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From:Continuous extension of a uniformly continuous function from a dense subset.

the answer from yunone has mentioned that

Since $f$ is uniformly continuous, there exists $\delta>0$ such that $d(p,q) < \delta$ implies $|f(p)-f(q)|<\epsilon$ for $p,q\in E$. Taking $n$ large enough so that $\frac{2}{n}<\delta$, then for $q,r\in V_n(p)$, $$ d(q,r)<d(q,p)+d(p,r)<\frac{2}{n}<\delta $$ so $|f(q)-f(r)|<\epsilon$. Thus $f(V_n(p))$ is bounded in $\mathbb{R}$

I'm trying to figuring out the boundedness of $f(V_n(p)$. I think the reason why $f(V_n(p))$ is bounded in $\Bbb R$ is for any $p,r\in V_n(p)$, $|f(p)|<\epsilon+|f(r)|<2\epsilon+|f(p)|=M$, so $f(x)<M$ for all $f(q)\in f(V_n(p))$

Is that correct?

Appreciate any suggestions.

Steven Lu
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1 Answers1

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The way you wrote it does not quite make sense. Here we fix $p\in E$ and want to show that $f(V_n(p))$ is bounded. You don't want to get a bound for $f(p)$. Instead, we want to find a bound for $|f(q)|$ for $q\in V_n(p)$, and the bound should be independent of $q$. Or, equally good, find a lower bound and an upper bound for $f(q)$.

We already know that for any $q,r\in V_n(p)$ the inequality $|f(q)-f(r)|<\epsilon$ holds. In particular, by taking $r=p$ we get $|f(q)-f(p)|<\epsilon$ for all $q\in V_n(p)$. That's the same as saying $$f(p)-\epsilon < f(q) < f(p)+\epsilon\;.$$ So that gives both a lower bound and an upper bound for $f(q)$ and proves what you wanted.

PatrickR
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