Proof that if a complex sequence ($x_k$) converges to $a$ if and only if $a$ is an accumulation point of every subsequence of $x_k$ [duplicate]

Solution 1:

If you don't have $\lim_{k\to\infty}x_k=a$, then, for some $\varepsilon>0$, there are infinitely many $k$'s such that $|x_k-a|\geqslant\varepsilon$. Take a sequence $(x_{k_n})_{n\in\Bbb N}$ of such $x_k$'s, and that will give a subsequence of $(x_k)_{k\in\Bbb N}$ of which $a$ is not an accumulation point.

Solution 2:

If $\{x_n\}$ converges to $a$, then $\forall\varepsilon>0$, $\exists n_0\in\mathbb N$, such that $$ \forall n\ge n_0 \quad\Longrightarrow\quad |x_n-a|<\varepsilon. $$

If $\{x_n\}$ DOES NOT converge to $a$, then $\exists\varepsilon>0$, such that $\forall n_0\in\mathbb N$, $$ \exists n\ge n_0 \quad\text{\&}\quad |x_n-a|\ge\varepsilon. $$ This is equivalent to: There exists a $\varepsilon>0$, such that the set $$ S=\{n\in\mathbb N: |x_n-a|\ge \varepsilon\} $$ is infinite.

Since $S$ is infinite, then we can write the elements of $S$ as a strictly increasing sequence: $$ S=\{k_1<k_2<\cdots<k_n<k_{n+1}<\cdots\} $$ In particular, $k_1=\min S$ and recursively, $k_{n+1}=\min S\setminus\{k_1,\ldots,k_n\}$

Now the subsequence $\{x_{k_n}\}$ does not converge to $a$, since $$ |x_{k_n}-a|\ge\varepsilon, \quad\text{for all $n\in\mathbb N$}. $$