An infinite sequence is contained in (a, b) but the limit is not in (a, b). Prove the limit is either a or b.

Question

This seems really straigtforward, but I'm struggling with rigorously explaining why any the sequence can't converge to anything other than a or b, i.e. anything outside this interval.

Answer 1

Note that $[a, b]$ is closed. Closed sets have the property that, if a sequence $(x_n)$ is contained in the closed set and is convergent, then the limit is also contained in the closed set.

So, if we have a sequence $(x_n)$ in $(a, b)$ that converges, then it is also contained in $[a, b]$, so the limit must be in $[a, b]$. By assumption, the limit is not in $(a, b)$, so it must lie in $[a, b] \setminus (a, b) = \lbrace a, b \rbrace$. That is, the limit is $a$ or $b$.

Answer 2

A limit of a sequence is a number so that every open neighborhood of the limit contains a tail of the sequence.

This implies for your instance that every neighborhood of the limit must contain terms from $(a,b)$, and thus meet $(a,b)$. But the limit is also not in $(a,b)$ and so the neighborhoods also meet points of the complement. Thus the limit point satisfies the definition of a boundary point. But $\partial\big( (a,b) \big) = \{ a,b\}$.

Answer 3

Hint (assuming that the sequence is assumed to in fact converge): If your sequence $\{x_n\}$ converges to some value $b'>b$, then for any $\epsilon>0$, there must exist some integer $n$ so that, for all $m>n$, $|x_m-b'|<\epsilon$. Can you find an $\epsilon$ so that this condition cannot hold for any $x_m$?