NoteNextra-origin/content/Math4111/Math4111_L10.md

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)