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

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 $$x \in D$$.

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

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.

• 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). Dec 13, 2020 at 15:35
• 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 Dec 13, 2020 at 15:36
• @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. Dec 13, 2020 at 15:37
• 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 $x$ Dec 13, 2020 at 15:41
• Oh wait, is it because the subsequence is either monotone increasing or monotone decreasing? So it approaches $x$ either from right or left? Dec 13, 2020 at 15:42