Question

Prove that S = {0} union {1/n : n in natural numbers} is a compact set in R with the usual topology (w/o using Heine Borel)

Answer 1

Wow, this is crazy stuff.

Answer 2

{1/n : n in natural numbers} has only one acumulation point. And it is {0}. Sincé S includes this point , it means it is closed. Just left to show it's bounded. But this is trivial. So : since it is closed and bounded , it's compact

Answer 3

What about if you had to use the definition of a compact set? Not Heine Borel?

Answer 4

your job is to prove that every open cover contains a finite subcover

Answer 5

on account of it says "without heine borel"

Answer 6

call your open cover something say \(O=\cup_{\alpha \in A}O_{\alpha}\)

Answer 7

then pick the one zeros lives in, say \(0\in O_{\alpha_1}\)

Answer 8

since \(O_{\alpha_1}\) is open, there is an open ball contained in that set
then argue that for all but a finite number of number of points live in that ball

Answer 9

hello joe!
@joemath314159 i am sure can explain better than i, but the idea is that any cover contains zero, and then all but a finite number of elements will be in the one that contains zero, therefore you can reduce to a finite subcover

Answer 10

your explanation is great!

Answer 11

thanks, shows i have some memory left