I am working through a theorem in Advanced Calculus, Fitzpatrick. Ch3, Theorem 3.23

Suppose $f: D \rightarrow \mathbb{R}$ is monotone. Prove that if $f(D)$ is an interval, then $f$ is continuous

I solved it a different way than the book, I am looking for confirmation that my method is valid. If not, where did I go wrong?

My attempt I will just do the case when $f$ is monotone increasing.

Assume $f(D)$ is an interval. Our goal is to prove $$\forall x \in D \quad \{x_n\} \text { converges to } x \rightarrow \{f(x_n)\} \text{ converges to } f(x)$$

Let $x$ be arbitrary element of $D$ and assume there exists a sequence $\{x_n\} \subset D$ that converges to $x$. There exists a subsequence of $\{x_n\}$, call it $\{x_{n_k}\}$, that is monotone increasing. It can be shown that since $\{x_n\}$ converges to $x$, then $\{x_{n_k}\}$ converges to $sup \{x_{n_k}\} = x$.

Since $f$ is monotonic increasing, $\{f(x_{n_k})\}$ is monotonic increasing. Note that this implies $\forall k \in \mathbb{N}, f(x_{n_k}) \leq f(x)$, since $x$ is the supremum of the subsequence. Therefore $\{f(x_{n_k})\}$ is a bounded monotonic sequence, which converges to $sup \{f(x_{n_k})\} = f(x)$.

Therefore we have shown $\{f(x_{n_k})\}$ converges to $f(x)$ for arbitrary $x \in D$.

  • $\begingroup$ There exists a subsequence that is monotone increasing. No, but the other possibility is: a subsequence that is monotone decreasing. $\endgroup$
    – GEdgar
    Dec 13, 2020 at 15:44

1 Answer 1


That proof cannot be correct because you did not use the fact that $f(D)$ is an interval (and the statement is surely wrong without that condition).

It is correct that $f(x_{n_k})$ increasing implies that $f(x_{n_k}) \to \sup \{f(x_{n_k})\}$, but $f(x_{n_k}) \le f(x)$ for all $k$ implies only that the supremum is $\le f(x)$. You have not shown that the limit is equal to $f(x)$.

Another (minor) error is that an arbitrary sequence converging to $x$ does not necessarily have an increasing subsequence.

You can use your approach to show that $f$ has a left and a right limit at $x$. Then assume that these are different, i.e. $$ \lim_{y \to x-} f(x) < \lim_{y \to x+} f(x) \, . $$ Now show that $f$ does not take values between those limits, contradicting the assumption that $f(D)$ is an interval.

  • $\begingroup$ Appreciate the help. I'm wondering now if I can rescue it, because if sup != f(x), then that defines an interval [sup, f(x)] in f(D). $\endgroup$ Dec 13, 2020 at 15:35
  • $\begingroup$ Regarding the increasing subsequence, there is a theorem 2.32 in the book that says "Every sequence has a monotone subsequence", which is what I used in that step $\endgroup$ Dec 13, 2020 at 15:36
  • $\begingroup$ @YetiMountainButter: It is correct that every sequence has a monotone subsequence. But that is not necessarily increasing. – I have updated the answer with a sketch about how your argument can be rescued. $\endgroup$
    – Martin R
    Dec 13, 2020 at 15:37
  • $\begingroup$ Thanks @Martin R, can you add a little more detail to your suggested approach? I don't see how we can show $f$ has a left and right limit at $x$ $\endgroup$ Dec 13, 2020 at 15:41
  • $\begingroup$ Oh wait, is it because the subsequence is either monotone increasing or monotone decreasing? So it approaches $x$ either from right or left? $\endgroup$ Dec 13, 2020 at 15:42

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