Consider a sequence $\{x_n\}$ in $S$. Where $S$ is a metric subspace. Given that every convergent subsequence of $\{x_{k(n)}\}$ converges to the same point say $x$. Prove that if S is compact, show that $\{x_n\}$ converges to $x$. Is my answer correct?
Proving the sequence is convergent
4
$\begingroup$
sequences-and-series
convergence-divergence
-
1@Mathematics: I think you should edit your question, because it is otherwise misleading in my opinion. – 2012-11-04
2 Answers
4
I'll assume we're in a metric space with metric $d$.
Suppose $\{x_n\}$ does not converge to $x$. Then $\exists \, \epsilon > 0$ and a sequence of positive integers $m_1,m_2,\ldots$ such that $d(x, x_{m_i})\geq \epsilon$ for all positive integers $i$. The sequence $\{ x_{m_i}\}$ has a convergent subsequence because $S$ is compact. This subsequence converges to $x$, but that's a contradiction because no term in this subsequence is within $\epsilon$ of $x$.
-
0I think it should be a metric space – 2012-11-04
1
Every subsequent converges to the same point if S is compact and hence every subsequence are converges and converge to the same points and hence the sequence also convergent to x. Is the proof valid?
-
0No. You really do need to use the argument that littleO gave and I repeated in an earlier comment. – 2012-11-05