2009 August « Thoughts on my Mind

A Fix on the Mistake August 17, 2009

Posted by putnam120 in Uncategorized.
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So here is one proof for the question from the previous post. In case you forgot I shall now restate it:

Assume that in a metric space $(X,d)$ you are given 2 closed disjoint sets $A$ and $B$ . Prove that if one of them is compact then $\inf d(a,b)>0$ with $a\in A$ and $b\in B$ .

proof: Without loss of generality (WLOG) assume that $A$ is compact. Then define $f:A\to{\mathbb{R}}$ as $f(a)=\inf d(a,B)$ . Otherwise, define it to be the infimum of the distance from the point in $A$ to a point in the set $B$ . Now $f$ is continuous and thus attains its infimum. But this must be positive, because if it were not we would have contradicted the fact that the sets were disjoint and closed.

This proof works, but I would also like to find one that relies on a set being compact if any open cover has a finite subcover. Sure it might be more work than necessary, but I feel that it should be good exercise.

Sum Function on Countable Set August 9, 2009

Posted by putnam120 in Uncategorized.
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Here is an interesting problem that I saw a few days ago. I don’t really know why I found this more interesting that other problems/solutions I’ve seen but it just is.

Here is the problem statement (generalized). Consider a function $f:X\to\mathbb{R}$ , and $f(x)\ge{0}$ , such that $f(x_1)+f(x_2)+\cdots+f(x_n)\le{1}$ for any sequence $\{x_n\}\in{X}$ . Prove that the set of points where $f$ is not zero is countable.

Proof:

Suppose for the sake of contradiction that the set is not countable, thus uncountable. Define $E_n=\{x\in{X}:f(x)>\frac{1}{n}\}$ . It should be obvious that $\displaystyle\bigcup_{n=1}^\infty E_n=X$ . So one of the $E_n$ must be uncountable. Thus if we choose points from this set (one of the uncountable sets) we can make $f(x_1)+f(x_2)+\cdots+f(x_n)$ as large as we want. Hence we must have that the set of points wehre $f(x)>0$ is at most countable.