# Abstract Nonsense

## Thoughts about compactness (Pt. III, compactness and mappings)

Up until now we’ve talked quite extensively about necessary and sufficient conditions for a topological space to be compact, but we’ve neglected to speak about how compactness interacts with mappings of topological spaces into another topological spaces. Probably the most widely used theorem is the following:

Theorem 1: The continuous image of a compact topological space is compact, or phrased differently if $f:X\mapsto Y$ be continuous and $X$ compact, then $f(X)$ is a compact subspace of $Y$.

Proof: Let $\left\{O_j\right\}_{j\in\mathcal{J}}$ be an open cover of $f(X)$ then $O_j=E_j\cap f(X)$ for some open set $E_j$ in $Y$. Clearly we have that $\left\{E_j\right\}$ covers $f(X)$ and so $\displaystyle X=f^{-1}\left(\bigcup_{j\in\mathcal{J}}E_j\right)=\bigcup_{j\in\mathcal{J}}f^{-1}\left(E_j\right)$ and since $f$ is continuous and $E_j$ open in $Y$ we see that $\left\{f^{-1}\left(E_j\right)\right\}_{j\in\mathcal{J}}$ is an open cover of $X$ and thus by assumption we have that there exists some $\left\{f^{-1}\left(E_{j_1}\right),\cdots,f^{-1}\left(E_{j_m}\right)\right\}\subseteq\left\{f^{-1}\left(E_j\right)\right\}_{j\in\mathcal{J}}$ which is also an open cover. Therefore,$\displaystyle X=f^{-1}\left(E_{j_1}\right)\cup\cdots f^{-1}\left(E_{j_m}\right)$ $\displaystyle =f^{-1}\left(\bigcup_{k=1}^{m}E_{j_k}\right)\implies f(X)=ff^{-1}\left(\bigcup_{k=1}^{m}E_{j_k}\right)\subseteq\bigcup_{k=1}^{m}E_{j_k}$. Therefore, $\left\{E_{j_1}\cap f(X),\cdots,E_{j_m}\cap f(X)\right\}=\left\{O_{j_1},\cdots,O_{j_m}\right\}\subseteq\left\{O_{j}\right\}_{j\in\mathcal{J}}$ is an open cover for $X$. The conclusion follows. $\blacksquare$

Corollary : If $f:X\mapsto Y$ is a surjective continuous mapping and $X$ is compact, then $Y$ is compact.

Corollary : If $X\approx Y$ (that’s what I use for homeomorphic) then $X$ is compact if and only if $Y$ is compact.

With this theorem we can prove one of the most highly used, and least thought about, theorems in elementary calculus.

Theorem 2: Let $f:X\mapsto\mathbb{R}$ be continuous, where $X$ is a compact topological space. Then $\sup\left\{f(x):x\in X\right\},\inf\left\{f(x):x\in X\right\}\in f(X)$.

Proof:

Lemma: A compact subspace of a metric space $X$ is bounded.

Proof: Let $E\subseteq X$ be compact. Consider the class $\Omega=\left\{B_{1}(e)\right\}_{e\in E}$. Clearly $\Omega$ is an open cover for $X$ and so there exists some finite subcover $\left\{B_{1}(e_1),\cdots,B_{1}(e_n)\right\}$. Clearly then, $\displaystyle \text{diam }E\leqslant\text{diam }\bigcup_{j=1}^{n}B_1(e_j)\leqslant \sum_{j=1}^{n}2=2n$. The conclusion follows. $\blacksquare$.

Now, we know that $f(X)$ is compact, and combining the above with the earlier proof that ever compact subspace of a Hausdorff space is closed we see that $f(X)$ is closed and bounded. Thus, we have that $\sup\left\{f(x):x\in X\right\},\inf\left\{f(x):x\in X\right\}$ exist and it’s a relatively easy exercise in the basic topology of the real line that if $C\subseteq\mathbb{R}$ is closed then $\inf\text{ }C,\sup\text{ }C\in C$. It follows that $\sup\left\{f(x):x\in X\right\},\inf\left\{f(x):x\in X\right\}\in f(X)$. $\blacksquare$

Corollary 1: A continuous function $f:[a,b]\mapsto\mathbb{R}$ assumes a maximum and minimum value on $[a,b]$

Corollary 2: If $f:X\mapsto\mathbb{R}^+$ is continuous and $X$ compact, then $\inf\text{ }f(X)\geqslant 0$

Also, using corollary two  you can prove the Heine-Borel theorem in more elementary, albeit more complex, terms.

Theorem (Heine-Borel): If $E\subseteq\mathbb{R}$ is closed and bounded, then $E$ is compact.

Proof: Just as last time we note that since $E$ is bounded that it is contained in some interval $[a,b],\text{ } a. And so, if we can prove that $[a,b]$ is compact then $E$‘s compactness follows since it is a closed subspace of a compact space. So now we need to prove that $[a,b]$ is compact, but we even have to prove something simpler thanks to the above theorem and the following lemma.

Lemma: Let $[a,b],[c,d]\subseteq\mathbb{R}$ and $a then $[a,b]\approx[c,d]$

Proof: If we can find a linear function $\ell:[a,b]\mapsto[c,d]$ we are done. So assume that one exists and is of the form $\ell(x)=mx+n$. Then we need $\ell(a)=ma+n=c,\ell(b)=mb+n=d$. Solving this gives $\displaystyle \ell(x)=\frac{d-c}{b-a}x+\frac{bc-ad}{b-a}$ which clearly satisfies the conditions of a homeomorphism. $\blacksquare$

With this lemma in mind it remains to prove that any closed interval is compact. We choose, of course, for convenience $I=[0,1]$. So let $\Omega$ be an open cover of $[0,1]$, and assume that it admits no finite subcover. Clearly then either no finite subclass of $\Omega$ can cover either $\displaystyle \left[0,\frac{1}{2}\right],\left[\frac{1}{2},1\right]$. Let $[a_1,b_1]$ be one of these subintervals. Clearly $[a_1,b_1]$ has length $\displaystyle \frac{1}{2}$. Now, using the same logic we see that no finite subclass of $\Omega$ can cover either $\displaystyle \left[a_1,\frac{a_1+b_1}{2}\right],\left[\frac{a_1+b_1}{2},b_1\right]$. Continuing in this manner we have a sequence of closed intervals $\left\{[a_n,b_n]\right\}_{n\in\mathbb{N}}$. It is clear that $a_n\leqslant a_{n+1}\leqslant b_n\leqslant b_{b+1}$. Clearly then each $b_n$ is an upper bound for the set $\{a_n\}_{n\in\mathbb{N}}$ and so let $\displaystyle \alpha=\sup_{n\in\mathbb{N}}a_n$ and $\displaystyle \beta=\sup_{n\in\mathbb{N}}b_n$. Clearly by prior comment $\alpha\leqslant\beta$ but by definition $a_n\leqslant \alpha\leqslant\beta\leqslant b_n \text{ }(1)$ and by the contrivance of our sequence $(1)$ implies that $\beta-\alpha\leqslant\frac{1}{2^n}$ for all $n$ and so $\alpha=\beta$. Since $\Omega$ covers $I$ and $\alpha=\beta\in I$ where exists some $\omega\in \Omega$ such that $a\in\omega$. Now, since $\omega$ is open there exists some $\varepsilon>0$ such that $B_{\varepsilon}(a)\subseteq \omega$.

The Archimedean principle now furnishes us with some $N\in\mathbb{N}$ such that $\displaystyle \frac{1}{2^N}<\varepsilon$. Thus, $b_N-a_N<\varepsilon$. Now, $\alpha=\beta\in\left[a_N,b_N\right]$. Therefore $\alpha-a_N<\frac{1}{2^N}<\varepsilon$ and $b_N-\beta<\frac{1}{2^N}$. Consequently, $[a_N,b_N]\subseteq B_{\varepsilon}(q)\subseteq\omega$. Therefore, we have that $[a_N,b_N]$ may be covered by a finite subclass of $\Omega$. Boom. $\blacksquare$

Another theorem which proves useful has to do with mappings from compact spaces into Hausdorff spaces, in particular:

Theorem: Let $X$ be compact and $Y$ Hausdorff, then if $f:X\mapsto Y$ is continuous and bijective, then it is a homeomorphism.

Proof: We must merely show that the mapping is open to conclude that it’s a homeomorphism, but since $f$ is bijective we know that given any $E\subseteq X$ that $f\left(E'\right)=\left[f\left(E\right)\right]'$ and so $f$ is open if and only if it’s closed. Thus, we must merely show that $f$ is closed. To do this we merely note that since $X$ is compact, that any closed $E\subseteq X$ must necessarily be compact. And by our first theorem then $f(E)$ is a compact subspace of $Y$. But, by previous problem compact subspaces of Hausdorff spaces are closed and so $f(E)$ is closed. The conclusion follows. $\blacksquare$

This theorem has some awfully amazing applications when trying to prove that two spaces are homeomorphic. It saves a lot of hassle. Our next theorem has to do with the graph of a continuous mapping from a compact space into a Hausdorff space, and although they’re are easier methods to prove this, I believe that a little elbow grease is worth the reward. That said, we need to prove some theorems not related to compactness.

Theorem: Let $f:X\mapsto Y$ be continuous and define $f\times\iota_Y=X\times Y\mapsto Y\times Y$ by $(x,y)\mapsto (f(x),y)$. Then, this function is continuous.

Proof:

Lemma: $\left(f\times\iota_Y\right)^{-1}\left(U\times V\right)=f^{-1}(U)\times V$

Proof: Let $x\in \left(f\times \iota_Y\right)^{-1}(U\times V)$, then $\left(f\times \iota_Y\right)(x,y)=(f(x),y)\in U\times V$ and so $f(x)\in U$ and $y\in V$ and so $x\in f^{-1}(U)$ and $y\in V$ and so $(x,y)\in f^{-1}(U)\times V$. Conversely, let $(x,y)\in f^{-1}(U)\times V$, then $f(x)\in U$ and $y\in V$. Therefore, $(f(x),y)=\left(f\times \iota_Y\right)(x,y)\in U\times V$ and so $(x,y)\in\left(f\times\iota_Y\right)^{-1}\left(U\times V\right)$. The conclusion follows.$\blacksquare$

Now, since we must only prove that the inverse image of any basic open set in $Y$ is open in $X\times Y$, but this follows immediately since any basic open set is of the form $U\times V$ where both $U$ and $V$ are open in $Y$. And since $f:X\mapsto Y$ is continuous we see that $f^{-1}(U)$ is open and so $\left(f\times \iota_Y\right)^{-1}\left(U\times V\right)=f^{-1}(U)\times$ is the product of two open sets in $X$ and $Y$ and thus basic open in $X\times Y$ and so clearly open. The conclusion follows. $\blacksquare$

Next, we prove a fact that is used incessantly as a counter example when dealing with Hausdorff spaces.

Theorem: Let $\Delta=\left\{(y,y):y\in Y\right\}$ (this is called the diagonal in $Y\times Y$). Then $\Delta$ is closed in $Y\times Y$.

Proof: We prove that $\Delta$ is closed by proving that $\Delta'$ is open. So, let $(x,y)\in\Delta'$, then $x\ne y$ and since both $x$ and $y$ are in the Hausdorff space $Y$ there exists open sets $O_x$ and $O_y$ such that $x\in O_x$ and $y\in O_y$ and $O_x\cap O_y=\varnothing$. Clearly then $latext (x,y)\in O_x\times O_y$ which is open in $Y\times Y$ and since $O_x\cap O_y=\varnothing$ we see that $O_x\times O_y\cap\Delta=\varnothing$. The conclusion follows. $\blacksquare$

We prove our second to last theorem before we get to the relevant part.

Theorem: Let $f\times\iota_Y$, $\Delta$,  $Y$, and $X$ be as above, then $\left(f\times\iota_Y\right)^{-1}\left(\Delta\right)=\Gamma_f$ where ($\Gamma_f=\left\{(x,y)\in X\times Y:y=f(x)\right\}$).

Proof: Let $(x,y)\in \left(f\times \iota_Y\right)^{-1}\left(\Delta\right)$, then $\left(f\times \iota_Y\right)(x,y)=(f(x),y)\in \Delta$ but this clearly implies that $f(x)=y$. Therefore, $(x,y)\in\Gamma_f$

Conversely, let $(x,f(x))\in\Gamma_f$, then $\left(f\times\iota_Y\right)(x,f(x))=(f(x),f(x))\in\Delta$. The conclusion follows. $\blacksquare$

From this we quickly derive the following fact:

Theorem: Let $f:X\mapsto Y$ be continuous and $Y$ Hausdorff, then $\Gamma_f$ is a closed in $X\times Y$.

Proof: Since $f:X\mapsto Y$ is continuous we know that $f\times\iota_Y:X\times Y\mapsto Y\times Y$ is continuous, and since $\Delta$ is closed in $Y\times Y$ it follows that $\left(f\times\iota_Y\right)^{-}\left(\Delta\right)=\Gamma_f$ is closed. The conclusion  follows. $\blacksquare$

We now return to the actual theorem we wished to prove.

Theorem: If $f:X\mapsto f(X)$ is continuous and $X$ is compact, then $\Gamma_f$ is a compact suspace of $X\times f(X)$.

Proof: It follows from out last theorem that $\Gamma_f$ is a closed subspace of $X\times f(X)$, but since $X$ is compact we know that $f(X)$ is compact and since (see next post) the product of compact spaces is compact, we know that $X\times f(X)$ is compact. Thus, $\Gamma_f$ is a closed subspace of a compact space, and thus compact. The conclusion follows. $\blacksquare$