Lecture 22 - Intro to Baire's Theorem

See also Finite Subcover.

Example

Consider the open interval $(0,1)$. For each point $x\in (0,1)$ let $O_x=(\frac{x}{2},1)$. Taken together $\{O_x:x\in (0,1)\}$ forms an open cover for the interval $(0,1)$. Notice that it is impossible to find a Finite Subcover. Given any proposed finite subcollection:
$\{O_{x_1},O_{x_2},\dots ,O_{x_n}\}$
then let $x^{\prime }=min\{x_1,x_2,\dots ,x_n\}$ and observe that we can construct a number $y$ satisfying $0<y\le \frac{x^{\prime }}{2}$. Then $y\notin \bigcup _{i=1}^{\mathrm{\infty }}{O}_{x_i}$.

Now consider a similar subcover for the closed interval $[0,1]$. For $x\in (0,1)$ the set $O_x$ defined above works covering $(0,1)$ but in order to have an open cover of the closed $[0,1]$ interval then we need the endpoints. Let $\epsilon >0$ and let $O_0=(-\epsilon ,\epsilon )$ and $O_1=(1-\epsilon ,1+\epsilon )$. Then the collection:
$\{O_0,O_1,O_x:x\in (0,1)\}$
is an open cover for $[0,1]$. But this time notice that there is a finite subcover. This is because the addition of $O_0$ is here, we can choose $x^{\prime }$ so that $\frac{x^{\prime }}{2}<\epsilon$. It follows that our collection
$\{O_0,O_{x^{\prime }},O_1\}$is a finite subcover for $[0,1]$.

See also Open Cover.

For examples see Open Cover#Example.

Proof

Since Characterization of Compactness in R gives $(1)⟺(2)$ then all we have to show is $(3)\Rightarrow (2)$ and then $(3)\Leftarrow (2)$.

($\Rightarrow$): Suppose every open cover of $K$ has a finite subcover. To show that $K$ is bounded, we construct an open cover for $K$ by defining $O_x$ to be an open interval of radius 1 around each point $x\in K$. In the language of Epsilon-Neighborhoods, $O_x=V_1(x)$. The open cover $\{O_x:x\in K\}$ then must have a finite subcover (see the supposition) $\{O_{x_1},\dots ,O_{x_n}\}$. Because $K$ is contained in a finite union of bounded sets (ie: $K\subseteq \bigcup _{i=1}^n{O}_{x_i}$) then $K$ itself must be bounded (just take the right boundary of the largest $x_i$).

To show that $K$ is closed, there's more work. We'll argue it by contradiction. Let $(y_n)$ be a Cauchy Sequence contained in $K$ where $y_n\to y$. To show that $K$ is closed it is enough to show that $y\in K$.

Assume for contradiction that $y\notin K$. Then $\mathrm{\forall }x\in K$, then each $x$ is some positive distance away from $y$ (otherwise then $x=y\Rightarrow y\in K$). Thus, we can try to construct an open cover by taking $O_x$ to be an interval of radius $\frac{|x-y|}{2}$ around each point $x\in K$. Because we suppose (3), the resulting open cover $\{O_x:x\in K\}$ must have some finite subcover $\{O_{x_1},\dots ,O_{x_n}\}$. The contradiction comes from noticing that this finite subcover cannot contain all of the elements of the sequence $(y_n)$. To show this, set:
${\epsilon }_0=min\{\frac{|x_i-y|}{2}:1\le i\le n\}$
Because $(y_n)\to y$ and ${\epsilon }_0>0$ (remember, all the distances are positive), then $\mathrm{\exists }N\in \mathbb{N}$ such that:
$|y_n-y|<{\epsilon }_0=min\{\frac{|x_i-y|}{2}:1\le i\le n\}$
for $n\ge N$. Namely just choosing $n=N$ here works, because notice that:
$x\in O_{x_i}⟺\frac{|x_i-x|}{2}<\frac{|x_i-y|}{2}$
Notice the inequality never holds for $y$ so then $y\notin O_{x_i}$ for all $1\le i\le n$. Thus, $y_N\notin O_{x_1},\dots ,O_{x_n}$, so then:
$y_N\notin \bigcup _{i=1}^n{O}_{x_i}$
So our cover doesn't actually cover all of $K$, which is a contradiction. Thus $y\in K$ and thus $K$ is closed and bounded.

($\Leftarrow$): We'll use the following lemma:

Proof

(1) If $x\in {\left(\bigcup _{\lambda \in \mathrm{\Lambda }}{E}_{\lambda }\right)}^c$ then $x\notin \bigcup _{\lambda \in \mathrm{\Lambda }}{E}_{\lambda }$, meaning $x\notin E_{\lambda }$ for any $\lambda \in \mathrm{\Lambda }$. This implies that $x\in E_{\lambda }^c$ for all $\lambda$. So then $x\in \bigcap _{\lambda \in \mathrm{\Lambda }}{E}_{\lambda }^c$ Thus giving $\subseteq$ for (1).

For $\supseteq$, let $x\in \bigcap _{\lambda \in \mathrm{\Lambda }}{E}_{\lambda }^c$. Then $x\notin E_{\lambda }$ for all $\lambda$ meaning $x\notin \bigcup _{\lambda \in \mathrm{\Lambda }}{E}_{\lambda }$ so then $\supseteq$.

(2) Comes from doing a very similar argument to (1).

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We'll show closure inductively. Let $F=F_1\cup F_2$. Let $x_n\in F$ and $x=\underset{n\to \mathrm{\infty }}{lim}{x}_n$. Let $(x_{n_k})$ be a subsequence of $(x_n)$ fully contained in $F_1$ or $F_2$. The subsequence must also converge to $x$, so $x\in F_1\cup F_2=F$. This is the base case.

For the inductive step, let $F=\bigcap _{\lambda \in \mathrm{\Lambda }}{F}_{\lambda }$. Then:
$F^c=\bigcup _{\lambda \in \mathrm{\Lambda }}{F}_{\lambda }^c$
Each $F_{\lambda }^c$ is closed, thus $F^c$ is open, so then $F=(F^c{)}^c$ is closed.

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Lecture Proof

We did the proof of this in class, but I prefer to use the textbook's proof of the idea (it's just easier to follow). If you want to view the proof in another context though see the following.

Scratch:

For ($2\to 3$) it's kind-of a mess. But the idea is that we can get a suprema $s$ and then actually say that we could go even futher past the suprema to indicate a finite subcover.

Proof

We'll show $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (1)$.

(1 $\to$ 2): See Characterization of Compactness in R.

(2 $\to$ 3): Let's prove a useful lemma really quick:

Proof

For $n\in \mathbb{N}$ choose $a_n\in A$ such that $supA-\frac{1}{n}<a_n\le supA$. Using the Squeeze Theorem then we can show $a_n\to supA$. The same can be done for the $inf$ by using $infA\le b_n<infA+\frac{1}{n}$.

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Let's start the proof. Suppose $K$ is closed and bounded. Then all limit points of $K$ are in $K$ and any sequence $(x_n)\subseteq K$ is bounded. If $K$ is empty then we are done so suppose $K\ne \mathrm{\varnothing }$. Because of this then $supK$ and $infK$ exist, so then let $a=supK$ and $b=infK$. Therefore $K\subseteq [a,b]$. Note that $a,b\in K$ since $K$ is closed.

Now let $\{U_{\lambda }|\lambda \in \mathrm{\Lambda }\}$ be an open cover of $K$. Let $S=\{c\in [a,b]|K\cap [a,c]\text{ has a finite subcover}\}$. Notice that since $S\subseteq [a,b]$ then $S$ is bounded. Further $S\ne \mathrm{\varnothing }$ since $a\in S$ for example. Thus we can take $s=supS$. Since $K$ is closed, then $s\in K$ (see the Lemma we just proved). Thus then $\mathrm{\exists }{U}_{{\lambda }_0}$ such that $s\in U_{{\lambda }_0}$. Therefore $\mathrm{\exists }ϵ>0$ such that $s\in V_{ϵ}(s)\subseteq U_{{\lambda }_0}$.

Since $s-ϵ<s$ then $\mathrm{\exists }{s}_1\in S$ such that $s-ϵ<s_1\le s$. Therefore $\mathrm{\exists }$ a finite subcover of $K\cap [a,s_1]$. Let's call that subcover $U_{{\lambda }_1},\dots ,U_{{\lambda }_n}$. Then taking:
$U_{{\lambda }_0},U_{{\lambda }_1},\dots ,U_{{\lambda }_n}$
certainly still covers $K\cap [a,s]$. Therefore $s\in S$.

Now assume for contradiction that $s<b$. Then $\mathrm{\exists }{s}_2$ such that $s<s_2\le b$ and $s_2<s+ϵ$ and $K\cap [a,s_2]$ is covered by $U_{{\lambda }_0},\dots ,U_{{\lambda }_n}$. Therefore $s_2\in S$ is a contradiction $s_2>s$ which contradicts $s=supS$.

(3 $\to$ 1): Suppose every open cover of $K$ has a finite subcover. Now if you consider:
$\bigcup _{n\in \mathbb{N}}(-n,n)=\mathbb{R}\supseteq K$
Thus we found a finite subcover $(-n_1,n_1),\dots ,(-n_{\ell },n_{\ell })$. Let $M=max(n_i|i=1,\dots ,\ell )$. Then $M$ is a bound for $K$, so $\mathrm{\forall }x\in K$ then $|x|\le M$.

Now let $(x_n)\subseteq K$. By the Balzano-Weierstrass Theorem then $\mathrm{\exists }$ a subsequence $x_{n_k}$ where $x_{n_k}\to x$.

We need to show that $x\in K$ to get (1), so assume for contradiction that $x\notin K$. For $y\in K$ let:
$U_y=V_{\frac{|y-x|}{2}}(y)$
forms an open over of $K$, so then $\mathrm{\exists }$ a finite subcover $U_{y_1},\dots ,U_{y_n}$. Now let $ϵ=min\{\frac{|y-x|}{2}|i=1,\dots ,n\}$. None of the $U_y$ epsilon-neighborhoods get into distance of $x$, since every $y\in K$ is a distance $\ge ϵ$ from $x$. This contradicts $x$ being a limit.

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