@FortuonPaendrag I know, but proving that is maybe tougher than using the criteria on monotone bounded sequences.
Since the completeness of $\Bbb R$ is axiomatic, but continuity isn't
It obivously seems harder to resort to my solution, but from a mathematical point of view, the tools used are more basic, and one is proceeding with care.
@PeterTamaroff That's a question that is on topic for MSE. Like Arturo (see his profile), generally I cannot accept direct questions. If I did, I'd be so swamped that I could not achieve my teaching goals. Generally you'll get better more well-rounded answers by posting questions to MSE rather than by directly contacting individual members. Attempting to short-circuit that process makes the site functon less optimally.
@MattN Yes, just now. The answer is that a nbhd of a point $p$ is any set that contains $p$ in its interior, so if you want to talk about an open nbhd of $p$, you have to say so.
@MattN When you say ‘Let $U$ be an open nbhd of $p$’, you’re specifying two things: $p\in U$, and $U$ is open. When you say ‘Let $N$ be a nbhd of $p$, you’re specifying two things, one of which is different: $p\in N$, and there is some open set $U$ such that $p\in U\subseteq N$. You’re no longer specifying that the nbhd itself is open.
@BrianMScott I know. But what I'm asking is: why not say "open set containing $p$" instead of "open neighbourhood" because every open set is automatically a nbhd. (of course not the other way around)
@robjohn $g$ is a unit and $u$ is a unit and we cannot have $g = u$. (if we have $g = u$ then the $u$ ball only contains the zero function $f=0$ but that's not a unit)
@MattN Now that I think about it, there is also a way of defining topogies that takes neighborhood as the primitive concept and develops open and closed sets and so on from that. In that approach open nbhd of p does come more naturally than *open set containing p$.
@MattN perhaps I am missing something, but $\frac1u$ and $\frac1g$ can both exist and be different, but does that prevent $g-u$ or $g+u$ from being $0$ somewhere?
@robjohn Well $g$ and $u$ are both positive by assumption so $g+u$ can't be $0$. And $g-u$ can't be zero because if it were we'd have $g=u$ but then that wouldn't be a closed ball $B_{g,u}$ because all it could contain would be $f=0$ but that's not a unit so it would be the empty ball.
@PeterTamaroff I think Europe is the cradle of all mathematics. : ) Gauss and Euler aren't very modern but of course there probably are many important contemporary European mathematicians. : )
