Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that a closed subset of a complete normed vector space is complete.

My attempt,

Let S be a subset of $(V,\parallel . \parallel)$. Since S is closed, for all $x\in S$ there exists $\langle X_n \rangle_{n=1}^{\infty} \in S$ that converges to x. Therefore, $\forall \epsilon \gt 0 $ there exists $N \gt 0$ such that $\parallel x_n - x\parallel \lt \epsilon , n \ge N$. Therefore, every point in S corresponds to a bounded sequence in S, and since every bounded sequence has a convergent subsequence, every convergent sequence in S corresponds to a point in S.

share|improve this question
    
You should look up what "closed" and "complete" actually mean. –  Chris Eagle Feb 13 '13 at 16:39
    
@bobdylan Please reject my answer –  El Angel Exterminador Feb 14 '13 at 5:53

2 Answers 2

up vote 1 down vote accepted

Your proof is wrong. The fact that every convergent sequence has a limit does not imply that every Cauchy sequence has a limit. Furthermore it is not true that every bounded sequence has a convergent subsequence, this is true for $\mathbb R$, and some metric spaces, but not all.

For example take the unit sphere in an infinite dimensional vector space, and find a sequence of isolated points. The sequence is bounded, but it has no convergent subsequence.

The proof, however, is not very different. Suppose that $S$ is a closed subset of $V$, and let $\{x_n\}$ be a Cauchy sequence in $S$. Since $V$ is complete there is some $x\in V$ such that $x$ is the limit of $x_n$.

We want to show that $x\in S$. For every $\varepsilon>0$ there is some $N$ such that for all $n>N$, $\|x-x_n\|<\varepsilon$, and therefore for every $\varepsilon>0$ we can find some point $x'$ from $S$ such that $\|x-x'\|<\varepsilon$, and since $S$ is closed this means that $x\in S$, as wanted.

share|improve this answer

Let $x_n$ be a Cauchy sequence in $S$ closed.

Since $V$ is complete, $x_n$ convegerges to $x$ in $V$.

Since $S$ is closed, $x$ belong to $S$.

So $x_n$ converges to $x$ in $S$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.