Uniform continuity of a subset of $R$

Question

Let $A \subset R$ and $f:A\to R$ be given by $f(x) = x^2$. Then $f$ is uniformly continuous if

1. $A$ is bounded subset of $R$.
2. $A$ is dense subset of $R$.
3. $A$ is unbounded and connected subset of $R$.
4. $A$ is unbounded and open subset of $R$.

I think that some information is missing from this question.

Suppose $A =\{ 1,2,3\}$ then one can not speak of uniform continuity on $A$. This is because limit and continuity at a point $x$ is always defined on a neighborhood of that point. So (1) fails.

Take $A = R$ then (2), (3) and (4) fail.

Answer 1

Note that the restriction of a uniformly continuous function is uniformly continuous.

Assuming that $A$ is bounded, then there exists $M>0$ such that $A\subseteq [-M,M]$. Notice that $g:[-M,M]\to \mathbb{R}$ given by $g(x):=x^2$ is continuous on a compact subset and, in turn, is uniformly continuous. Notice that $g|_A=f$ and hence, $f$ is a uniformly continuous function.

So, (1) holds. I agree that (2), (3), and (4) all fail since, like you mentioned, letting $A=\mathbb{R}$ gives a counterexample.

Answer 2

The right answer is 1.

It makes perfect sense to define continuity and uniform continuity in isolated point. In particular uniform continuity is never defined by means of limit. It is defined as: $$ \forall \varepsilon>0\quad \exists \delta>0 \colon |x-y|<\delta \implies |f(x)-f(y)|<\varepsilon. $$ which makes perfect sense whatever is the domain of $f$.

Most text don't define limits in isolated point, but they surely define continuity on isolated point by saying that a function is continuous by definition in such points. As a matter of fact if you take the definition of continuity given as: $$ \lim_{x\to x_0} f(x) = f(x_0) $$ and plug in the definition of limit, you get: $$ \forall \epsilon>0\quad \exists\delta>0\colon 0<|x-x_0|<\delta \implies |f(x)-f(x_0)| < \varepsilon $$ which makes perfect sense also in isolated point (and is always true in such points). Notice also that you can remove the condition $0<|x-x_0|$ since for $x=x_0$ the statement is anyway true.