Let X be a metric space, and let (x$_n$) be a sequence in X that has no convergent subsequences. Consider a subset A = {x$_n$ : n $\in$ N} of X.

a). Show that A'= $\emptyset$, where A' is the set of all limit point of A

I've been trying to figure out this problem for a while now, here's what I'm trying

Let A' be the set of limit points of A and let a $\in$ A'. Then, $\exists$ a sequence (x$_n$) in A where x$_n$ $\neq$ a for all n $\ge$ 0 such that x$_n$ $\to$ a.

Also I know since x$_n$ has no convergent subsequences then x$_n$ is not bounded

I was trying to think of a way to show that no such a exists. I believe my main fault is that I'm having trouble using the fact that X is a metric space in this proof, I don't really understand what a metric space by this definition of it,

A metric space is a set X that has a notion of the distance d(x, y) between every pair of points x, y ∈ X.

Any advice would be appreciated.

  • $\begingroup$ What is your definition of a convergent subsequence and what is your definition of a limit point. If they are the usual this seems like it is straight forward. If $a$ is a limit point of $A$ then for each open ball $B(a,\epsilon)$ there is some $x_n\neq a$ in the ball $B(a,\epsilon)$. Now to get your convergent subsequence just take $\epsilon_i=d(a,x_{n_i})$ and iterate. $\endgroup$ – DRF Apr 13 '17 at 5:57
  • $\begingroup$ Welcome to MSE. The question's title is too long! Please make it shorter :) $\endgroup$ – Arman Malekzadeh Apr 13 '17 at 6:16
  • $\begingroup$ Thank you so much for your advice, I really appreciate it :) $\endgroup$ – Niko L Apr 14 '17 at 20:46

Good start! Assume there is an $ a\in A' $, then for every ball $ B(a,1/j) $ for each $ j \in \mathbb{N} $ we can pick an $ t_j\in B(a,1/j) $. By construction this is a subsequence of $ \{x_n\}_{n\in\mathbb{N}} $ which converges to $ a $ which is a contradiction.

  • $\begingroup$ Thank you so much, this helped me understand the proof and construct it $\endgroup$ – Niko L Apr 14 '17 at 20:46

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