Limit of a bounded increasing sequence

Question

Let $(a_n)_{n\in\mathbb{N}}$ an increasing and bounded sequence. Then $(a_n)_n$ is convergent.

I'm not sure if this proof is right: by hypothesis $(a_n)_n$ is increasing, so for the increasing sequences theorem we know that the limit of $(a_n)_n$ exists.

Let $l$ be the limit of $(a_n)_n$, by now we just know that $l\in\mathbb{R} \ \cup \ \{-\infty,+\infty\}$ so we can't say that the sequence is convergent yet; but by hypothesis $(a_n)_n$ is bounded, so exists $M\in\mathbb{R}$ such that $|a_n|\leq M$ for all $n\in\mathbb{N}$.

So we have that $-M\leq a_n \leq M$ and then, letting $n\to+\infty$ in the inequalities, we have that $$-M \leq \lim_{n\to +\infty} a_n \leq M$$ Hence $-M\leq l \leq M$, so $l\in\mathbb{R}$ and then $(a_n)_n$ is convergent.

Answer 1

You haven't really proven anything here. A valid argument would be something like this:

Since $\{a_n : n=1,2,\ldots\}$ is bounded above, $a:=\sup\{a_n:n=1,2,3,\ldots\}$ exists. Hence for any $\varepsilon>0$, we may choose a positive integer $N$ such that $a-a_N<\varepsilon$. Since $a_n$ is increasing, it follows that $a-a_n\leqslant a-a_N<\varepsilon$ for $n\geqslant N$, and therefore $\lim_{n\to\infty}a_n=a$.

Answer 2

Suppose: $k:\mathbb N\to\mathbb R$

Look at $(k_n)_{n\in\mathbb N}\in[-x,x]\subset\mathbb R$.

Choose an arbitrary closed subinterval $[a,b]\subseteq[-x,x]$ that contains almost all the terms (apart from finitely many of them).

We're dealing with nested closed intervals $[a_i,b_i],a_{i+1}\ge a_i\;\&\;b_{i+1}\le b_i,\;\forall n\in\mathbb N$. Now, what is the relation between$$[-x,x]\bigcap_{i=1}^{\infty}\;[a_i,b_i]$$ and $\lim k_{p(n)}$, where $k_{p(n)}$ is a subsequence of $k_n$?