Showing that a limit point $L$ of $(x^{(n)})^∞_{n=m}$ and a subsequence of $(x^{(n)})^∞_{n=m}$ which converges to L are equal

Question

Let $(x^{(n)})^∞_{n=m}$ be a sequence of points in a metric space $(X, d)$, and let $L ∈ X$. Then the following are equivalent:

$1$. $L$ is a limit point of $(x^{(n)})^∞_{n=m}$.

$2$. There exists a subsequence $(x^{(n_j)})^∞_{j=1}$ of the original sequence $(x^{(n)})^∞_{n=m}$ which converges to $L$.

Definitions:

Let $(x^{(n)})^∞_{n=m}$ be a sequence in $(X,d)$

$(a)$ We call the sequence $(x^{(n_j)})^∞_{j=1}$ a subsequence of $(x^{(n)})^∞_{n=m}$

$(b)$ Let $L\in X,$ we say that $L$ is a limit point of $(x^{(n)})^∞_{n=m}$ if and only if $\forall N\ge m$ and $\forall \epsilon>0$, $\exists n\ge N $ such that $d((x^{(n)}, L) \le \epsilon$

Question $1$: In order to show that the two are equivalent, do I need to show that $1\implies2$ and $2\implies 1$, or is one direction sufficient?

My attempt so far:

$1\implies2$

Assume that $L$ is a limit point of $(x^{(n)})^∞_{n=m}$

Then $\forall N\ge m$ and $\forall \epsilon>0$, $\exists n\ge N $ such that $d((x^{(n)}, L) \le \epsilon$

If there is a certain number $L$ which results in $d((x^{(n)}, L) \le \epsilon$ then our sequence converges in our space $(X,d)$

If our sequence converges in $(X,d)$ then by definition every subsequence $(x^{(n_j)})^∞_{j=1}$ of $(x^{(n)})^∞_{n=m}$ also converges to $L$

Therefore there does exist a subsequence $(x^{(n_j)})^∞_{j=1}$ of the original sequence $(x^{(n)})^∞_{n=m}$ which converges to $L$.

Question $2$: Am I on the correct path at all? And if not, can I please get some help?

Thank you!

Answer 1

"If there is a certain number $L$ which results in $d(x^{(n)},L)\leq \epsilon$ then our sequence converges in our space $(X,d)$". That would only be true if it were true for all $n$ big enough. But if that were true then every subsequence, the sequence itself include, would converge to $L$. So you haven't actually proven anything.

By the definition you give, if $L$ is a limit point then no matter how far along the sequence (big $N$), we can always find a later point within $\epsilon$ of $L$. It does not say that every later point is within $\epsilon$ of $L$, which is what you appear to be assuming.

Proof 1 -> 2 Fix some sequence $\{\epsilon_j\}$, with $\epsilon_j > 0$. By the definition of limit point there exists some $n_1> m$ for which $d(x^{(n_1)}, L) <\epsilon_1$. Now defining inductively: there exists some $n_{j+1} > n_j$ so that $d(x^{(n_{j+1})},L)<\epsilon_{j+1}$.

If we require that the sequence $\{\epsilon_j\}$ also satisfies $\epsilon_j \to 0$ then the resulting subsequence $\{x^{(n_j)}\}$ has the property $$0 \leq \lim_{j\to\infty} d(x^{(n_j)}, L) \leq \lim_{j\to\infty} \epsilon_j = 0$$ Hence the subsequence converges to $L$.

2->1 You will have to prove this case to conclude equivalence. It shouldn't be too hard to see why a subsequence that converges to $L$ allows the sequence to satisfy the definition of limit point that you have given.