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For my proof, I had to inverse the definitions for uniform continuity and sequence convergence, so it feels a bit sketchy to me. I'd also appreciate suggestions if you have any.

Let $f:(a,b) \rightarrow \mathbb{R}$ be uniformly continuous. For the sake of contradiction, assume $f$ is unbounded. Let the sequences $a_n \rightarrow a$, $b_n \rightarrow b$, and $x_n \rightarrow x_0 \in (a,b)$. Since $f$ is continuous, then $f(x_n) \rightarrow f(x_0) \in \mathbb{R}$. Since $f$ is unbounded, then either of the sequence $f(a_n),f(b_n)$ must diverge, otherwise $f$ would be bounded. Without loss of generality, assume the sequence $b_n$ diverges. Since $f$ is uniformly continuous, then for $\epsilon > 0$, there exists $\delta > 0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$ for $x,y \in (a,b)$. Let $\delta > 0$. Since $b_n \rightarrow b$, then there exists some $N \in \mathbb{N}$ where $|b-b_n|<\delta$ for $n \geq N$. Restrict $x \in (a,b)$ such that $b_n<x<b$. Then $|x-b_n|<\delta$. Since $f(b_n)$ diverges, then there exists some $\epsilon > 0$ and $n \in \mathbb{N}$ with $n \geq N$ such that $|f(x)-f(b_n)| \geq \epsilon$. Therefore we can write for $|x - b_n|<\delta$ and $|f(x)-f(b_n)| \geq \epsilon$. Therefore $f$ is not uniformly continuous. This is a contradiction since $f$ is uniformly continuous. Therefore $f$ is bounded.

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  • $\begingroup$ Why must the function diverge on the interval limits, and not somewhere inbetween? $\endgroup$ – B.Swan Jun 10 at 23:20
  • $\begingroup$ Since is unif cont you can extend the function continuously to $[a,b]$. Since this is a compact set your function is bounded. $\endgroup$ – Julian Mejia Jun 10 at 23:20
  • $\begingroup$ @B.Swan If the function diverged at $x_0 \in (a,b)$, then the function wouldn't be defined at $x_0$. $\endgroup$ – Spencer Kraisler Jun 10 at 23:24
  • $\begingroup$ What about the function $f:(-1, 1) \rightarrow \Bbb{R}, f(x)=\frac{1}{x} \forall x \neq 0, f(0)=0$? $\endgroup$ – B.Swan Jun 10 at 23:27
  • $\begingroup$ @B.Swan that function is not continuous however. $\endgroup$ – Spencer Kraisler Jun 10 at 23:27
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Your argument doesn't work. Consider the function $\frac{\sin\left(\frac1x\right)}x$. It is unbounded near $0$. However, if $a_n=\frac1{n\pi}$, then $\lim_{n\to\infty}f(a_n)=0$.

Here's a proof. There is a $\delta>0$ such that $\lvert x-y\rvert<\delta\implies\bigl\lvert f(x)-f(y)\bigr\vert<1$. Now, consider points $a_1,a_2,\ldots,a_N\in(a,b)$ such that $a<a_1<a_2<\cdots<a_N<b$ and that $a_1-a,a_2-a_1,\ldots,b-a_M<\delta$. Then, if $x,y\in(a,b)$ and $x<y$, there are numbers $k<l\leqslant N$ such that $a_{k-1}<x\leqslant a_k$ and $a_{l-1}<y\leqslant a_l$. But then$$\lvert f(x)-f(y)\bigr\vert\leqslant\overbrace{\bigl\lvert f(x)-f(a_k)\bigr\rvert}^{<1}+\overbrace{\bigl\lvert f(a_k)-f(a_{k+1})\bigr\rvert}^{<1}+\cdots+\overbrace{\bigl\lvert f(a_l)-f(y)\bigr\rvert}^{<1}\leqslant l-k+1.$$

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    $\begingroup$ Or, you could use a previous theorem to conclude that $f$ is bounded on $[a + \delta, b - \delta]$, and then every member of $(a,b)$ is within $\delta$ of some element of $[a + \delta, b - \delta]$... (Or in case $\delta > \frac{b-a}{2}$, then every member of $(a,b)$ is within $\delta$ of $\frac{a+b}{2}$.) $\endgroup$ – Daniel Schepler Jun 10 at 23:25
  • $\begingroup$ Is that function also uniformly continuous? $\endgroup$ – Spencer Kraisler Jun 10 at 23:26
  • $\begingroup$ @SpencerKraisler Which function? $\endgroup$ – José Carlos Santos Jun 10 at 23:28
  • $\begingroup$ @JoséCarlosSantos My bad. I meant $\frac{\sin\left(\frac{1}{x}\right)}{x}$. This function doesn't seem to be uniformly continuous. Is it? $\endgroup$ – Spencer Kraisler Jun 10 at 23:28
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    $\begingroup$ In your proof, you wrote “Since $f$ is unbounded, then either of the sequence $f(a_n),f(b_n)$ must diverge, otherwise $f$ would be bounde.” My example shows that that assertion is false. $\endgroup$ – José Carlos Santos Jun 10 at 23:35
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Take $a=0,\ b=1$ for convenience. Set $\epsilon_n=1/n$ and find $\delta_n>0$ such that $|x-y|<\delta_n\Rightarrow |f(x)-f(y)|<1/n.$ We may assume $\delta_{n-1}>\delta_n\to 0.$

The idea is to extend $f$ to a continuous function on $[0,1]$. To this end, choose $(x_n)$ such that $x_n\in (1-\delta_n,1)$ and $x_n>x_{n-1}$. Then, $x_n\to 1$ and $(f(x_n))$ is Cauchy because $|f(x_m)-f(x_n)|<1/\min(n,m)$. Therefore, $f(x_n)\to y$ as $x_n\to 1.$

If $(z_n)$ is another sequence such that $z_n\to 1$ then, since $(w_n)=f(x_1),f(z_1),f(x_2),f(z_2),\cdots,\ $ is Cauchy, $f(z_n)\to y$ and it makes sense to define

$F(x)=\begin{cases}f(x)\quad a<x<1 &\\ y\quad \quad x=1\end{cases}$

$F$ is a continuous extension of $f$ by construction.

Similarly, we extend $F$ to a continuous function $G$ at $x=0.$ Now, $G$ is continuous on the compact set $[0,1]$ and so is bounded there. And since $G$ agrees with $f$ on $(0,1)$, it follows that $f$ is bounded.

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Suppose $f$ were unbounded on $(a,b).$ Let $x_1$ be any member of $(a,b).$ For $n\in \Bbb N$ let $x_{n+1}\in (a,b)$ with $|f(x_{n+1})|\ge 1+|f(x_n)|.$

Observe that this implies that $|f(x_m)-f(x_n)|\ge 1$ when $m\ne n.$

Let $(j(n))_{n\in \Bbb N}$ be some sub-sequence of $\Bbb N$ such that $(x_{j(n)})_{n\in \Bbb N}$ converges to some $x\in [a,b].$

Note that $m\ne n$ implies that $j(m)\ne j(n),$ which implies $|f(x_{j(m)})-f(x_{j(n)})|\ge 1.$

Then there does NOT exist $\delta>0$ such that $\forall u,v\in (a,b)\;(|u-v|<\delta \implies |f(u)-f(v)|<1).$

Because for any $\delta>0 $ the interval $I_{\delta}=(a,b)\cap (x-\delta/2,x+\delta/2)$ contains $x_{j(n)}$ for all but finitely many $n,$ so there exist $m,n$ with $m\ne n$ and $\{x_{j(m)},x_{j(n)}\}\subset I_{\delta}.$ And for such $m,n$ we have $$|x_{j(m)}-x_{j(n)}|<\delta \;\text { but }\; |f(x_{j(m)})-f(x_{j(n)})|\ge 1.$$

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