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I read this answer: Continuity of an inverse function. and it got me thinking about the end points. Will the following proof suffice provided f cts inj and we're in $\mathbb{R}$? Injective $ \implies$ $ \uparrow or \downarrow$ so WLOG assume $ f \uparrow$. Then let the following be true: $$ y_{0} = f(a), x \in f(I), \forall \epsilon>0$$ Then if $a-\epsilon < a < a+\epsilon$ clearly $f(a-\epsilon)<f(a)<f(a+\epsilon)$by continuity of f so let a $\delta$ be chosen to be the $\min (f(a)-f(a-\epsilon),f(a+\epsilon)-f(a))$. Then necessarily $\forall a, f(a-\epsilon)<f(a)-\delta$ and $f(a+\epsilon)>f(a)+\delta$ . Then $\forall y$ such that $f(a)-\delta<y<f(a)+\delta$, clearly $ f(a-\epsilon)<y<f(a+\epsilon) \implies a-\epsilon<f^{-1}(y)<a+ \epsilon$ and then $\lvert y-f(y_{0}) \rvert < \delta \implies \lvert f^{-1}(y)-f^{-1}(y_{0})\rvert< \epsilon $.

Christheyankee
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  • okay what was the downvote for...? – Christheyankee Nov 25 '19 at 05:39
  • I'm not the one who downvoted this, but I think you got a downvote because what you wrote is confusing. 1) What is "$y = f(a), x \in f(I), \forall \epsilon>0$" supposed to mean? 2) Next line, you write "if $a-\epsilon < a < a+\epsilon$" : isn't that always the case? You should double check your post! :) – Olivier Roche Nov 25 '19 at 08:33
  • @OlivierRoche I was just making some notation I was going to take advantage of explicit – Christheyankee Nov 25 '19 at 08:55
  • @OlivierRoche I edited the y's to make what I was saying more clear. – Christheyankee Nov 25 '19 at 09:00
  • Your goal is to show that $\forall y_0 , \forall \epsilon > 0 , \exists \delta>0, , \forall y \ |y-y_0|<\delta \Rightarrow |f^{-1}(y) - f^{-1}(y_0)|<\epsilon$. Hence, the first thing to do is to pick an arbitrary $y_0$ and an arbitrary $\epsilon>0$. Once you've picked $y_0$ and $\epsilon$, your next task is to exhibit some $\delta > 0$ that makes things work. – Olivier Roche Nov 25 '19 at 09:35
  • Erratum : please read "$\forall y_0 \in f(I), \forall \epsilon > 0 , \exists \delta>0, , \forall y \in f(I) \ |y-y_0|<\delta \Rightarrow |f^{-1}(y) - f^{-1}(y_0)|<\epsilon$ ". – Olivier Roche Nov 25 '19 at 09:49

1 Answers1

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Let $I,J$ be intervals of $\mathbb{R}$ whose interior isn't empty and $f : I \mapsto J$ a continuous and injective function. Let's show that $f^{-1} : f(I) \mapsto I$ is continuous. I leave you as an exercise to prove each of the claims bellow.

$f$ is either strictly increasing or strictly decreasing. WLOG $f$ is strictly increasing.
Let $y_0 \in f(I)$ and $\epsilon > 0$, say $y_0 = f(a)$. We distinguish 3 cases (the endpoints correspond to case 2 and case 3) :

Case 1. $y_0$ belongs to the interior of $f(I)$. Then, there are $c,d \in I$ whith $a - \epsilon < c < a < d < a + \epsilon$. Take $\delta_1 := \min (y_0 - f(c);f(d) - y_0)$.

claim : $\forall y \in f(I) \ |y-y_0|<\delta_1 \Longrightarrow |f^{-1}(y) - f^{-1}(y_0)| < \epsilon$

Case 2. $y_0 = \max f(I)$. Then, $a = \max I$. Since the interior of $I$ isn't empty, there's $c \in I$ with $a - \epsilon < c < a$. Take $\delta_2 := y_0 - f(c)$.

claim : $\forall y \in f(I) \ |y-y_0|<\delta_2 \Longrightarrow |f^{-1}(y) - f^{-1}(y_0)| < \epsilon$

Case 3. $y_0 = \min f(I)$. Then, $a = \min I$. Since the interior of $I$ isn't empty, there's $d \in I$ with $a < d < a + \epsilon$. Take $\delta_3 := f(d) - y_0$.

claim : $\forall y \in f(I) \ |y-y_0|<\delta_3 \Longrightarrow |f^{-1}(y) - f^{-1}(y_0)| < \epsilon$

Olivier Roche
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