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


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


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.


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?

Thanks in advance, Lucas!

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

1 Answer 1


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 $L$.

  • 1
    $\begingroup$ You have to show that $|f(x)| \le a$ for all $x \in A$, how do you conclude that statement from your last inequality ? $\endgroup$
    – Falcon
    Jan 2, 2021 at 23:21
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    $\begingroup$ 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. $\endgroup$
    – Falcon
    Jan 2, 2021 at 23:28
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    $\begingroup$ 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. $\endgroup$
    – Falcon
    Jan 2, 2021 at 23:39
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    $\begingroup$ Ok I will edit my answer with the full argument. $\endgroup$
    – Falcon
    Jan 2, 2021 at 23:46
  • 1
    $\begingroup$ You can find above a more precise argument. $\endgroup$
    – Falcon
    Jan 3, 2021 at 0:04

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