2.7 KiB
Math 4111 Exam 2 review
E is open if \forall x\in E, x\in E^\circ (E\subset E^\circ)
E is closed if E\supset E'
Then E closed \iff E^c open \iff \forall x\in E^\circ, \exists r>0 such that B_r(x)\subset E^c
\forall x\in E^c, \forall x\notin E
B_r(x)\subset E^c\iff B_r(x)\cap E=\phi
Past exam questions
S,T is compact \implies S\cup T is compact
Proof:
Suppose S and T are compact, let \{G_\alpha\}_{\alpha\in A} be an open cover of S\cup T
(NOT) \{G_\alpha\} is an open cover of S, \{H_\beta\} is an open cover of T.
...
QED
K-cells are compact
We'll prove the case k=1 and I=[0,1] (This is to simplify notation. This same ideas are used in the general case)
Proof:
That [0,1] is compact.
(Key idea, divide and conquer)
Suppose for contradiction that \exists open cover \{G_a\}_{\alpha\in A} of [0,1] with no finite subcovers of [0,1]
Step1. Divide [0,1] in half. [0,\frac{1}{2}] and [\frac{1}{2},1] and at least one of the subintervals cannot be covered by a finite subcollection of \{G_\alpha\}_{\alpha\in A}
(If both of them could be, combine the two finite subcollections to get a finite subcover of [0,1])
Let I_1 be a subinterval without a finite subcover.
Step2. Divide I_1 in half. Let I_2 be one of these two subintervals of I_1 without a finite subcover.
Step3. etc.
We obtain a seg of intervals I_1\subset I_2\subset \dots such that
(a) [0,1]\supset I_1\supset I_2\supset \dots
(b) \forall n\in \mathbb{N}, I_n is not covered by a finite subcollection of \{G_\alpha\}_{\alpha\in A}
(c) The length of I_n is \frac{1}{2^n}
By (a) and Theorem 2.38, \exists x^*\in \bigcap^{\infty}_{n=1} I_n.
Since x^*\in [0,1], \exists \alpha_0 such that x^*\in G_{\alpha_0}
Since G_{\alpha_0} is open, \exist r>0 such that B_r(x^*)\subset G_{\alpha_0}
Let n\in \mathbb{N} be such that \frac{1}{2^n}<r. Then by (c), I(n)\subset B_r(x^*)\subset G_{\alpha_0}
Then \{G_{\alpha_0}\} is a cover of I_n which contradicts with (b)
QED
Redundant subcover question
M is compact and \{G_\alpha\}_{\alpha\in A} is a "redundant" subcover of M.
\exists \{G_{\alpha_i}\}_{i=1}^n is a finite subcover of M.
We define S be the x\in M that is only being covered once.
S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)
We claim S is a closed set.
G_{\alpha_i}\cap G_{\alpha_j} is open.
\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right) is closed
S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right) is closed.
So S is compact, we found another finite subcover yeah!