March 13, 2018 - Shawn Zhong

$Theorem 2.34 • Statement ○ Compact subsets of metric spaces are closed • Proof ○ Let K be a compact subset of a metric space X ○ We shall prove that the complement of K is open ○ Let p∈K^c, q∈K ○ Let V_q=N_r (p),W_q=N_s (q) where r,s 1/2 d(p,q) ○ Since K is compact, ∃q_1,q_2,…q_n∈K s.t. ○ K⊂W_(q_1 )∪W_(q_2 )∪…∪W_(q_n )=W ○ Let V=V_(q_1 )∩V_(q_2 )∩…∩V_(q_n ) ○ Then V is a neighborhood of p that does not intersect W ○ V⊂K^c⇒p is an interior point of K^c ○ So K^c is open and therefore K is closed Theorem 2.35 • Statement ○ Closed subsets of compact sets are compact • Proof ○ Let X be a metric space ○ Suppose F⊂K⊂X, where F is closed, and K is compact ○ Let {V_α } be an open cover of F ○ Consider {V_α }∪{F^c }, where F^c is open ○ Then {V_α }∪{F^c } is an open cover of K ○ Since K is compact, K has a finite subcover Φ ○ If F^c∈Φ, then Φ∖{F^c } is still finite and covers F ○ So we have a finite subcover of {V_α } ○ Therefore F is compact • Corollary ○ If F is closed and K is compact , then F∩K is compact • Proof ○ K compact ⇒K is closed ○ We know F is closed, so F∩K is closed ○ F∩K⊂K, and K is compact ○ So F∩K is compact Theorem 2.36 (Cantor s Intersection Theorem) • Statement ○ If {K_α } is a collection of compact subsets of a metric space X s.t. ○ The intersection of every finite subcollection of {K_α } is nonempty ○ Then ⋂_α▒K_α is nonempty • Proof ○ Fix K_1∈{K_α } and let G_α=K_α^c \,∀α ○ Assume no point of K_1 belongs to every K_α ○ Then {G_α } is an open cover of K_1 ○ Since K_1 is compact, K_1⊂G_(α_1 )∩G_(α_2 )∩…∩G_(α_n ) ○ Where α_1,α_2,…,α_n is a finite collection of indices ○ Then K_1∩G_(α_2 )∩…∩G_(α_n )=∅ ○ This is a contradiction, so no such set K_1 exists ○ The result follows • Corollary ○ If {K_n } is a sequence of nonempty compact sets s.t. K_n⊃K_(n+1),∀n∈N ○ Then ⋂24_(n=1)^∞▒K_n is nonempty Theorem 2.37 • Statement ○ If E is an infinite subset of a compact set K ○ Then E has a limit point in K • Proof ○ If no point of K were a limit point of E ○ Then ∀q∈K, ∃N(q) s.t. no point of E other than q ○ i.e. N(q) contains at most one point of E (namely, q, if q∈E) ○ So no finite sub-collection of {N(q)} can cover E, and thus not K ○ This is a contradiction, so E has a limit point in K$

Shawn Zhong

Shawn Zhong

Math 521 – 3/12

Search

WeChat Account