# Showing Convergent Subsequence Exists

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1. Feb 5, 2017

### transmini

1. The problem statement, all variables and given/known data
Consider the space $([0, 1], d_1)$ where $d_1(x, y) = |x-y|$. Show that there exists a sequence $(x_n)$ in $X$ such that for every $x \epsilon [0, 1]$ there exists a subsequence $(x_{n_k})$ such that $\lim{k\to\infty}\space x_{n_k} = x$.

2. Relevant equations
N/A

3. The attempt at a solution
We had no idea where to even begin with this one. It doesn't seem to make sense in the first place when you take the theorem:

Let $(x_n)$ be sequence, and let $x \epsilon X$. Prove that if $\lim{n\to\infty}\space x_n = x$, then for every subsequence $(x_{n_k})$ we have $\lim{k\to\infty} \space x_{n_k} = x$

With the problem saying that there is specific sequence such that for any limit point we can find a subsequence, this implies there a multiple subsequences with different limits, which contradicts what this previous theorem says.

Any suggestions on where to begin with this one?

2. Feb 5, 2017

### Ray Vickson

There are no contradictions in the problem statement. The main sequence $\{x_n\}$ does not have a limit; it has numerous limit points, meaning that it has numerous subsequences converging to their own, individual limits. The problem asserts that there is a sequence such that any $x \in [0,1]$ is a limit of some subsequence. Believe it or not, there is a very easy solution.