Prove that for a convergent sequence $(a_n)$ of real numbers $ inf(a_n) \le \lim_{n\to \infty} a_n = a \le sup(a_n)$ (not by contradiction)

Question

I was able to prove that by contradiction, showing that (for example) if $a > sup(a_n) = s$, then also $s$ is a limit of the sequence $(a_n)$.

I was wondering if it can be proved in some other way.

I actually proved it starting from the cases $a < a_n$ and $a \ge a_n$ for all $n$, but I think I cannot do that as $(a_n)$ has infinitely many elements and in some sense the argument is circular.

Answer 1

Here's a more general argument: If $a_n\le M$ for all $n$ and $a_n\to a$, then $a\le M$ as well. We'll give a direct proof.

Given any $\epsilon>0$, we know that there is an $N$ so that $|a_n-a|<\epsilon$ for all $n\ge N$. So this means that $-\epsilon<a_n-a<\epsilon$ for all $n\ge N$. In particular, we find that $$a-\epsilon<a_N\le M.$$ This tells us that $a<M+\epsilon$. But $\epsilon>0$ is arbitrary, and so $a\le M$.

Answer 2

Starting with

$\tag 1 \text{inf}(a_n) \le \text{sup}(a_n)$

and

$\tag 2 \lim_{n\to+\infty} a_n = a $

we need direct proofs of the following

$\tag 3 \text{inf}(a_n) \le a$

and

$\tag 4 a \le \text{sup}(a_n)$

$\text{(3)}$: Let $\alpha = \text{inf}(a_n)$ and $b_n = a_n-\gamma$. Once we directly prove that

$\tag 5 \lim_{n\to+\infty} b_n = a - \gamma$

it is easy to argue that $\text{(3)}$ is true.

$\text{(4)}$: Similar to finding a direct argument that $\text{(3)}$ is true.

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This should help.

If $(a_n)$ is any sequence such that for all $n$ $a_n \ge 0$ and $b \lt 0$ then the sequence $(a_n)$ can't converge to $b$.

Reason: For sufficiently small $\varepsilon \gt 0$ the open interval $(b-\varepsilon, b+\varepsilon)$ is contained in $(-\infty,0)$ but

$$\quad (-\infty,0) \cap \{a \, | \, a = a_n \text{ for some } n\}= \emptyset$$