# Thread: Uniformly Cont. Functions

1. ## Uniformly Cont. Functions

Consider the following statements:

1.) If $f : A \rightarrow \mathbb{R}$ is uniformly continuous, then $f(A)$ is bounded.

2.) If $f : A \rightarrow \mathbb{R}$ is uniformly continuous, and also if $A$ is bounded, then $f(A)$ is bounded.

Either prove, or disprove, the statements above.

2. Originally Posted by alikation0
Consider the following statements:
1.) If $f : A \rightarrow \mathbb{R}$ is uniformly continuous, then $f(A)$ is bounded.
2.) If $f : A \rightarrow \mathbb{R}$ is uniformly continuous, and also if $A$ is bounded, then $f(A)$ is bounded.
a) Is obviously false. Consider $f:\mathbb{R} \mapsto \mathbb{R}\;,\;f(x) = x.$

Hints for b)
Uniformly continuous functions map Cauchy sequences into Cauchy sequences.
If E is closed the result in immediate.
You may try proof by contradiction.

3. Originally Posted by Plato
a) Is obviously false. Consider $f:\mathbb{R} \mapsto \mathbb{R}\;,\;f(x) = x.$

Hints for b)
Uniformly continuous functions map Cauchy sequences into Cauchy sequences.
If E is closed the result in immediate.
You may try proof by contradiction.
That hint for b helped a lot! Thanks a lot, I figured that out. But, for a) (AKA 1), I still don't see how that disproves the statement. Mind elaborating?

4. Originally Posted by alikation0
That hint for b helped a lot! Thanks a lot, I figured that out. But, for a) (AKA 1), I still don't see how that disproves the statement. Mind elaborating?
the real-valued function f(x) = x is uniformly continuous, but if we take the domain of that function to be the reals, the range is unbounded