Proof.Exercise in Topology: A Categorical Approach

Lemma. $U$$U$ is open $⟺$$\iff$ $U$$U$ contain a neighbourhood ${\mathcal{N}}_x$$\mathcal N_x$ for each $x\in U$$x\in U$.

Proof

If $U$$U$ is open, then $U$$U$ is the neighbourhood for all $x\in U$$x\in U$.

Conversely, If $U$$U$ contain a neighbourhood ${\mathcal{N}}_x$$\mathcal N_x$ for each $x$$x$, $\mathrm{\exists }$$\exists$ an open set $U_x$$U_x$, $x\in U_x\subseteq {\mathcal{N}}_x$$x\in U_x\subseteq\mathcal N_x$.

Then $U=\bigcup _{x\in U}\{x\}\subseteq \bigcup _{x\in U}{U}_x\subseteq U$$U=\bigcup_{x\in U}\{x\}\subseteq\bigcup_{x\in U} U_x\subseteq U$. Therefore $U=\bigcup _{x\in U}{U}_x$$U=\bigcup_{x\in U} U_x$, $U$$U$ is open. $Q.E.D.$$Q.E.D.$

Theorem 2.12

In any locally path-connected topological space, the connected components and path components are the same.

Proof

Let $U$$U$ be a connected component in $X$$X$. For any $x\in U$$x\in U$, let $U(x):=\{y\in U|\mathrm{\exists }path(t),path(0)=x,path(1)=y\}$$U(x):=\{y\in U|\exist path(t),path(0)=x,path(1)=y \}$.

Firstly, $U(x)$$U(x)$ is an open set. Since for all $y\in U(x)$$y\in U(x)$, $\mathrm{\exists }$$\exists$ a path connected neighborhood $V_y$$V_y$. By definition, $V_y\subseteq U(x)$$V_y\subseteq U(x)$.

Let $B=U-U(x)$$B=U-U(x)$, that is, there is not a single path connected $z\in B$$z\in B$ and $x$$x$.

Then $B$$B$ is open as well. Since for all $z\in B$$z\in B$, there exists a path connected neighbourhood $V_z$$V_z$ and again, by definition, $V_z\subseteq B$$V_z\subseteq B$.

Otherwise, if there exists $v\in V_z\notin B,$$v\in V_z\notin B,$ then we could have a path from $x$$x$ to $v$$v$, and another path $v$$v$ to $z$$z$.

Since $U$$U$ is connected and $U=U(x)\cup B$$U=U(x)\cup B$. Thus $B=\mathrm{\varnothing }$$B=\emptyset$. $Q.E.D.$$Q.E.D.$

Theorem 2.14

$X$$X$ is Hausdorff $⟺$$\iff$ the diagonal map $\mathrm{\Delta }:X\to X\times X,x⟼(x,x)$$\Delta:X\to X\times X,x\longmapsto(x,x)$ is closed in $X\times X$$X\times X$.

Proof

$⟹$$\Longrightarrow$

If $X$$X$ is Hausdorff, then for $x\ne y,\mathrm{\exists }{U}_x,U_y,U_x\cap U_y=\mathrm{\varnothing }$$x\ne y,\exists U_x,U_y,U_x\cap U_y=\empty$. Then $U_x\times U_y\cap \mathrm{Im}(\mathrm{\Delta })=\mathrm{\varnothing }$$U_x\times U_y\cap\mathrm{Im}(\Delta)=\empty$

Since $U_x\times U_y$$U_x\times U_y$ generate the $X\times X-\mathrm{Im}(\mathrm{\Delta })$$X\times X-\mathrm{Im}(\Delta)$, thus $X\times X-\mathrm{Im}(\mathrm{\Delta })$$X\times X-\mathrm{Im}(\Delta)$ is open, thus $\mathrm{I}\mathrm{m}(\mathrm{\Delta })$$\rm Im(\Delta)$ is closed.

$⟸$$\Longleftarrow$

If $\mathrm{I}\mathrm{m}(\mathrm{\Delta })$$\rm Im(\Delta)$ is closed, then $X\times X-\mathrm{Im}(\mathrm{\Delta })$$X\times X-\mathrm{Im}(\Delta)$ is open. If$x\ne y$$x\ne y$, Let $U_x\times U_y\subseteq X\times X-\mathrm{Im}(\mathrm{\Delta })$$U_x\times U_y\subseteq X\times X-\mathrm{Im}(\Delta)$.

Then $(x,x)\notin U_x\times U_y$$(x,x)\notin U_x\times U_y$, $U_x\cap U_y=\mathrm{\varnothing }$$U_x\cap U_y=\emptyset$. $Q.E.D.$$Q.E.D.$

Theorem 2.15 If X is compact and $f:X\to Y$$f : X → Y$ is continuous, then $fX$$fX$ is compact.

Proof

Suppose $\mu$$\Large\mu$ is an open covering of $fX$$fX$, then $f^{-1}\mu$$f^{-1}\Large\mu$ is an open covering of $X$$X$.

Since $X$$X$ is compact, then $f^{-1}\mu$$f^{-1}\Large\mu$ has a finite sub-open covering $f^{-1}\mu$$f^{-1}\mu$, then $\mu$$\mu$ is a finite sub-open covering of $\mu$$\Large\mu$. $Q.E.D.$$Q.E.D.$

Theorem 2.16 A space $X$$X$ is compact if and only if every collection of closed subsets of $X$$X$ with the FIP has nonempty intersection.

Proof

$X$$X$ is compact $⟺$$\iff$ every open covering has a finite sub-covering.

i.e.

$$\begin{array}{}\text{(1)}& \bigcup \mu =X⟹\mathrm{\exists }\bigcup _{i=1}^n{\mu }_i=X⟺\bigcup \mu \ne X⟸\mathrm{\forall }\bigcup _{i=1}^n{\mu }_i\ne X⟺\bigcap {\mu }^c\ne \mathrm{\varnothing }⟸\mathrm{\forall }\bigcap _{i=1}^n{\mu }_i^c\ne \mathrm{\varnothing }\end{array}$$

$Q.E.D.$$Q.E.D.$

Corollary 2.18.4

Continuous functions from compact spaces to $\mathbb{R}$$\R$ have both a global maximum and a global minimum.

By Theorem 2.15, $fX\subseteq \mathbb{R}$$fX\subseteq\R$ is compact as well. Since $fX$$fX$ is compact if and only if $fX$$fX$ is closed and bounded.

$fX$$fX$ is bounded and $fX\subseteq \mathbb{R}$$fX\subseteq\R$ implies $fX$$fX$ has $sup$$\sup$ and $inf$$\inf$. $fX$$fX$ is closed implies $sup,inf\in fX.$$\sup,\inf\in fX.$ $Q.E.D.$$Q.E.D.$

Exercise 2.2

A map$X\to Y$$X → Y$ is locally constant if for each $x\in X$$x ∈ X$ there is an open set $U$$U$ with $x\in U$$x ∈ U$ and $f{|}_U$$f|_U$ constant. Prove or disprove: if $X$$X$ is connected and $Y$$Y$ is any space, then every locally constant map $f:X\to Y$$f : X → Y$ is constant.

Proof

Suppose $f$$f$ is not constant, pick $a\in X$$a\in X$,

then define $A=\{x\in X|f(x)=f(a)\}$$A=\{x\in X|f(x)=f(a)\}$, $B=\{x\in X|f(x)\ne f(a)\}$$B=\{x\in X|f(x)\ne f(a)\}$. $A$$A$ is not empty since $a\in A$$a\in A$.

Then $A$$A$ is open set since if $j\in A$$j\in A$, then $\mathrm{\exists }$$\exists$ an open neighbourhood $U_j$$U_j$, $f{U}_j=a$$fU_j=a$.

$B$$B$ is also open-set since if $i\in B$$i\in B$, then $\mathrm{\exists }$$\exists$ an open neighbourhood $U_i,f{U}_i\ne a$$U_i,fU_i\ne a$. $X=A\cup B$$X=A\cup B$ is connected implies $B=\mathrm{\varnothing }.$$B=\empty.$ $Q.E.D.$$Q.E.D.$

Exercise 2.18

Show that the product of Hausdorff Spaces is Hausdorff. Give an example to show that the quotient of a Hausdorff space need not be Hausdorff.

Proof

Suppose $X,Y$$X, Y$ is Hausdorff Spaces.

$X\times Y$$X\times Y$ is Hausdorff $⟺\mathrm{\Delta }:X\times Y⟶X\times Y\times X\times Y$$\iff \Delta:X\times Y\longrightarrow X\times Y\times X\times Y$ is closed.

Since $X\times Y\times X\times Y\cong X^2\times Y^2.$$X\times Y\times X\times Y\cong X^2\times Y^2.$

Since ${\mathrm{\Delta }}_X:X⟶X\times X,{\mathrm{\Delta }}_y⟶Y\times Y$$\Delta_X:X\longrightarrow X\times X,\Delta_y\longrightarrow Y\times Y$ is closed. Thus $\mathrm{\Delta }\cong {\mathrm{\Delta }}_X\times {\mathrm{\Delta }}_Y$$\Delta\cong\Delta_X\times\Delta _Y$ is closed. $Q.E.D.$$Q.E.D.$

Example

Let $\begin{array}{c}f:\mathbb{R}\to \{0,1\}:=\{\begin{array}{l}1,x\in \mathbb{Q}\\ 0,x\in \mathbb{R}-\mathbb{Q}\end{array}\end{array}$$f:\mathbb R\to\{0,1\}:=\begin{cases}1,x\in\mathbb Q\\0,x\in\mathbb R-\mathbb Q\end{cases}$ , give $\{0,1\}$$\{0,1\}$ the quotient topology, i.e. trivial topology. It is not Hausdorff Space.