{x\in A_0}U_{x}\in\cali{N}(A)$. Then we can take $U=\cups{}{x\in A_0}U_{x}\in\cali{N}(A)$ and $V=\caps{}{x\in A_0}V_{x}\in\cali{N}(B)$ since

$\square$

Proof.

1. (1)

   For each $y\in Y$, let $U_{y}\in\cali{N}_{X}(x)$ and $V_{y}\in \cali{N}_{Y}(y)$ such that $(x,y)\in U_{y}\ti V_{y}\suf U$, then $\curl{V_y:y\in Y}$ is an open cover of $Y$. By compactness, there exists $\curl{y_j}_{1}^{n}$ such that $\cups{n}{j=1}V_{y_j}=Y$. Let $V=\caps{n}{j=1}U_{y_j}$, then $V\in\cali{N}_{X}(x)$ and $V\ti Y\suf U$.

2. (2)

   For each $x\in A$, there exists $V_{x}\in\cali{N}_{X}(x)$ such that $V_{x}\ti Y\suf U$. Let $V=\cups{}{x\in A}V_{x}$, then $V\in\cali{N}_{X}(A)$ and $V\ti Y\suf U$.

$\square$

Proof. By Proposition 3.6.2 it suffices to prove the assertion for compact spaces. Suppose that $X$ is compact then for each $i\in I$, $X_{i}=\pi_{i}(X)$ is compact by Proposition 3.10.3(3). Conversely suppose $X_{i}$ is compact for each $i\in I$. Let $\scr{U}\suf \cali{P}(X)$ be an ultrafilter. For each $i\in I$, $\pi_{i}(\scr{U})$ is an ultrafilter base. Since $X_{i}$ is compact, there exists $x_{i}\in X_{i}$ such that $\pi_{i}(\scr{U})\to x_{i}$. By Axiom of Choice, there exists $x\in X$ such that $\pi_{i}(\scr{U})\to \pi_{i}(x)$. Therefore $\scr{U}\to x$.$\square$