66 lines
3.8 KiB
Markdown
66 lines
3.8 KiB
Markdown
# 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 $K$.
|
|
|
|
QED
|
|
|
|
#### 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 $r=min\{\frac{1}{2}d(p_i,q_i)\}$, $\bigcap_{i=1}^n V_{q_i}=B_r(p)$ |