Limit of a sequence inside a set Let $A$ be a bounded set, and let $(a_n)_{n=1}^{\infty}$ be a convergent sequence, with a limit $L$, such that $a_n\in A$ for all $n$. I want to prove that $\inf{A}\leq L\leq\sup{A}$.
Intuitively, it is clear - since $a_n$ converges, I can find elements arbitrarily close to $L$.
Since $a_n$ is convergent, then for all $\varepsilon > 0$, for some $N\in\mathbb{N}$, for all $n>N$, $|a_n-L| < \varepsilon$. Suppose $L < \inf{A}$. Then let $\varepsilon = \inf{A} - L>0$, and then let $N$ be such that for all $n>N$, $|a_n - L| < \inf{A} - L$.
Then $L-\inf{A} < a_n- L < \inf{A} - L$, and from this, we get that $a_n < \inf{A}$, a contradiction.
Should I care about the left inequality? Is the proof for $\sup{A}$ the same, just flip $|a_n-L|\leftrightarrow |L-a_n|$?

Your proof is correct - essentially what you're saying is "The limit is $\varepsilon$ away from  $\inf A$. My terms are eventually all close to $L$, so in particularly they must be within $\varepsilon$ of it, and therefore far away from $\inf A$". The $\sup$ argument is essentially the same.   
You would have used the left inequality if you'd written $|a_{n}-L|$ as $|L-a_{n}|$ (they are the same). I'm confused by your comment at the end - why would you swap those terms round?
