Show that every subsequence converging to the same limit implies limit of sequence exists.

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.