Thread: Closed Cauchy sequence proof

A set F being a subset of Reals is closed iff every Cauchy sequence contained in F has a limit that is also an element of F.

Let F be closed. Then F contains its limit points.
This means x=lim an are elements of F.

I know a sequence is Cauchy if abs value(am-an)<epsilon

I just am confused on where to go with this one.

I find to OP post to be equivalent to ‘Prove that a closed set is closed’.
What is there to prove there?
In the real numbers every Cauchy sequence has a unique limit. Every closed set contains all its limit points. So what is there to prove?

Ok that helps understanding wise, but i guess the formal proof is what is getting me.

Every closed set contains all of its limit points. Then there is a a neighborhood(x) such that the intersection of F is not equal to {x}. I guess I don't see how to get to the Cauchy sequence from there.

Other direction: Let an be cauchy sequence such that an has a limit contained in F.
an is a Cauchy sequence if abs value(an-am)<epsilon
A limit exists for an if abs value(an-a)<epsilon
we can rewrite an-am as (an-a)+(a-am)
Then by the triangle inequality, abs value(an-am)<abs value(an-a)+abs value(a-am)
I'm stuck going from there to the set being closed

Ok I feel like I understand the 2nd direction after thinking about it, but I don't see how to do the first part. I don't see how to we get from F being closed to every Cauchy sequence contained in F having a limit that is also an element of F. I just don't see how the whole idea of Cauchy sequences come in.

Thinking about it more, we have some neighborhood of limit points such that (x-epsilon,x+epsilon) intersect F is not equal to x.