Mathematics - 2023-05-30 (page 2 of 2)

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9:04 PM

Let $S \subseteq\mathbb{R}$, with $S \ne \emptyset$ and bounded above. Show that if $\max S$ exists, then $\sup S \in S$.

I tried this: by definitions of maximum and supremum, we have $s \le \max S$ for each $s \in S$ and $\max S \le \sup S$. By the characterization of the supremum, for each $\epsilon>0$ there exists $s_\epsilon \in S$ such that $\sup S -\epsilon < s_\epsilon$. Hence $\sup S - \epsilon < s_\epsilon \le \max S \le \sup S$ for each $\epsilon>0$, that is $\max S = \sup S$. Being $\max S \in S$, we have $\sup S \in S$.

Does this work?

noballpointpen

Seems ok to me.

Ted Shifrin

But from the first inequality, since $\max S$ is an upper bound, you have $\sup S\le \max S$.

noballpointpen

So we do not need the rest of the argument :)

Right?

Sonozaki

9:25 PM

@TedShifrin Oh, you're right. I didn't see that. Is my proof still valid, even if over complicated? Thank you.

noballpointpen

You already proved it in the first sentence and then just confirmed that the property of supremum holds.

Ted Shifrin

You skipped the key point. Your argument with $\epsilon$ shows precisely the inequality I stated. But you jumped right to equality.

Are you thinking of a squeeze principle?

Sonozaki

@TedShifrin Exactly, I'm saying that since $\sup S-\epsilon< \max S \le \sup S$ holds for each $\epsilon>0$, it must be $\sup S=\max S$.

Ted Shifrin

Why? Are you not using $x-\epsilon<y$ for all $\epsilon>0$ implies $x\le y$?

Sonozaki

9:42 PM

Yes, exactly. From $\sup S-\epsilon < \max S$ for each $\epsilon>0$ I obtain $\sup S \le \max S$. On the other hand, from the supremum definition $\max S \le \sup S$. Hence $\sup S \le \max S \le \sup S$, that is $\sup S=\max S$. Am I missing something?

noballpointpen

9:53 PM

I think I have constructed some necessary steps in proving continuity. I continued with my sequential approach.
The set $E\subseteq X$ is dense in this context.
Let $p\in X$ and $x_n \to p$ $(x_n \neq p)$ be any. We can split the sequence in two subsequences: one, $(x_{e_n})$ consists of only points of $E$ and the other, $(x_{k_n})$ of the limits of $E$.
We have that $g(x_{e_n})\to g(p)$.
For each $x_{k_n}$ we can have a Cauchy sequence $(y_n)$ in $E$ such that $y_n\to x_{k_n}$ and $g(y_n)\to g(x_{k_n})$.

I didn't cover the case when $(x_n)$ has only finitely many such diverse points in this brief draft.

Ted Shifrin

@Sonozaki Nope, not missing anything. I just wanted you to be explicit in your argument.