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Let $\{x_n\}$ denote a sequence for $n \in \Bbb N$. Prove that $\{x_n\}$ is convergent if and only if it is bounded and has a single subsequential limit.


Let $P$ denote "a sequence is convergent" and $Q$ denote "is bounded and has a single subsequential limit". We want to prove two things: $$ P\implies Q \tag1 $$ $$ Q \implies P \tag2 $$


$\Box$ Start with $(1)$. Bu definition of a limit: $$ \lim_{n\to\infty} x_n = a \stackrel{\text{def}}{\iff} \forall \epsilon>0\ \exists N_\epsilon \in \Bbb N:\forall n > N_\epsilon \implies |x_n - a| < \epsilon $$ Take for instance $\epsilon = 1$: $$ |x_n - a| < \epsilon, \forall n > N_\epsilon $$

But also: $$ |x_n| - |a| < \epsilon \iff |x_n| < |a| + \epsilon $$

Now take some $M$: $$ M = \max\{1+|a|, x_1, x_2, \dots, x_N\} $$

That would give: $$ \forall n \in \Bbb N: |x_n| \le M $$

Which proves the sequence must be bounded. We also need to show that it has a uniq limit, which is pretty straightforward by contradiction (not posting it here) $\Box$


The second part is where I got stuck. I need to somehow show that $Q\implies P$, but not sure if the below is correct. Here are some thoughts.

$\Box$ Suppose that there is a single subsequential limit. Denote the subsequence as $\{x_{n_k}\}$: $$ \lim_{{n_k}\to \infty} x_{n_k} = L $$

We need to show that: $$ \lim_{n\to\infty} x_n = L $$

Suppose the opposite: $$ \lim_{n\to\infty}x_n = C \ne L $$

We know that $x_n$ is bounded, therefore there are such $a$ and $b$ that all the terms of $x_n \in [a, b]$.

Split $[a, b]$ into two equal parts. At least one of them contains infinitely many terms of $x_n$. Denote this interval as $[a_1, b_1]$. Let $x_{n_1} \in [a_1, b_1]$.

Split $[a_1, b_1]$ into two parts. One of them contains infinitely many terms. Denote this interval $[a_2, b_2]$. Chose $x_{n_2} \in [a_2, b_2]$ such that $n_2 > n_1$. Repeating that infinitely many times we have a set of intervals $[a_k ,b_k]$ and a set of terms from $x_n$ such that $x_{n_k}\in [a_k, b_k]$ and $n_{k_1} > n_{k_2}$ when $k_1 > k_2$.

From the above if follows that $x_{n_k}$ satisfies the definition of a subsequence of $x_n$. Also: $$ b_k - a_k = \frac{b-a}{2^k} \to 0\ \text{as} \ k \to \infty $$

Which means there exist at least one point $C$ that belongs to all intervals. Taking the limit and using squeeze theorem for $a_k < b_k$ we get: $$ \lim_{k\to\infty}a_k = \lim_{k\to\infty}b_k = C \\ a_k \le x_{n_k} \le b_k $$

And thus: $$ \lim_{k\to\infty}a_k = \lim_{k\to\infty}x_{n_k} = \lim_{k\to\infty}b_k = C $$

Which contradicts the assumption that only one subsequential limit exists. Hence indeed: $$ Q \implies P $$ $\Box$

I would appreciate if someone could approve/reject the reasoning above. Thank you!

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  • $\begingroup$ What you call the opposite isn't the opposite at all. $\endgroup$ – Michael Hoppe Dec 11 '18 at 20:09
  • $\begingroup$ @MichaelHoppe Well, then i believe the whole argument for at least $(2)$ is wrong $\endgroup$ – roman Dec 11 '18 at 20:11
  • $\begingroup$ Does the single substantial limit mean that all the subsequences of $a_n$ are convergent to one limit? $\endgroup$ – Mostafa Ayaz Dec 11 '18 at 23:02
  • $\begingroup$ @MostafaAyaz yes, that’s true $\endgroup$ – roman Dec 12 '18 at 6:01

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