Theorem 4.20(c) in Baby Rudin: Is every continuous function whose domain is an unbounded subset of $\mathbb{R}$ uniformly continuous?

Question

Here is Theorem 4.20 in the book Principles of Mathematical Analysis by Walter Rudin, third edition:

Let $E$ be a non-compact set in $\mathbb{R}^1$. Then

(a) there exists a continuous function on $E$ which is not bounded;

(b) there exists a continuous and bounded function on $E$ which has no maximum.

If, in addition, $E$ is bounded, then

(c) there exists a continuous function on $E$ which is not uniformly continuous.

Thus Theorem 4.20(c) can be stated as follows: On a bound subset non-compact of $\mathbb{R}$, there exists a continuous function which is not uniformly continuous. Or, equivalently, if every continuous function on a non-compact subset $E$ of $\mathbb{R}$ is also uniformly continuous, then $E$ is unbounded.

Now my question is as follows:

Let $E$ be a non-compact subset of $\mathbb{R}$ such that $E$ is not bounded. Then can we conclude from Theorem 4.20(c) in Baby Rudin that every continuous function $f \colon E \to \mathbb{R}$ is also uniformly continuous on $E$? If so, then how to give a proof of this fact?

And, is there a way of generalising this result to maps between arbitrary metric spaces, along the lines of Theorem 4.19 in Baby Rudin?

Answer 1

This is not true. Take $E=\mathbb R$ for example. Then your "theorem" would imply that every continuous function $\mathbb R\to\mathbb R$ is also uniformly continuous, which is wrong. Consider $f(x)=x^2$ as a counterexample.

Answer 2

assertion c need not to be always true for unbounded non compact set. AS we have counter example of set of integers. f(x)=x^2 is uniformly continous on set of integers. In fact assertion c is not always wrong for unbounded non compact set but we have a counter example.