Help with the proof that a sequence is convergent iff it is bounded and has a single subsequential limit.

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!

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