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Here @Ian says there is a particular property of $\mathbb R$ and intervals that prevents a hypothetical continuous injective function from $\mathbb R$ onto $[-1, 1]$ from having a discontinuous inverse. What is this property?

  • Since there is no such continuous injective onto function, that seems a bit hypothetical – Hagen von Eitzen Aug 19 '20 at 14:00
  • @HagenvonEitzen If you notice, Ian uses it specifically to argue that such a function doesn't exist. Do you have some alternative justification for why there cannot be such a continuous injective function from $\mathbb R$ onto $[-1, 1]$? –  Aug 19 '20 at 14:01
  • We are talking about compactness. The image of a compact set under a continuous function is compact again. One can show that $[-1,1]$ is compact, but $\mathbb{R}$ is not. – Severin Schraven Aug 19 '20 at 14:03
  • @SeverinSchraven Sure. But $\mathbb R$ is not compact here while $[-1, 1]$ is. We can directly conclude that $\mathbb R$ cannot be the continuous image of $[-1, 1]$, but I don't see any direct argument that $[-1,1]$ cannot be the continuous image of $\mathbb R$. –  Aug 19 '20 at 14:05
  • Assume w.l.o.g., that $f(0) < f(1)$ (if not replace $f$ by $-f$). It follows from the intermediate value theorem that $f(x) < f(y)$ whenever $x < y$. But, if $f$is onto, there is some $x$ with $f(x) = 1$, but then what is $f(x + 1)$? – Rob Arthan Aug 19 '20 at 14:11

2 Answers2

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A continuous injective map $f$ between two real intervals is monotonous. If $f$ is also onto, the direct image of any open interval is an open interval (for the induced topology on $[-1,1]$).

Hence the inverse image of any open interval under $f^{-1}$ is open. Proving that $f^{-1}$ is continuous.

mathcounterexamples.net
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  • Interesting! Can you point me to a proof of the fact that the continuous injective map between two real intervals is monotonous? –  Aug 19 '20 at 14:06
  • This is a direct consequence of the intermediate value theorem and a proof by contradiction. See here. – mathcounterexamples.net Aug 19 '20 at 14:11
  • Hint: try to do the proof yourself. draw a picture of three points $a,b,c$ on the graph of $f$ such that $f(a)<f(b)$ and $f(b)>f(c)$ and apply IVT – Matematleta Aug 19 '20 at 14:19
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Such $f$ cannot exist. Consider $f(x)=1$, let $a<x<b$, suppose $f(a)<f(b)$, $f([a,x])$ is an interval since the image of a connected set by a continuous map is connected, it contains $f(a)$ and $f(x)=1$, since $f(a)<f(b)<1$, it contains $f(b)$. There exists $c\in [a,x]$ such that $f(x)=f(b)$. contradiction.

If $f(b)<f(a)$, $f([b,x])$ is an interval it contains $f(a)$ contradiction.

Tsemo Aristide
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