Proposition: Let $(a_n)^\infty_0$ be a sequence of real numbers, and let $L$ be a real number. Then the following two statements are logically equivalent:
(a) $L$ is a limit point of $(a_n)^\infty_0$
(b) There exists a subsequence of $(a_n)^\infty_0$ which converges to $L$.
I am trying to show that (b) implies (a). I am confused how to relate the different epsilons and Ns together to serve the proof:
Proof:
Let $(b_n)^\infty_0$ be a subsequence that converges to $L$:
$\forall \epsilon> 0$ there exists an $N$ such that $$|b_n − L|\leq\epsilon,\forall n\geq N$$
For $L$ to be a limit point of the sequence $(a_n)^\infty_0$, we need:
$$\forall \epsilon >0, \forall N\geq0, \exists n\geq N,\text{ such that }|a_n − L| ≤\epsilon$$
I am confused how to continue. Should I fix an epsilon and relate the Ns?
Best Answer
Let $X \subset \mathbb{R}$ and $L \in \mathbb{R}$.
$(b)\implies (a)$
Consider $\lim x_n = L$, where $x_n \in X - \{L\}$. For every $n_0 \in \mathbb{N}$ the set $\{x_n ; n> n_0\}$ is infinite, because if it wasn't there would be a $x_{n_1}$ repeating itself infinite times and this would give us a constant sequence such that $\lim x_{n_1} \neq L$. Then by definition of limit we have that $(L-\epsilon, L + \epsilon)$ centered at $L$ has an infinity points of $X$, so $L$ is a limit point of $X$, that is $(L-\epsilon, L + \epsilon) \cap X - \{L\} \neq \emptyset$.