# Math Help - !Cluster Point <=> Convergent

1. ## !Cluster Point <=> Convergent

Suppose that the sequence $\{ x_n \}$ is in $\mathbb {R}$.

Prove that $\{ x_n \}$ converges if and only if it has a unique cluster point.

My proof:

Given $\epsilon > 0$, find the cluster point $x \in \mathbb {R}$ such that there exist infinite values n with $| x_n - x | < \epsilon$

I know that I need to find a $N \in \mathbb {N}$ such that whenever $n \geq N$, we have $| x_n - x | < \epsilon$.

But which N do I pick that so I can be sure that all the n afterward would get the sequence going into x and not going somewhere else? (In which I assume since the sequence only has one cluster point, the other point cannot go anywhere but x)

For the converse, let $\epsilon > 0$, find $N \in \mathbb {N}$ such that whenever $n \geq N$, we have $|x_n - x | < \epsilon$.

Then there exist infinitely values of n such that $| x_n - x | < \epsilon$, which means x is a cluster point.

Now, suppose to the contrary that there exist another point $y \in \mathbb {R}, y \neq x$ such that there exist infinitely values of n with $| x_n - y | < \epsilon$.

I want to find an N such that the sequence would be converging both to x and y, which is impossible, but how should I go about choosing it?

Thank you!

2. Use $\varepsilon = \frac{{\left| {x - y} \right|}}{4} > 0$ at both x & y.
The definition of convergence will fail.

3. So here is what I'm doing:

I want to show that, if a sequence in real numbers converges, then it has a unique cluster point.

Proof.

Suppose that the sequence $\{ x_n \} \rightarrow x$. Then by definition, $\forall \epsilon > 0$, there exist $N_1 \in \mathbb {N}$ such that whenever $n \geq N_1$, we have $| x_{n} - x | < \epsilon$. Then x is a cluster point of $\{ x_n \}$, since there exist an infinitely values of n such that $| x_n - x | < \epsilon$

Assume to the contrary that there exist another cluster point y.

Now, pick $\epsilon = \frac { | x- y | } {4}$, then I have $| x_n - x | < \frac { | x- y | } {4}$, and there exist $N_2 \in \mathbb {N}$ such that whenever $n \geq N_2$, I have $| x_n - y | < \epsilon = \frac { | x- y | } {4}$

4. Suppose that $\left( {x_n } \right) \to a\,,\,\left( {x_n } \right) \to b\,\& \,a \ne b$.
There if $\varepsilon = \frac{{\left| {a - b} \right|}}{4}$ then $\left( {\exists N} \right)\left[ {n \ge N \Rightarrow \left| {x_n - a} \right| < \varepsilon \,\& \,\left| {x_n - b} \right| < \varepsilon } \right]$.

Now for the contradiction: $\left| {a - b} \right| \le \left| {a - x_N } \right| + \left| {x_N - b} \right| < \varepsilon + \varepsilon < \left| {a - b} \right|$.

5. A question of definition: consider the series $\{ x_n\}$ where $x_n= 1/n$ if n is even, n if n is odd. What are the "cluster points" of that sequence?

6. I was wondering the same thing. limit points, cluster points, points of accumulation all seem to mean different things to different people.

I was under the impression that a cluster point is any point for which a subsequence converges to that point. Or equivalently I presume a point for which any open neighborhood about it contains an infinite number of points of the sequence (certainly in a metric space this has to be true).

If this is the case the $\Leftarrow$ direction is certainly false as your example demonstrates, that sequence has a subsequence converging to 0, but clearly does not converge as the sequence is not even bounded.

The forward direction is clear simply from noting the fact that $\mathbb{R}$ is Hausdorff and a sequence can converge to at most one point. Only that point is a cluster point, or point of accumulation, or any of those type of thing because you can be sure that ALL but a finite number of the points in the sequence are in any neighborhood of the limit point, and you can seperate this point from any other possible cluster point with a disjoint open set (this is by Hausdorff).

7. My interpretation: A set $A \subseteq (S, d)$ is said to cluster at a point $p \in S \Longleftrightarrow$ every ball $B_p$ about $p$ contains infinitely many points of $A$.

8. That is certainly how I would define "cluster point". Specifically, "infinity" is NOT a cluster point and so the example I gave has one cluster point but does not converge. So there appear to be at least three of us who say that the theorem the OP is trying to prove is NOT true!

Of course, it is easy to prove that if a sequence converges, then its limit is the only cluster point. But the converse is not true.