prove {1/n : n is a natural} U {0} is compact using the definitions

Question

anonymous · Answer

i'll do the easy step. Let {G_alpha} be an open cover

ganeshie8 · Answer

you want to prove every open cover contains a finite subcover

anonymous · Answer

yes, im stuck on that step

anonymous · Answer

maybe contradiction?

anonymous · Answer

Suppose there is a cover that has no finite subcover

zzr0ck3r · Answer

Do you know that a closed bounded set in $\mathbb{R}$ is compact?

anonymous · Answer

I'm not supposed to use Heine-Borel theorem. That's why it says use the definition

zzr0ck3r · Answer

oh sorry didn't read that part.

zzr0ck3r · Answer

So open cover implies finite open cover?

anonymous · Answer

finite *subcover*

zzr0ck3r · Answer

right...

ganeshie8 · Answer

contradiction looks like a nice idea

zzr0ck3r · Answer

ok so we know the sequence converges, so we can grab some N so that everything is with in some eps of 0. I think that this N could stand for our desired N. Then go by way of contradiction and assume it is not

anonymous · Answer

Let's not use sequence since this lesson is covered before that of sequence.

zzr0ck3r · Answer

Well are you ok with this?

$S=\cup_{n=1}^{\infty}\{\frac{1}{n} \mid n\in \mathbb{N}\}$

zzr0ck3r · Answer

where $S$ is your set.

zzr0ck3r · Answer

even with t he bad notation...

anonymous · Answer

I was gonna say S a set :)

anonymous · Answer

I was thinking it has something to do with limit point

anonymous · Answer

|dw:1418535182810:dw|

It looks like that is the suggested subcover and it's finite

anonymous · Answer

:O i think that must be it!

anonymous · Answer

|dw:1418535579796:dw|