# Solution check: Uniform continuity

While reading Bartle's book, i came across the following:

Problem:

Prove that if $$f$$ is uniformly continuous on a bounded subset $$A$$ of $$\mathbb{R},$$ then $$f$$ is bounded on $$A$$.

Attempt:

Given $$A \subset \mathbb{R}$$ such that $$A$$ is bounded and $$f: A \rightarrow \mathbb{R}$$ an uniformly continuous function, we will prove that $$f$$ is bounded.

Given $$(x_{n})$$ a sequence of points in $$A$$ one can obtain a convergent subsequence $$(x_{nk})$$, by using the Bolzano-Weierstrass theorem. Since $$f$$ is uniformly continuous and $$(x_{nk})$$ is a Cauchy sequence (since it is convergent), $$(f(x_{nk}))$$ is also Cauchy.

Finally, since every Cauchy sequence is bounded, there exists $$a$$ and $$b$$ in $$\mathbb{R}$$ such that: $$a \leqslant f\left(x_{n k}\right) \leqslant b$$

Therefore, $$f$$ is bounded.

Questions:

1. Is the solution correct?

2. If the solution is indeed correct, is it well written?

3. If the solution is incorrect, can someone gently me explain why and provide a solution?

• How do you conclude that $f$ is bounded? Jan 2, 2021 at 23:04
• You need to say something about the choice of $x_n$ to be able to conclude. So, as is, 1. is not true. Jan 2, 2021 at 23:08
• I am afraid i might be missing something simple, but why do i need to say something about the choice of $x_{n}$? Jan 2, 2021 at 23:13

It seems that you have the main idea but I disagree with your conclusion, you cannot conclude that $$|f(x)| \le a$$ for all $$x \in A$$ with what you wrote. Let us try by contradiction: suppose that $$f$$ is not bounded, $$i.e.$$ it exists $$(x_n)_n \subset A$$ such that $$\forall K > 0, \exists N_K \in \mathbb N~\text{ s.t. }~ \forall n \ge N_K: |f(x_n)| \ge K.$$ As you wrote above we can find a subsequence $$(x_{n_k})_k$$ such that $$(|f(x_{n_k})|)_k$$ converges to a certain $$L \in \mathbb R^+$$. Since the function $$\mathbb N \to \mathbb N: k \mapsto n_k$$ is strictly increasing, we find that $$k \le n_k$$ (it is a well known result about strictly increasing function from $$\mathbb N$$ to $$\mathbb N$$, try to show it). Moreover, for $$K = L + 1$$ (for example), there is $$N_K$$ such that for all $$k \ge N_K$$, $$|f(x_{n_k})| > K$$ which contradicts the fact that $$(|f(x_{n_k})|)_k$$ converges to $$L$$.
• You have to show that $|f(x)| \le a$ for all $x \in A$, how do you conclude that statement from your last inequality ? Jan 2, 2021 at 23:21
• You showed that that every sequence $(y_n)_n = (f(x_n))_n \subset f(A)$ has a bounded subsequence. It is not enough to ensure that each $y \in f(A)$ is bounded. Jan 2, 2021 at 23:28
• Remember that $f(x_n) \to \infty$ iff $\forall K > 0,$ there exists $N \in \mathbb N$ such that $\forall n \ge N$ $f(x_n) \ge K$. Hence you cannot have $f(x_{n_k}) \to L <+\infty$ and $f(x_n) \to \infty$ at the same time. Jan 2, 2021 at 23:39