0568 #62

Forum for the GRE subject test in mathematics.
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Joined: Sun Jul 19, 2009 12:29 pm

0568 #62

Post by breezeintopl » Thu Nov 05, 2009 8:49 pm

there is one answer http://www.mathematicsgre.com/viewtopic ... p=287#p287

but i don't think it is right:

"for (2), K is bounded is obvious ( let f: K------->K with f(x)=x, them K is bounded)

Now prove that K is closed. Suppose that K is not closed, then R^n\K is not open : since K is bounded so K must have form :
so let F : K--------> R F(x1,x2,....,xn)=[1/(x1-a)]x2....xn then f is continuous and f is not bounded - contradiction

so K <R^n is closed and bounded then K is compact"

Why a bounded but not open set can have the form (a1,b1)x(a2,b2]x....x(an,bn)?
I think open set is really different from open interval,although we have the open set construct theorem,ie every open set are the combination of some countable open intervals. But it is conuntable,not n(finite).

Is there any other proof?

thank you ~~

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Joined: Tue Jul 19, 2011 6:49 pm

Re: 0568 #62

Post by flubadub » Tue Jul 19, 2011 7:33 pm

How about this?

Let $$x_n$$ be a sequence in $$K$$ such that $$x_n \rightarrow a \in \mathbb{R}^n$$.

Then suppose that a is not in K, i.e. $$a \in \mathbb{R}^n \setminus K$$.

Then define a map $$f:K\rightarrow \mathbb{R}$$ by $$f(x)=\frac{1}{|x-a|}$$. Then this is continuous since the denominator will never equal zero (since a is not in K). But this function is unbounded since as$$x_n \rightarrow a, f(x_n) \rightarrow \infty$$. So this function can be made arbitrarily large on the sequence values and is therefore unbounded, so we have a contradiction, i.e. every convergent sequence in K must have its limit in K and so K is closed.

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