# How to prove that the inverse of a continuous strictly monotone increasing function is continuous? (Terence Tao Analysis 1, Proposition 9.8.3)

I'm having some problems proving the inverse is continuous. The hint in the book is to use the standard epsilon-delta definition of continuity. I believe the easiest route is a proof by contradiction, but with all of the quantifiers in the statement, I may be incorrectly negating the statement I am trying to prove. Also, I have at my disposal the intermediate value theorem, which most of my proof relies on. Below is the proposition:

Let $$a < b$$ be real numbers, and let $$f:[a, b] \to \mathbb{R}$$ be a function which is both continuous and strictly monotone increasing. Then $$f$$ is a bijection from $$[a, b]$$ to $$[f(a), f(b)]$$, and the inverse $$f^{-1}: [f(a), f(b)] \to [a, b]$$ is also continuous and strictly monotone increasing.

Below is my attempt at a proof:

Let $$x_1, x_2 \in [a, b]$$ be real numbers such that $$f(x_1) = f(x_2)$$. From the trichotomy of the real numbers, we have that exactly one of the following is true: $$x_1 = x_2$$, $$x_1 < x_2$$, or $$x_1 > x_2$$. Suppose $$x_1 \not = x_2$$. Then, by definition of strictly increasing monotone functions, we have that $$f(x_1) \not = f(x_2)$$. Thus, $$x_1 = x_2$$, and $$f$$ is injective. Now let $$y \in [f(a), f(b)]$$ be a real number. Then, by the intermediate value theorem, there exists a real number $$c \in [a, b]$$ such that $$f(c) = y$$. Thus, $$f$$ is a surjection from $$[a, b]$$ to $$[f(a), f(b)]$$. Since $$f$$ is both injective and surjective, we can conclude that $$f$$ is a bijection from $$[a, b]$$ to $$[f(a), f(b)]$$. To show that $$f^{-1}$$ is strictly monotone increasing, let $$y_1, y_2 \in [f(a), f(b)]$$ be real numbers such that $$y_1 < y_2$$. Then,by the intermediate value theorem, there exist $$x_1, x_2 \in [a, b]$$ such that $$f(x_1) = y_1$$ and $$f(x_2) = y_2$$. Since $$f$$ is strictly monotone increasing, we have $$x_1 < x_2$$. Using the definition of an inverse, we have\begin{align*}f^{-1}(y_1) &= f^{-1}(f(x_1)) \\&= x_1 \\&< x_2 \\&= f^{-1}(f(x_2)) \\&=f^{-1}(y_2) \text{,}\end{align*}showing that $$f^{-1}$$ is strictly monotone increasing. Finally, we will show that $$f^{-1}$$ is continuous. Let $$y_0 \in [f(a), f(b)]$$ be a real number, and let $$\epsilon > 0$$ be a real number. As before, there exists a real number $$x_0 \in [a, b]$$ such that $$f(x_0) = y_0$$. Likewise, for any real number $$y \in [f(a), f(b)]$$, the intermediate value theorem tells us that there exists a real number $$x \in [a, b]$$ such that $$f(x) = y$$. We want to show that there exists a $$\delta > 0$$ such that $$| f^{-1}(y) -f^{-1}( y_0) | < \epsilon$$ for all $$y \in [f(a), f(b)]$$ such that $$|y - y_0| < \delta$$. This is equivalent to showing that there exists a $$\delta > 0$$ such that $$| x - x_0 | < \epsilon$$ for all $$f(x) \in [f(a), f(b)]$$ such that $$|f(x) - f(x_0)| < \delta$$. Written in the order we are more accustomed to, this is equivalent to showing that there exists a $$\delta > 0$$ such that $$|f(x) - f(x_0)| < \delta$$ for all $$x \in [a, b]$$ such that $$|x - x_0| < \epsilon$$. Suppose, for the sake of contradiction, that $$f^{-1}$$ is not continuous. That is, suppose for all $$\delta > 0$$, there exists an $$\epsilon > 0$$ such that $$|f(x) - f(x_0)| \ge \delta$$ for all $$x \in [a, b]$$ such that $$|x - x_0| < \epsilon$$.

I am not really sure where to go from here, and I'm not certain I correctly negated the statement that the inverse of $$f$$ is continuous. Any help is greatly appreciated.

P.S. This is not for any homework, just self study. I've never taken a class in analysis, so please feel free to point out anything I am doing wrong (or that is less than rigorous).

• Let me get this straight. You're not trying to prove that the inverse of a continuous monotone increasing function is continuous, you're tryinf to prove that it has a continuous inverse? Is that right? You've got a continuous monotone increasing function $f$ and you want to prove that the inverse of $f$ has a continuous inverse?
– bof
Jun 14, 2020 at 15:11
• Quick note about your answer: You assumed the image of $f$ is $[f(a), f(b)]$ but that was not an assumption in the question. It is true but technically you have to prove it. To simplify the proof of continuity, note that definition of continuous on point $a$ is the same as: for all $\epsilon > 0$ there exists $\delta$ s.t if $x \in (a - \delta, a + \delta)$ then $f(x) \in (f(a)-\epsilon, f(a) + \epsilon)$. Hint: This function and it's inverse map intervals to intervals
– MBW
Jun 14, 2020 at 15:17
• @MBW I believe that is a corollary of the intermediate value theorem, which also appears to be assumed here. They do both need to be proved. Jun 14, 2020 at 15:36
• @bof, I am trying to prove that $f$ has a continuous inverse. And it is strictly increasing (as you noted, this is necessary for $f$ to be injective). Jun 14, 2020 at 15:43
• @MBW, you are correct that I would need to prove that. That is a corollary of the IVT, and I proved it before getting to this point (so I should have stated that it is one of our assumptions). Jun 14, 2020 at 15:45

$$\newcommand{\ep}{\epsilon}$$ $$\newcommand{\de}{\delta}$$ $$\newcommand{\f}{f^{-1}}$$ $$\newcommand{\ga}{\gamma}$$

Use this formulation of the problem:

For any $$y_0\in (f(a),f(b))$$ and any $$\ep>0$$, $$\exists \de>0$$ s.t. $$\f(y_0-\de,y_0+\de)\subset (\f(y_0)-\ep,\f(y_0)+\ep)$$

(the case where $$y_0=f(a)$$ or $$f(b)$$ is similar to the following, and just require you ignore either the left or right half of the intervals involved)

Set $$\ga=\textrm{min}(\ep,\f(y_0)-(a),b-\f(y_0))$$. Note here that $$\ga\leq \ep$$. It is easy to see that the set $$(\f(y_0)-\ga,\f(y_0)+\ga)$$ lies in $$[a,b]$$.

Now consider $$(f(\f(y_0)-\ga),f(\f(y_0)+\ga))$$. Because $$\f$$ is strictly increasing, it is easy to see that this interval is mapped under $$\f$$ into $$(\f(y_0)-\ga,\f(y_0)+\ga)$$.

Finally, simply set $$\de=\min(f(\f(y_0)+\ga)-y_0,y_0-f(\f(y_0)-\ga)]$$. The set $$(y_0-\de,y_0+\de)$$ is a subset of $$(f(\f(y_0)-\ga),f(\f(y_0)+\ga))$$, and so is sent into $$(\f(y_0)-\ga,\f(y_0)+\ga)$$ by $$\f$$.

Because $$\ga\leq \ep$$, we then have that $$\f(y_0-\de,y_0+\de)\subset(\f(y_0)-\ep,\f(y_0)+\ep)$$, as desired.

• Can you please clarify what you mean by "mapped under $f^-1$? I am not quite sure what you mean there. Jun 14, 2020 at 19:13
• @Shrodinger149 Are you familiar with the idea of the image of a set? If so, when I say something like 'set $A$ is mapped under $f^{-1}$ into $B$' or 'set $A$ is mapped by $f^{-1}$ into $B$', I am saying $f^{-1}(A)\subset B$. This is also what I mean by a set being 'sent' by $f^{-1}$ (or $f$). Jun 14, 2020 at 20:39
• I understand. Everything looks good, and makes sense (though it took me a little while to wrap my head around it). Is this your own approach, or have you seen this proof before? I'd really like to know the process of thinking up this proof, as it does not seem intuitive to me (if there is any way to answer that, because I understand sometimes things will just "come to you"). Thank you for the answer. Jun 14, 2020 at 21:04
• @Shrodinger149 I appreciate your question. This approach is motivated by my (rather basic) understanding of point-set topology. One of the main advantages to studying analysis-related results under the framework of topology is its generality. The definition of continuity I used in my answer is a special case of this general definition. That said, the stuff you'll find in an introductory topology will not really be useful for analysis at a higher level, as differentiation/ integration are never touched upon. (1/2) Jun 14, 2020 at 21:24
• @Shrodinger149 This is because to do calculus (or something 'like' calculus) on a space, that space needs to have certain specific properties (which an average topological space lacks). There is anl approach to calculus that is very general and uses a few concepts from topology (see books like 'Calculus on Manifolds'), but most of topology does not deal with such things. So basically, topology helps with results relating to continuity, limits, etc. but not analysis per se. So it is not really an appropriate way to get an understanding of analysis (except in a first course maybe). (2/2) Jun 14, 2020 at 21:31