Thread: If f is a uniformly continuous function on a bounded set S, then f is a bounded on S

1. If f is a uniformly continuous function on a bounded set S, then f is a bounded on S

Here's my attempt:

Assume to the contrary that $f$ is a uniformly continuous function on a bounded set $S$ but $f$ is unbounded on $S$. Then there exists a sequence $(s_{n})\in S$ such that $|f(s_{n})|\ge n$ for all $n\in \mathbb{N}$. Since $(s_{n})$ is a bounded sequence, there exists a subsequence $(s_{n_{k}})$ that is a Cauchy sequence. Since $f$ is uniformly continuous and $(s_{n_{k}})$ is a Cauchy sequence in $S$, then $(f(s_{n_{k}}))$ is a Cauchy sequence. So $(f(s_{n_{k}}))$ converges to some real number. But $|f(s_{n_{k}})|\ge n_{k}$ for all $n_{k}\in \mathbb{N}$ as well, which is a contradiction. Q.E.D.

Is this correct? Thanks.

2. Originally Posted by Pinkk
Here's my attempt:

Assume to the contrary that $f$ is a uniformly continuous function on a bounded set $S$ but $f$ is unbounded on $S$. Then there exists a sequence $(s_{n})\in S$ such that $|f(s_{n})|\ge n$ for all $n\in \mathbb{N}$. Since $(s_{n})$ is a bounded sequence, there exists a subsequence $(s_{n_{k}})$ that is a Cauchy sequence. Since $f$ is uniformly continuous and $(s_{n_{k}})$ is a Cauchy sequence in $S$, then $(f(s_{n_{k}}))$ is a Cauchy sequence. So $(f(s_{n_{k}}))$ converges to some real number. But $|f(s_{n_{k}})|\ge n_{k}$ for all $n_{k}\in \mathbb{N}$ as well, which is a contradiction. Q.E.D.

Is this correct? Thanks.
Looks good to me.

Another way to look at it might be as follows:

Since $f$ is unif. cont. we may choose a $\delta$ such that $|x-y|<\delta\implies |f(x)-f(y)|<1$.

Now notice that since $S$ is bounded we have that $S\subseteq [-M,M]$ for some $M>0$. But, this latter set is compact and thus we may cover it with finitely many intervals of the form $I_1,\cdots,I_n$ such that $\text{diam }I_k<\frac{\delta}{2}$. Let $I_{\alpha_1},\cdots,I_{\alpha_m}$ be those that intersect $S$. Choose $x_k\in I_{\alpha_k}\cap S$ then $I_{\alpha_k}\subseteq B_{\frac{\delta}{2}}(x_k)$. It follows that $B_{\frac{\delta}{2}}(x_1),\cdots,B_{\frac{\delta}{ 2}}(x_m)$ cover $S$.

So, let $x\in S$. Then $x\in B_{\frac{\delta}{2}}(x_k)$ for some $k$ and so $|f(x)|=|f(x)-f(x_k)+f(x_k)|\leqslant |f(x)-f(x_k)|+f(x_k)\leqslant 1+\max_{1\leqslant j\leqslant m}f(x_j)$

3. Hmm, thanks. The teacher's assistant for my real analysis class talks about using compactness in proofs for uniform continuity but our professor and curriculum doesn't cover compactness (at least we haven't touched on it at all). Thanks for the feedback!

4. Originally Posted by Pinkk
Hmm, thanks. The teacher's assistant for my real analysis class talks about using compactness in proofs for uniform continuity but our professor and curriculum doesn't cover compactness (at least we haven't touched on it at all).
That's a shame. Compactness is one of the most important parts of analysis. A lot of important theorems are just statements about compactness and continuous maps (e.g. the extreme value theorem)

Thanks for the feedback!
Any time

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if f is a uniformly continous on a bounded set then

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