Question

5- Recall that a set KCR is said to be compact if every open cover for K has a finite subcover 5-1) Use the above definition to prove that if A and B are two compact subsets of R then AUB is compact induction to show that a finite union of compact subsets of R is compact. 5-2) Now use 5-3) Let A be a nonempty finite subset of R. Prove that A is compact 5-4) Give an example of a countable union of compact sets that is not compact. (Hint: Recall that every compact subset of R is closed and bounded. The set N is not bounded and so it is not compact. However, N U{n}. Note that it follows from what you proved in the previous item that for each n E N, the singleton {n} is compact.)

Answer 1

5.1. IREN JKER Set 3 Uxl be an open cover of AUB. Then. Ix nA is an open cover of A and ? Van Ble is an open cover of B. Henco by compactness, I finite subcover 2 and anal and in B. a no it such a of A and B respectively.. Then, AUB C U U U U t (K-! 3) AUB compact u bus bon a finite colec 5.2. Let 2A; Sist in be a finire collection of compact subsets of, I will a Then by 5.1 AOA is compact. Let us suppose for nors i,,2 ALUAU. VAI is compact (1)

Then ANA₂ U- - UA; U Air = (A, UAGU... UA;) U Ait is compact by (1) and 5.1 . Hena by induction, finite union of compact subsets of IR is compact Bonnot Arane 5.3. It is sufficient to show that sugk-tow set subset hag e R is compact. Then by 5.2 finite we can conclude that a non empty finite to subscto of IR is compact. As if A=1 an 0 -- and then A= la sua go upang set Value be an open cover of ras. Then, I some xiel such that a E Uxi thin which is the finite subcover.. ras is compact langus y finite union of compact set is wh! co f rinite subert of IR is compact Olovy o bunto 5.4. For NEIN , In ń a compact subset of R. Uang = N as in is not closed at NE IN bomded. Hence in is not compact in R. A.S a compact subsets of R is closed and bonded. And ekarly it tol u ing b a countable unrou WEIN of compact sub rets.