Lecture 10

Review

Recall: If $K\subset \cup_{\alpha\in A} G_{\alpha}$, then we say $\{G_\alpha\}_{\alpha\in A}$ is a cover of $K$. If, in addition, each set $G_{\alpha}$ is open, then we say $\{G_{\alpha}\}_{\alpha\in A}$ is an open cover of $K$. If $\alpha_1,...,\alpha_n\in A$ are such that $K\subset \bigcup _{i=1}^n G_{\alpha_i}$, then we say $\{G_{\alpha_i}\}_{i=1}^n$ is a finite subcover.

Let $X=\mathbb{R}$. Come up with some examples of covers of $\mathbb{R}$. Try to find a few satisfying each of the following:

1. A cover of $\mathbb{R}$ which is not an open cover. $\{[x,x+1]:x\in \mathbb{Z}\}$

2. An open cover of $\mathbb{R}$ which does have a finite subcover. $\{\mathbb{R}\}$ is an open cover with finite subcover, itself $\{\mathbb{R}\}$. AND, $\{\mathbb{Q},\mathbb{R}\backslash\mathbb{Q}\}$ is not a subcover of $\{\mathbb{R}\}$ since we need to select subcover from cover set. And not taking the element of sets in the open cover.

3. An open cover of $\mathbb{R}$ which does not have a finite subcover. $\{(x,x+2):x\in \mathbb{Z}\}$ No finite subcover, infinitely many sets.

   Proof: we proceed by contradiction, suppose we take $\{(n_i,n_i+2):i=1,...,k\}$. The union does not contain $max\{n_1,...,n_k\}+2$

   $\{\{x\in\mathbb{R}:x<n\}:n\in \mathbb{Z}\}$

   And some stupid set we have is $\{\{x\in\mathbb{R}:x<n\}:n\in \mathbb{Z}\}\cup \{\mathbb{R}\}$ with finite subcover $\{\mathbb{R}\}$

New

Compact set

$K$ is compact if $\forall$ open cover, $\exists \{G_{\alpha_i}\}_{i=1}^n$ that is a finite subcover.

$\mathbb{R}$ is not compact since we can build a open cover $\{(x,x+2):x\in \mathbb{Z}\}$ such that we cannot find a finite subcover of $\mathbb{R}$

$\{1,2\}$ is compact let $${G_{\alpha}}_{\alpha\in A}$be an open cover of${1,2}$

Ironically, $[0,1]$ is compact. This will follow from Theorem 2.40.

Theorem 2.33

Let $(X,d)$ be a metric space, and $K\subset Y \subset X$. Then $K$ is compact relative to $X$ ($K$ is open in $X$ ) $\iff K$ is compact relative to $Y$. This implies that compactness is an absolute property

Proof:

$\implies$ Suppose $K$ is compact relative to $X$.

Let $\{V_{\alpha}\}_{\alpha\in A}$ be an open cover of $K$ relative to $Y$.

By Theorem 2.30, for each $\alpha$, $\exist G_{\alpha}$ open in $X$ such that $V_{\alpha}=G_{\alpha}\cap Y$. Then $\{G_\alpha\}_{\alpha}$ is an open cover of $K$ relative to $X$. Since $K$ is compact relative to $X$, $\{G_{\alpha}\}_{\alpha}$ has a finite subcover $\{G_{\alpha_i}\}_{i=1}^n$. Then $\{V_{\alpha_i}\}^n_{i=1}$ is a finite subcover of $\{V_{\alpha}\}_{\alpha}$ of $K$

$\impliedby$ Suppose $K$ is compact relative to $Y$. Let $\{G_\alpha\}_{\alpha\in A}$ be an open cover of $K$ relative to $X$. By Theorem 2.30, $\{G_\alpha\cap Y\}_\alpha$ is an open cover of $K$ relative to $X$

Since $k$ is compact relative to $Y$, $\{G_\alpha\cap Y\}_\alpha$ has a finite subcover $\{G_{\alpha_i}\cap Y\}_{i=1}^n$. Then $\{G_{\alpha_i}\}_{i=1}^n$ is a finite subcover of $\{G_\alpha\}_{\alpha}$ of $K$.

EOP

Theorem 2.24

Let $(X,d)$ be a metric space, $K\subset X$ is compact $\implies K$ is closed in $X$.

Proof:

Suppose $K$ is compact. We'll show that $K^c$ is open, i.e $\forall p\in K^c$, $\exists r>0$ such that $B_r(p)<K^c$. Let $p\in K^c$ ($p$ is fixed for the remainder of the $q$)

For each $q\in K$, Let $V_q=B_{\frac{1}{2}d(p,q)}(p),W_q=B_{\frac{1}{2}d(p,q)}(q)$ By triangle inequality, $V_q\cap W_q=\phi$

Since $K\subset \bigcup_{q\in K}W_q$ and $\forall q\in K, V_q\cap W_q=\phi$. By set theory, $\left(\bigcap_{q\in K}V_q\right)\cap K=\phi$, and $\left(\bigcap_{q\in K}V_q\right)\subset K^c$

Since $\{W_q\}_{q\in K}$ is an open cover of $K$, so it has a finite cover $\{W_{q_i}\}_{i=1}^n$. So $\left(\bigcap_{q\in K}V_q\right)=\left(\bigcap_{i=1}^n V_{q_i}\right)\subset K^c$.

Then let $r=min\{\frac{1}{2}d(p_i,q_i)\}$, $\bigcap_{i=1}^n V_{q_i}=B_r(p)$