Let $A$ be a nonempty set of real numbers and let $f : A \rightarrow [0,\infty)$ be given by $f(x) = x ^2$. Prove or disprove that if $A$ is open and $f$ is uniformly continuous, then $A$ is bounded.
I claim that the statement is false. Is the following counterexample correct?
Let $A = \bigcup_{n=1}^{\infty} I_n$, where $I_n = \left(n, \sqrt{n^2 + \frac{1}{n}} \right)$. Since $A$ is the union of open sets, $A$ is open. But, $A$ is not bounded.
We want to show that $f$ is uniformly continuous. Let $\varepsilon > 0$ be given. There exists some $N>0$ such that $\frac{1}{N} < \varepsilon$.
Since the $I_n$s are decreasing in size, the shortest distance between any two intervals is the distance between $I_1 = (1, \sqrt{2})$ and $I_2 = (2, \sqrt{4+1/2})$ which is $2 - \sqrt{2}$. Taking $\delta \leq 2- \sqrt{2}$, we guarantee that for any $x_1, x_2 \in A$, when $|x_1 - x_2| < \delta$, both $x_1$ and $x_2$ are in the same $I_n$.
Let $n > N$ and let $x_1, x_2 \in I_n$. Since $f$ is monotone increasing, the image of $I_n$ is $f(I_n) = (n^2, n^2 + \frac{1}{n})$. Since $x_1^2$ and $x_2^2$ are contained in $f(I_n)$, \begin{align*} \left| x_1^2 - x_2^2\right| &< \left|n^2 - \left(n^2 + \frac{1}{n}\right)\right| \\&= \left|\frac{1}{n}\right| < \frac{1}{N} < \varepsilon. \end{align*}
For $n \leq N$ and $x_1, x_2$ in $I_n$, we have $$ \left| x_1^2 - x_2^2\right| = |x_1 + x_2| |x_1 - x_2|. $$ Since $x_1$ and $x_2$ are both less than $\sqrt{N^2+\frac{1}{N}}$, $$ |x_1 + x_2| < 2\sqrt{N^2 + \frac{1}{N}}. $$ Then set $$\delta = \min\left( \frac{\varepsilon}{2\sqrt{N^2 + \frac{1}{N}}}, 2- \sqrt{2} \right).$$ So if $|x_1 - x_2| < \delta$, then $$ |x_1 + x_2| |x_1 - x_2| < 2\sqrt{N^2 + \frac{1}{N}} |x_1 - x_2| < \varepsilon. $$