# How to prove such a sequence exists

• Sep 6th 2011, 07:49 AM
paupsers
How to prove such a sequence exists
Let A be a non-empty subset of Y, and let $x\in y$.
If $d(x,A)=0$
then there is a sequence $x_n$ of points in A such that $x_n$ converges to x.

Can anyone give me a starting point here? Should I try to prove this by contradiction? I'm just unsure of how to prove a sequence exists...
• Sep 6th 2011, 08:11 AM
Plato
Re: How to prove such a sequence exists
Quote:

Originally Posted by paupsers
Let A be a non-empty subset of Y, and let $x\in y$.
If $d(x,A)=0$
then there is a sequence $x_n$ of points in A such that $x_n$ converges to x.

If $x\in A$ what sequence would work? (note the terms do not have to be distinct)

If $x\notin A$ then $\left( {\exists n \in \mathbb{Z}^ + } \right)\left( {\exists x_n \in A} \right)\left[ {d(x,x_n ) < \frac{1}{n}} \right]$.
Now you need to explain why that is true.
• Sep 6th 2011, 08:49 AM
paupsers
Re: How to prove such a sequence exists
Thanks... I had actually found a proof that works right after I posted this! I was coming back to say that but you had already commented. :-)