For $A \subset \mathbb{R}$, the function $f: A \to \mathbb R$, $f(x)=x^2$ is a uniform continuous function if
(A) A is bounded
(B) A is unbounded
(C) A is compact
(D) A is the set $\mathbb Z$ of all integers.
My attempt:
(A) If $A$ is a bounded set then from the mean value theorem, we will have that $|f(x)-f(y)|\le M|x-y|$. Hence, $f(x)$ is uniformly continuous.
(C) If $A$ is compact, then $f(x)$ is uniformly continuous.
For options (B) and (D), using definition of uniform continuity, $$|x^2-y^2|\le |x-y||x+y|$$ But if $A$ is unbounded of $\mathbb Z$, then there's no bound for $|x+y|$, so in my opinion only (A) and (C) are the correct options.
I don't know how to check for options (B) and (D).
The answer is: (A), (B), (C), (D)
For (D), the subspace topology on $\mathbb Z$ is discrete. So $\delta = \frac{1}{2}$ will ensure that $$|x-y| < \delta \implies x=y \implies |f(x) - f(y)| = 0 < \epsilon$$
I think (B) is inconclusive. In light of (D), $f$ could be uniformly continuous on an unbounded set. But $f$ is definitely not uniformly continuous on $\mathbb R$, another unbounded set.
I started to prove that $f$ couldn't be uniformly continuous on $A$ if $A$ was unbounded. Here is how that proof went: we want to show that there exists $\epsilon > 0$ such that for all $\delta > 0$ there exists $x$ and $y$ in $A$ with $|x - y | < \delta$ but $|f(x) - f(y) | \geq \epsilon$. Let $\epsilon = 1$. Given $\delta > 0$, there exists $x \in A$ such that $|x| > \frac{1}{\delta}$. Notice that for all $y$, $$ |f(x) - f(y)| = |f'(c)||x-y| = 2|c| |x-y| $$ for some $c$ between $x$ and $y$. If there exists $y \in A$ such that $|y| > |x|$ and $\frac{\delta}{2} < |x-y| < \delta $, we are done, because $$ 2|c| |x-y| > 2 |x| |x-y| > 2 \cdot \frac{1}{\delta} \cdot \frac{\delta}{2} =1 $$ Since $A$ is unbounded, there definitely exists $y \in A$ with $|y| > |x|$. But the second condition isn't guaranteed, and fails for discrete sets like $\mathbb Z$.
Like @zhw says in the comments, since it's not true that $f$ is uniformly continuous on every unbounded set $A$, the statement “If $A$ is unbounded, then $f$ is uniformly continuous on $A$” is false. But in light of (D), the statement, “If $A$ is unbounded, then $f$ is not uniformly continuous on $A$” is also false.