Let A be a bounded set in $R^n$. Prove that cl (A) is compact.
Please, help.

Question

Let A be a bounded set in $R^n$. Prove that cl (A) is compact.
Please, help.

loser66 · Answer

cl (A) is a closed set, so, remain is showing cl (A) is bounded. But how?

loser66 · Answer

@wio

anonymous · Answer

is this what you need? http://www.math.utah.edu/~bobby/3220/HW1-3solns.pdf

loser66 · Answer

Thanks for the link but it is not what I need. :)

terenzreignz · Answer

Look who it is... haha

terenzreignz · Answer

Quick question... what does cl(A) even mean?

(honest question, actually... I really don't know what it means :3 )

loser66 · Answer

closure of A = interior of A $\cup$ boundary of A

loser66 · Answer

it's advanced calculus stuff, I am tired with it. :(

terenzreignz · Answer

urgh

terenzreignz · Answer

Me no know this stuff haha

loser66 · Answer

:)

terenzreignz · Answer

What's the definition of 'boundary' ?

loser66 · Answer

It's easy, hehehe I can answer it: if A = (0,1) then bd (A) = {0,1}

loser66 · Answer

If A = { (x,y) in R^2 \{0}} , then bd (A) ={0}

loser66 · Answer

Since you said you no nothing about it, keep making question, I can answer (If I know), by that way, I review, you get a new concept, both us win (something) hahaha....

ganeshie8 · Answer

are you trying to prove closure of a bounded set is bounded ?

loser66 · Answer

Yes

ganeshie8 · Answer

i think you can use it as a proven fact, your main goal is proving compactness of cl(A) right ?

ganeshie8 · Answer

either way see this for proving closure of bounded set is bounded http://math.stackexchange.com/questions/322948/show-that-the-closure-of-a-subset-is-bounded-if-the-subset-is-bounded

loser66 · Answer

Oh, boy!! many things to do with this simple stuff. Thanks for the link @ganeshie8  It is a big help.

ganeshie8 · Answer

np :) the proof for bounded is straightforward :
show that any limit point in cl(A) is at fnite distance away from all other points in the set

loser66 · Answer

I prefer that 3rd answer on the page. It is neatly and no compute at all. Just logic.

ganeshie8 · Answer

me too

ganeshie8 · Answer

next you can appeal to Heine Borel theorem i think

loser66 · Answer

Sure, since it is a part of the whole , just combine the given condition and state that "to Heine Borel theorem,......."  to get the final conclusion. :)

ganeshie8 · Answer

i guess so, my analysis knowledge is rusty so don't trust me too much ;)