Show that every subsequence converging to the same limit implies limit of sequence exists. I need to prove that if $\exists$ subsequences $a_{n_{k_{1}}}$ and $a_{n_{k_{2}}}$ (of $a_{n}$) that converge to different limits, then the sequence $a_{n}$ does not converge. I'm not sure how to do this. If I suppose $a_{n_{k_{1}}}$ converges to some limit $l_{1}$, then $\forall \epsilon >0$, $\exists n_{k_{1}}>N$ s.t. $|a_{n_{k_{2}}}-l_{1}|<\epsilon$, and the same thing for $a_{n_{k_{2}}}$ and its limit $l_{2}$. 
$a_{n}$ not converging means $\exists \epsilon_{0}$ s.t. $\forall N\in \mathbb{N}$, $\exists n_{k}\geq N$ s.t. $|a_{n}-l|\geq \epsilon_{0}$. 
But, I'm not sure how to put them together in order to prove the implication that I want.
 A: Another way:
Suppose
$(a_{j(i)})$
and
$(a_{k(i)})$
are two subsequences of
$(a_n)$
that converge to different limits.
Informally:
If the original sequence
converged to a limit,
this limit would have to be
close to both of the
limits of the subsequences,
which is impossible.
Here's a more formal way to
do this:
Let
$a_n \to L$,
$(a_{j(i)}) \to L_j$
and
$(a_{k(i)}) \to L_k$
where $L_j \ne L_k
$.
For any $c > 0$,
there is a $N(c)$
such that
$|a_n-L| < c$,
$|a_{j(i)})-L_j| < c$
and
$|a_{k(i)})-L_k| < c$
for all
$n, j(i), k(i) > N(c)
$.
Then
$|a_{j(i)}-L| < c$,
$|a_{j(i)}-L_j| < c$,
$|a_{k(i)}-L| < c$,
and
$|a_{k(i)}-L_k| < c$.
Therefore
$\begin{array}\\
|a_{j(i)}-a_{k(i)}|
&=|a_{j(i)}-L+L-a_{k(i)}|\\
&\le |a_{j(i)}-L|+|L-a_{k(i)}|\\
&< 2c
\end{array}
$
so that
$\begin{array}\\
|L_j-L_k|
&=|L_j-a_{j(i)}+a_{j(i)}-a_{k(i)}+a_{k(i)}-L_k|\\
&\le|L_j-a_{j(i)}|+|a_{j(i)}-a_{k(i)}|+|a_{k(i)}-L_k|\\
&\le 4c\\
\end{array}
$
But if we choose
$c < |L_j-L_k|/4$
we have a contradiction.
Therefore two subsequences
of a convergent sequence
can not converge
to different limits.
A: Suppose $a_n$ does converge to some value $\alpha$.
This means that $\forall \epsilon>0 \exists N>0$ s.t $\forall n >N$: $|a_n - \alpha| < \epsilon$.
Let $a_{n_k}$ be a subsequence of our original $a_n$. Since both sequences are infinite, and $N$ is a finite number, we can infer that there is some $\beta$ such that $\forall i > \beta$: $a_{n_i} \in \{a_n,a_{n+1},...\}$ where $n>N$.
we already know that every element in the set $\{a_n,a_{n+1},...\}$ agrees with $|a_i-\alpha|<\epsilon$, and that $a_{n_i} \in \{a_n,a_{n+1},...\}$, so in particular, $|a_{n_i} - \alpha| < \epsilon$.
So overall: for all $\epsilon >0 \exists \beta>0$ s.t $\forall i>\beta$: $|a_{n_i} - \alpha|<\epsilon$, so $\alpha$ is the limit of $a_{n_k}$.
Result: if $a_n$ converges to $\alpha$, then ANY infinite subsequence of $a_n$ also converges to $\alpha$.
So if two subsequences are converging to different limits, then that means the entire sequence does not converge.
A: Suppose $a_{n_{k_1}}\to a$ and $a_{n_{k_2}}\to b$, and $a<b$. Then $$\liminf_{n\to\infty}a_n \leqslant a < b \leqslant \limsup_{n\to\infty}a_n, $$
so $\lim_{n\to\infty} a_n$ does not exist. 
