I want to prove that if $x$ is a point of convergence, then it is also a cluster point. By exhibiting the proof, I want to understand where the proof gets stuck the other way around, i.e., that a cluster point implies a convergence point.
To get the definitions clear, given a sequence $x_n(n\in\mathbb{N})$:
$x$ is a point of convergence if $\forall \epsilon>0$ $\exists N \in \mathbb{N}$ $\forall n\geq N (|x-x_n|<\epsilon)$.
$x$ is an cluster point if $\forall \epsilon>0$ $\forall N\in \mathbb{N}$ $\exists n\geq N$ ($|x_n - x|<\epsilon)$.
Showing that convergence $\Rightarrow$ cluster is straightforward(?):
Proof:
- ($\Rightarrow$) Let $\epsilon>0$ and $N\in \mathbb{N}$ be arbitrary. We need to construct an $n\geq N$ such that $|x-x_n|<\epsilon$. Our hypothesis of convergence allows us to assume some $\tilde{N}$ such that for all $n\geq\tilde{N}$: $|x-x_n|<\epsilon$. We can construct the necessary $n$ by choosing $n:=\max(N,\tilde{N})$. In this case, obviously, $n\geq N$, and $n\geq \tilde{N}$, concluding the proof.
- ($\Leftarrow$) I need some guidence as to why this way of the proof ought to get stuck. I assume some $\epsilon >0$ and I need to construct some $N=N(\epsilon)\in \mathbb{N}$ such that $\forall n\geq N(|x_n - x|<\epsilon)$. In my hypothesis it is true that $\forall N\in \mathbb{N} \ \exists n\geq N(|x-x_n|<\epsilon.)$ Any $N\in\mathbb{N}$ works, and I suspect the proof gets stuck in the fact that I do not operationally construct $N=N(\epsilon)$.
If the proof did not get stuck, then convergence point and cluster point would be equivalent concepts, which they are not; since a cluster point is a weaker notion than a convergence point.
The forward direction is indeed trivial: let $x$ be a convergence point (aka limit) of the sequence and let $\varepsilon>0$ and $N$ be given. Let $N_1$ be the promised natural number for this $\varepsilon$ and indeed define $n=\max(N,N_1) \ge N$. Then $|x-x_n| < \varepsilon$ as $n \ge N_1$. So we are done and $x$ is a cluster point.
The reverse is false: the sequence defined by $x_n=0$ for $n$ even and $x_n=1$ for $n$ odd has $0$ and $1$ as cluster points but it has no convergence points. The best you can do in this case is to show that a cluster point of $(x_n)$ is a convergence point of a subsequence of $(x_n)$; this fact holds in any metric space.