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A metric space $(X,d)$ is compact iff every real valued continuous function on $X$ is bounded.

I have got the solution to it which I dont get at all.

Since in a metric space compactness is equivalent to sequential compactness let us assume that $x_n$ is a sequence in $X$ which has no convergent subsequence. So each term of the sequence will occur finite number of times.

Now if we form a subsequence of that sequence $x_n$ taking only the distinct terms and name it $A$ we can for each $n\in\mathbb N$ find $r_n$ such that $B(x_n,r_n)\cap A=\{x_n\}$.

I cant understand it from here:

A function is taken $f_n(x)=n[1-\frac{2d(x,x_n)}{r_n}] ;d(x,x_n)\leq r_n$ otherwise it is taken to be zero.

Why is such a function chosen and what is the motivation?

Please help here .What to do next?

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  • Show that this function is continuous but not bounded. Then by contrapositive, you have one direction of your theorem. – Plutoro Sep 13 '15 at 05:47
  • Write down what this construction gives you in the simple case $X=\Bbb R$ (with the usual metric) and $x_n=n$ for all $n$. You should be able to graph the relevant function, $\sum_{n=1}^\infty f_n(x)$, and get some intuition for why it was chosen. – Greg Martin Sep 13 '15 at 05:56

1 Answers1

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The idea is naturally to construct a function which is continuous, but not bounded.

I will modify your construction (which seems to work, but is harder to analyze as the balls can intersect).

So we start with finding a sequence of non-intersecting balls. For $x_n$ and $r_n$ as in your answer, $B_n = B(x_n,r_n/2)$ are guaranteed to be disjoint. Now we construct a continuous function $f_n$ such that $f(x_n) = n$ and $f_n = 0$ outside $B_n$. This is easy, it is in fact constructed in your question, just replace $d(x,x_n)\le r_n$ with $d(x,x_n)\le r_n/2$. Finally, take $f(x) = \sum_{n\ge 1} f_n(x) \mathbf{1}_{B_n} (x)$. Then $f$ is continuous but unbounded.

zhoraster
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