Showing Convergent Subsequence Exists

[transmini]

Homework Statement

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 $\lim{k\to\infty}\space x_{n_k} = x$.

Homework Equations

N/A

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?

 

[Ray Vickson]

Homework Statement

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 $\lim{k\to\infty}\space x_{n_k} = x$.

Homework Equations

N/A

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?

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.