Prove that if there exists a subsequence of $a_{n}$ which converges to $L$ , then $L$ is a limit point of $a_{n}$. Let $(a_{n})_{n=0}^{\infty}$ be a sequence of real numbers, and let $L$ be a real number. Then the following statements are logically equivalent:
(a) $L$ is a limit point of $(a_{n})_{n=0}^{\infty}$.
(b) There exists a subsequence of $(a_{n})_{n=0}^{\infty}$ which converges to $L$.
MY ATTEMPT (EDIT)
I am mainly concerned with the implication $(b)\Rightarrow(a)$.
Let us start with the definition of convergent subsequence.
Let $\varepsilon > 0$. Then there exists a natural number $N\geq 0$ s.t.
\begin{align*}
f(n) \geq n \geq N \Rightarrow |a_{f(n)} - L| \leq \varepsilon
\end{align*}
where $f:\textbf{N}\to\textbf{N}$ is strictly increasing.
The definition of limit point which I was presented to is the following:
For every $\varepsilon > 0$ and every natural $N\geq 0$, there exists a natural $n\geq N$ such that $|a_{n} - L| \leq \varepsilon$.
But then I get stuck. Could someone help me with this?
 A: I am assuming you want to prove that $L$ is a limit point of the set $\{a_{n}\}_{n=0}^{\infty}$ which is actually the range of the function (called a sequence) that maps the natural number $n$ to the real number $a_{n}$. The definition of $L$ being a limit point of this set is that, for every $r>0$, there exists a real number $a_{n(r)}\in\{a_{n}\}_{n=0}^{\infty}\setminus\{L\}$ such that
\begin{equation} |L-a_{n(r)}|<r. \end{equation}
First of all, you need to assume (or it is probably written in the question) that $(a_{n})_{n=0}^{\infty}$ is not eventually constant because then its range would be finite and finite sets have no limit points (see this question). Assuming that $\{a_{n}\}_{n=0}^{\infty}$ is an infinite set, then consider the following argument.
Let $r>0$ be given and suppose that $(a_{n})_{n=0}^{\infty}$ has an infinite subsequence convergent to $L$. Choose $N$ such that $|L-a_{n_{k}}|<r$ for all $k\geq N$. If $a_{n_{N}}\neq L$, then you are done. If $a_{n_{N}}=L$, then choose the first $N'\geq N$ such that $a_{n_{N'}}\neq a_{n_{N}}$ (there is such an element because the subsequence is infinite). Then, $|L-a_{n_{N'}}|<r$ and $a_{n_{N'}}\neq a_{n_{N}}=L$ so you are done.
