# Convergence of a Sequence Via Convergence of Subsequences

In $\mathbb{R}$, a sequence converges if and only if each of its subsequences converges.

I wonder if it is still true in a topological space? Does this have something to do with possible nonuniqueness of the limit? When can the statement be true?

Thanks and regards!

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The sequence $a_n=(-1)^n$ does not converge but it has two convergent subsequences. You might want to rephrase. –  Asaf Karagila Feb 19 '12 at 0:50
@AsafKaragila: Thanks! Isn't the sequence itself a subsequence of itself? –  Tim Feb 19 '12 at 0:53
I take it what OP means is a sequence converges if and only if every subsequence converges. –  Gerry Myerson Feb 19 '12 at 0:54
I guess this is one of times using "any" is bad... :-( –  Asaf Karagila Feb 19 '12 at 0:59

It’s true in any topological space $X$ that a sequence $\sigma$ converges if and only if every subsequence of $\sigma$ converges. One direction is trivial: if every subsequence of $\sigma$ converges, then obviously $\sigma$ converges, since it’s a subsequence of itself. To prove the other direction, suppose that $\sigma=\langle x_n:n\in\omega\rangle$ converges to $x$, and let $\langle x_{n_k}:k\in\omega\rangle$ be any subsequence of $\sigma$. Let $U$ be any open nbhd of $x$. Then there is an $m\in\omega$ such that $x_n\in U$ whenever $n\ge m$. But $n_k\ge k$ for each $k\in\omega$, so $n_k\ge m$ whenever $k\ge m$, and therefore $x_{n_k}\in U$ whenever $k\ge m$. It follows that $\langle x_{n_k}:k\in\omega\rangle$ converges to $x$.
The argument doesn’t depend on uniqueness of the limit. If $\tau$ is a subsequence of a convergent sequence $\sigma$, then $\tau$ converges to each point to which $\sigma$ converges. –  Brian M. Scott Feb 19 '12 at 2:38