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I have a proof of the following theorem.

Let X be a metric space. "Every sequence in X has a Cauchy subsequence" implies that "$\forall\epsilon > 0, \exists $ a finite set T, s.t. $\forall x\in X, d(x,T)<\epsilon $."

Could you check that my proof is correct or not?

Proof) Suppose, by way of contradiction, $\exists\epsilon > 0,$ s.t. $\forall $ finite set T, $\exists x\in X, d(x,T)\geq\epsilon.$

pick any $x_1\in X.$

Let $T_1=\{x_1\}.$

Then, $\exists{\epsilon}_1 > 0,$ s.t. $\exists x_2\in X, d(x_2,T_1)\geq{\epsilon}_1.$

Let $T_2=\{x_1, x_2\}.$

Then, $\exists{\epsilon}_2 > 0,$ s.t. $\exists x_3\in X, d(x_3,T_2)\geq{\epsilon}_2.$

Let $T_3=\{x_1, x_2, x_3\}.$

Repeat this process to obtain a sequence ${x_i}$ in X.

Then this sequence doesn't have a cauchy subsequence, since $\exists\epsilon > 0$ for any i,j in N (the set of natural numers), $d(x_i,x_j)\geq\epsilon $, by letting $\epsilon$ = max $\{ {\epsilon}_1, {\epsilon}_2, \dots \}.$

Therefore, we have obtained a contradiction.

At the very last proof of the theorem, I'm not sure I could take the maximum to get the desired epsilon.. please help.

yhk
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1 Answers1

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you don't need different epsilons. negation of the statement say there's one epsilon that works for all sets - you've written it down yourself. the structure of the proof is ok other than that, you just continue picking $x_n$ such that $d(x_n, T_{n-1}) > \epsilon$ and the $\epsilon$ is fixed for all steps. if you wanted different $\epsilon$'s you'd actually need to take $\min$ of them which could turn out to be $0$ in case of infinite set, so that wouldn't work, you need to use your assumption of one epsilon working for all sets

mm-aops
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