Why does a uniformly continuous function on [a,b] in the reals need be closed AND bounded?

Question

I cannot come up with any counter examples as to why a function would need to be both closed and bounded to be uniformly continuous. Why is it not sufficient to just have one condition?

For example, if something is bounded, then it can't dramatically increase constantly and thus would be uniformly continuous no matter if it was closed or not.

Similarly, if something was closed but not bounded, a delta could be found where all differences between points would be within all epsilon greater than 0.

Answer 1

Let $f\colon[a,b]\longrightarrow\mathbb R$ be a continuous function. Then:

• If $C$ is a closed subset of $[a,b]$, the, since $[a,b]$ is compact, then $C$ is compact too. Therefore, $f(C)$ is compact and, in particular, it is closed.
• The function $f$ is bounded. This is a standard Real Analysis theorem.

Note that I did not need to assume that $f$ is uniformly continuous for this.

Answer 2

Take the interval $(0,1)$, which is bounded but not closed. The continuous function $f : (0,1) \to \mathbb R$ defined by $f(x) = \frac{1}{x}$ is not uniformly continuous on $(0,1)$.

Take the set $\mathbb R$, which is closed but not bounded. The continuous function $f : \mathbb R \to \mathbb R$ defined by $f(x) = x^2$ is not uniformly continuous on $\mathbb R$.

So, in order to conclude that a continuous function defined on a set $E$ is uniformly continuous, it is not enough that $E$ be assumed to have only one of the two properties: closed and bounded.