Prove that if every continuous function on a subset of $\Bbb R$ is uniformly continuous, then the set is closed.

Question

May I please ask how to prove or disprove the following statements:

1.If every continuous function on a subset of $\Bbb R$ is uniformly continuous, then the set is closed.

2.If every continuous function on a subset of $\Bbb R$ is uniformly continuous, then the set is bounded.

Answer 1

It's almost certainly easier to prove the contrapositive of the first statement.

1. If a set $S\subset \mathbb{R}$ is not closed, then there exists a continuous function $f : S\to \mathbb{R}$ such that $f$ is not uniformly continuous. As $S$ is not closed, we let $x\in \overline{S}\setminus S$ and $\{x_n\}\subset S$ be a strictly monotonic sequence such that $x_n\to x$ (the sequence doesn't have to be monotonic, but it's probably easier to visualize the resulting function if it is). Then, there is a continuous function that satisfies $f(x_n) = n$ (we could really pick the values of $f(x_n)$ to be almost anything, as long as they don't get too close together as $n$ increases). For example, we can let $f$ be linear between $x_{n-1}$ and $x_n$ and constant elsewhere. We can see that $f$ is not uniformly continuous, as $$\lim_{n\to \infty} \frac{\lvert f(x_{n+1})-f(x_n)\rvert}{\lvert x_{n+1}-x_n\rvert} = \lim_{n\to \infty} \frac{1}{\lvert x_{n+1}-x_n\rvert} = \infty$$ This tells us that for any $\delta > 0$, we can choose $n$ such that $\lvert x_{n+1}-x_n\rvert < \delta$ and $\lvert f(x_{n+1})-f(x_n)\rvert = 1$. Therefore, if every continuous function from $S$ to $\mathbb{R}$ is uniformly continuous, then $S$ must be closed.

The second statement, however, is not true.

2. There is an unbounded set $S$ such that any continuous function from $S$ to $\mathbb{R}$ is uniformly continuous. Consider $S = \mathbb{N}$ (or any discrete set with some minimum distance between points). Then, for $f : S\to \mathbb{R}$ a continuous function (which is really any function), for any $\epsilon > 0$, we can let $\delta < 1$. Then, for $x$, $y\in S$, if $\lvert x-y\rvert < \delta$, then $x = y$, and $\lvert f(x)-f(y)\rvert = 0 < \epsilon$.