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Theorem. Let $f\colon I \to J$, where $I$ is an interval and $J$ is the image $f(I)$, be a function such that:

  1. $f$ is strictly increasing on $I$.
  2. $f$ is continuous on $I$.

Then $J$ is an interval, and $f$ has an inverse function $f^{-1}\colon J\to I$ such that:

  1. $f^{-1}$ is strictly increasing on $J$.
  2. $f^{-1}$ is continuous on $J$.

I need to prove only that $f^{-1}$ is continuous on $J$.

Proof. Let $y\in J$, and (for simplicity) assume that $y$ is not an end-point of $J$. Then $y=f(x)$, for some $x\in I$, and we want to prove that $$y_n\to y\implies f^{-1}(y_n)\to f^{-1}(y)=x.$$ Thus we want to deduce that for each $\varepsilon>0$, there is some number $X$ such that $$x-\varepsilon<f^{-1}(y_n)<x+\varepsilon\qquad \forall n>X\qquad(1).$$ Since $f$ is strictly increasing, we know that $$f(x-\varepsilon)<f(x)<f(x+\varepsilon).$$ also, since $y_n\to y=f(x)$, there is some number $X$ such that $$f(x-\varepsilon)<y_n<f(x+\varepsilon), \qquad\forall n>X.$$ Then, applying the strictly increasing function $f^{-1}$ to these inequalities, we obtain (1).

Edit

I don't understand why, since $y_n\to y=f(x)$, there is some number $X$ such that $$f(x-\varepsilon)<y_n<f(x+\varepsilon), \qquad\forall n>X.$$ I would have written the last line as follows: $$f(x)-\varepsilon<y_n<f(x)+\varepsilon, \qquad\forall n>X.$$ Applying the $f^{-1}$ to my last line, which I think it's correct, I obtain $$f^{-1}[f(x)-\varepsilon]<f^{-1}[y_n]<f^{-1}[f(x)+\varepsilon], \qquad\forall n>X.$$ which is not (1).

What am I misunderstanding? Why the author of this proof wrote that last line and not my version?

Thank you.

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  • $\begingroup$ Thank you @Mauro, I added some lines to my question. I completely agree with your point in the the comment. I can't get the point I explained in my edit. $\endgroup$
    – Charlie
    Feb 11, 2014 at 11:15
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    $\begingroup$ You must use (check on your textbook) the result : Let $A$ be a non-empty subset of $\mathbb{R}$, and suppose $f : A \rightarrow \mathbb{R}$ is a monotone function such that the image $f(A)$ is an interval; then $f$ is a continuous function. Then you apply it to show that : if $I$ is an interval and $f : I \rightarrow \mathbb{R}$ is a strictly monotone function, then the inverse function $f^{-1} : f(I) \rightarrow I$ is continuous. $\endgroup$ Feb 11, 2014 at 11:23
  • $\begingroup$ What makes things unclear to You may be just the notation: if you set $\alpha:=f(x-\varepsilon)$ and $\beta:=f(x+\varepsilon)$, then you have $\alpha< f(x)<\beta$. Since $y_n\to f(x)$, it follows that you can indeed find $X$ such that $\alpha <y_n <\beta$ for all $n>X$. $\endgroup$
    – Etienne
    Feb 11, 2014 at 11:28
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    $\begingroup$ About your "misunderstanding", I think that you cannot assume that the two "epsilons" (i.e.$f(x - \epsilon)$ and $f(x) - \epsilon$)are the same. $\endgroup$ Feb 11, 2014 at 13:03
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    $\begingroup$ About my previous comment : "show that : if $I$ is an interval and $f : I \rightarrow \mathbb{R}$ is a strictly monotone function", the "trick" is to avoid "degenerate" cases like a function $f : [0,1] \rightarrow [0,1]$, such that $f(x)=x$, except for $x = 1/2$ where $f(x) = 10$. In this case, the isolated point, violating the condition of monotonicity, force the inverse function having a domain that is not an interval, and so it is not continuous. $\endgroup$ Feb 11, 2014 at 14:02

1 Answer 1

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I've found DJH Garling, A Course in Mathematical Analysis (2013) the following corollary of the The intermediate value theorem [page 163] :

If $f$ is a strictly monotonic function on an interval $I$ then $f^{-1} : f(I) \rightarrow I$ is continuous.

Proof. Suppose without loss of generality that $f$ is strictly increasing.

Suppose that $b \in f(I)$ and that $\epsilon > 0$.

Suppose that $a = f^{-1}(b)$ is an interior point of $I$.

There exist $c, d \in I$ with $a − \epsilon < c < a < d < a + \epsilon$.

Then $f(c) < f(a) = b < f(d)$; let $δ = \min\{b − f(c), f(d) − b\}$. If $|y − b| < δ$, then $f(c) < y < f(d)$, so that $c < f^{-1}(y) < d$, and $|f^{-1}(y) − f^{-1}(b)| < \epsilon$.

The case where a is an end-point of I is left to the reader.

Note that we do not require $f$ to be continuous.

Comment. The strict monotonicity of $f$ is needed because otherwise we can have a saw-tooth function that is continuous but, being not-monotone, its inverse is not defined, because the mapping $f^{-1}$ is not injective.

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  • $\begingroup$ Grazie Mauro, we have eluded the impasse. Nice Book! $\endgroup$
    – Charlie
    Feb 11, 2014 at 14:38
  • $\begingroup$ Why $f$ has to be defined on an interval? $\endgroup$
    – mxdxzxyjzx
    Jul 17, 2020 at 1:26
  • $\begingroup$ @mxdxzxyjzx, if we omit this assumption then it would be possible to apply the theorem to functions like: $$ f(x)=\begin{cases}x, & 0\leq x <1\\x-2, & 3\leq x\leq4 \end{cases}.$$ However, the inverse function of $f$ is not continuous at point $1$. Moreover, we need $I$ to be an interval in order to apply the intermediate value theorem. $\endgroup$
    – Philipp
    Feb 18, 2021 at 23:22
  • $\begingroup$ I have two questions: 1. It seems that the domain of $f$ (in Philipp's comment) is $[0,1)\cup[3,4]$, so I am having trouble seeing why $f$ is discontinuous at 1. 2. Is the following not necessarily true?: "Let $f:[0,\infty)\to[0,\infty)$ be continuous and strictly monotonically increasing. Then $f^{-1}$ is continuous." $\endgroup$ Jun 21, 2022 at 21:18

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