How to prove such a sequence exists

**paupsers** · September 6th 2011, 07:49 AM

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...

**Plato** · September 6th 2011, 08:11 AM

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.

**paupsers** · September 6th 2011, 08:49 AM

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. :-)

Thread: How to prove such a sequence exists

LinkBack

Thread Tools

Display

How to prove such a sequence exists

Re: How to prove such a sequence exists

Re: How to prove such a sequence exists

Similar Math Help Forum Discussions

Prove that exists m among ai whose sum is at least m

Show that there exists no sequence of functions satisfying the following

Prove the limit of this sequence exists

[SOLVED] How to prove that a limit exists?

limit, sequence, exists

Search Tags