# Analysis Limit Point Proof

1. Oct 18, 2009

### dancergirlie

1. The problem statement, all variables and given/known data
Let x $$\in$$ R, and let A $$\subset$$ R. Let (an) be a sequence with an $$\in$$ A and (an) $$\neq$$ x for all n $$\in$$ N, and assume
that x = lim (an.) Prove that x is a limit point of A.

2. Relevant equations

3. The attempt at a solution
Suppose that x=lim(an). Meaning there exists a N $$\in$$ N so that for n$$\geq$$N:
|an-x|<$$\epsilon$$

Which would make the $$\epsilon$$- neighborhood of (an)= (an- $$\epsilon$$, an+ $$\epsilon$$)

After this I don't know what to do. I need to show that every epsilon neighborhood of V(x) intersects A in some point other than x.

Any tips would be great!

2. Oct 18, 2009

### snipez90

Careful, your e-neighborhood is off. Remember |a_n - x| < e means x - e < a_n < x + e or a_n is in (x - e, x + e). Now you know that a_n =/= x for each n, so what can you conclude.

Of course this proposition is an if and only if, but this direction is probably clearer since "all but a finitely many points" certainly implies "at least one".

3. Oct 18, 2009

### dancergirlie

alright so if the epsilon neighborhood is (x-e,x+e) and (an) is unequal to x for all n in N would that mean that for some N in N when n>=N that (an) is contained in the epsilon neighborhood. Which would mean that x is the limit point of (an)?

4. Oct 18, 2009

### dancergirlie

oh wait, nevermind, i think I got it. I was trying to make it more complicated than it really was. I am finding this whole open sets, closed sets and limit points idea kind of confusing.