-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.