# Thread: Closure and accumlation point problem

1. ## Closure and accumlation point problem

Suppose that the sequence $\displaystyle \{ x_n \}$ converges to a point x, prove that $\displaystyle x \in \ cl \{ x_1, x_2, x_3, . . . \}$

In addition, when would x be an accumlation point.

Proof so far.

So since $\displaystyle x_n \rightarrow x$, $\displaystyle \forall \epsilon > 0$, there exist $\displaystyle N \in \mathbb {N}$ such that whenever $\displaystyle n \geq N$, we would have $\displaystyle d(x,x_n) < \epsilon$.

Now, I need to show that this x is in the closure, which is the intersection of all closed set containing $\displaystyle \{ x_1, x_2 , . . . , \}$.

So my idea is, for any closed set that contains the sequence, suppose that x is not in them, then x is not in the set that the sequence is in, then but the completeness property, that is impossible.

And x would be an accumlation point if the set is closed.

Is this right?

Suppose that the sequence $\displaystyle \{ x_n \}$ converges to a point x, prove that $\displaystyle x \in \ cl \{ x_1, x_2, x_3, . . . \}$
In addition, when would x be an accumulation point?
• No finite set can have an accumulation point.
• If x is in the closure of S but not in S then x is an accumulation point of S.
• If x is an accumulation point of S the there is a sequence of distinct points of S converging to x.

3. If x is an accumulation point of S the there is a sequence of distinct points of S converging to x.

Why is this statement true?

Thanks.

If x is an accumulation point of S the there is a sequence of distinct points of S converging to x. Why is this statement true?
It is easy to construct such. If a is an accumulation point of S any open set containing a contains a point of S different from a. If we call it b then there is an open set containing a but not b. Thus we can get a new point c of S different from a in the second set. So by a inductive process we can construct the sequence.

Warning: That is a very loose outline on how we prove that statement.

5. Here is what I have:

Proof.

Suppose that the sequence $\displaystyle \{ x_n \}$ is a finite set, then $\displaystyle x \in \{ x_n \}$, implies that $\displaystyle x \in cl \{ x_1 , x_2,... \}$.

So suppose that $\displaystyle \{ x_n \}$ is not finite and that x is not included in the sequence.
Claim: x is an accumulation point of the sequence.

Let U be any open set that contains the point x, then there exist $\displaystyle \gamma > 0$ such that $\displaystyle D(x, \gamma ) \subset U$.

Now, since the sequence $\displaystyle \{ x_n \} \rightarrow x$, pick $\displaystyle \epsilon = \frac { \gamma }{2}$, then there exist an $\displaystyle N \in \mathbb {N}$ such that whenever $\displaystyle n \geq N$, we have $\displaystyle d( x_n , x ) < \epsilon < \gamma \ \ \ \ \ \forall n$.

So each points in the sequence is contained in the set U, which means x is an accumulation point of the sequence, implies that x is in its closure.

Is this right? Thanks.