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Show that for a continuous, increasing function $f : \mathbb R \rightarrow [0,1]$ that is onto $(0,1)$ it's inverse $f^{-1}$ is also continuous.

What I've tried

My first approach:

$f^{-1}$ is continuous if $f$ maps open sets to open sets. Since $f$ is increasing it is a one-to-one mapping and it suffices to show that $f$ maps open intervals to open intervals. Let $V \subset [0,1]$ so that $f^{-1}(V)$ is open (since $f$ is continuous) and $f(f^{-1}(V))=V$ which is also open. All that remains to show is that every open interval $U = f^{-1}(V)$ for some open interval $V$. This is true since $f$ is a one-to-one mapping.

My second approach:

$f^{-1}$ is continuous if there exists a $\delta$ such that when:

$$|\alpha - \alpha'| < \delta$$

we have:

$$|f^{-1}(\alpha) - f^{-1}(\alpha')| < \epsilon$$

Call the first event $A$ and the second event $B$. To show $A \implies B$ it is sufficient to show that $B^c \implies A^c$. Let $\alpha = f(x),\, \alpha' = f(x')$ for some $x,x'$ so that we need to show for each $\epsilon$: $|x-x'| \geq \epsilon \implies |f(x) - f(x')| \geq \delta$ for some $\delta$. But this is clear since $f$ is strictly increasing.

I could use some feedback on my reasoning for both approaches.

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  • $\begingroup$ Why does this map even have an inverse? $\Bbb R$ is not compact, while $[0,1]$ is, so $f$ cannot possibly be a homeomorphism... $\endgroup$ – PrudiiArca Feb 13 at 18:46
  • $\begingroup$ I updated the question to specify that $f$ is onto $(0,1)$ not onto $[0,1]$. Does that resolve your point? $\endgroup$ – dmh Feb 13 at 18:51
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    $\begingroup$ Kind of. You still won't get a homeomorphism of cause, but you might have a retraction $r:[0,1] \rightarrow \Bbb R$ (I would refrain from calling it $f^{-1}$ though, since it is not an inverse as $fr \neq id_{[0,1]}$). $\endgroup$ – PrudiiArca Feb 13 at 18:58
  • $\begingroup$ I see, that's right it's not an inverse on the whole range, only on $(0, 1)$. Although in this case that's not essential to the question. $\endgroup$ – dmh Feb 13 at 19:00
  • $\begingroup$ I think your first argument that $f^{-1}$ is continuous iff $f$ is open does not work, as it is phrased like it holds for arbitrary continuous injections. It would be true, if we had a bijection, but as mentioned this is not the case here. Maybe it holds true for injective continous maps with dense image, but then this should be part of your argument... $\endgroup$ – PrudiiArca Feb 13 at 19:41
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I think the first argument is broken: it is not true that for a continuous map $m:X\rightarrow Y$ with a retraction function $r:Y \rightarrow X$ the map $r$ is continuous iff $m$ is open. The problem is that if $m$ is not surjective onto $Y$ preimages of open sets $U \subseteq X$ under $r$ might contain elements not in the image of $U$ under $m$. We have canonically $m(U) \subseteq r^{-1}(U)$, but the latter needs not be open.

I can give you a counterexample: Consider the topological spaces $X=\{\circ,\square\}$ and $Y=\{\circ,\square,\bullet\}$, where $\circ,\square$ denote open points, $\bullet$ is a closed point and we consider the topologies generated by this conditions. Then the obvious inclusion is a continous and open injection $m:X\rightarrow Y$ and we can define the retraction $r:Y \rightarrow X$ doing the obvious things and sending $\bullet$ to $\circ$. The latter is not continous though as the preimage $r^{-1}(\{\circ\}) = \{\circ,\bullet\}$ is not open in the topology on $Y$.

Edit I think the statement itself is wrong (if you use $[0,1]$). The function $\operatorname{arctan}:\Bbb R \rightarrow (-\frac{\pi}{2}, \frac{\pi}{2}) \subseteq [-\frac{\pi}{2},\frac{\pi}{2}]$ is a continuous strictly increasing function, which is not surjective. If it had a continuous retraction $r:[-\frac{\pi}{2},\frac{\pi}{2}]\rightarrow \Bbb R$ we would have in particular $r\vert_{(-\frac{\pi}{2},\frac{\pi}{2})} = \tan$ and continuity would force $$r(\frac{\pi}{2})=\lim \limits_{x \rightarrow \frac{\pi}{2}} r(x) = \lim \limits_{x \rightarrow \frac{\pi}{2}} \tan(x) = \infty \in \Bbb R,$$ which is absurd.

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  • $\begingroup$ I see, thanks. I'll need to go through this closely. But the point I took away is that the first claim in the first argument breaks down when $x \neq m(r(x))$. In particular I need to be careful with how I define the retraction function $r$ over the domain. I think if I restrict the range of $f$ to $(0, 1)$ then that handles the issue, right? $\endgroup$ – dmh Feb 13 at 20:30
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    $\begingroup$ Exactly. If the codomain of $f$ is $(0,1)$ you are speaking of a continuous bijection, in which case the inverse being continuous is equivalent to $f$ being open. $\endgroup$ – PrudiiArca Feb 13 at 21:40
  • $\begingroup$ Thanks! I'm pretty unsure about the second argument. Were you able to look through it? $\endgroup$ – dmh Feb 13 at 23:01
  • $\begingroup$ Well you got the $\varepsilon-\delta$ semi-correct. A function $g$ is continuous, if forall $\varepsilon>0$ there exists $\delta>0$ st forall $x,y$ it holds that $\vert x-y\vert<\delta$ implies $\vert g(x)-g(y)\vert<\varepsilon$. All of this (in particular the order of the quantifiers!) is important. Now of cause you might use the contrapositive implication $\vert g(x)-g(y)\vert>\varepsilon$ implies $\vert x-y\vert>\delta$. But again, if you want to invoke monotonicity of $f$ you need to know that any $x\in [0,1]$ is given by $x=f(t)$ for some $t\in \Bbb R$, which is not the case $\endgroup$ – PrudiiArca Feb 14 at 0:01

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