How to show a injective continuous function $f$ from $[0,1)$ to $\Bbb R$ has a continuous inverse?

Asked Dec 28 '20 at 12:44

Active Dec 28 '20 at 12:44

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Let $f:[0,1)\to \Bbb R$ be a one-to-one continuous function and let $A$ be the range of $f$. Show that the inverse function $f^{-1}:A\to [0,1)$ is continuous.

I know the case when $f:[0,1] \to \Bbb R$, but it is $[0,1)$. Hope someone could give me a hint how to prove it.

What I have done: Let $U$ be an open set in $[0,1)$, I need to show that $(f^{-1})^{-1}(U)=f(U)$ is open in $A$, but I don't know how to show that $f(U)$ is open in $A$.

asked Dec 28 '20 at 12:44

Steven Lu

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Does this help? – Aven Desta Dec 28 '20 at 12:49
1

Either $f$ extends to $[0,1]$ or it doesn't. Characterize the situations and proceed separately. – Andrea Mori Dec 28 '20 at 12:51
@AndreaMori Do you mean either $f$ has a continuous extension to $[0,1]$ or not? – Steven Lu Dec 28 '20 at 12:53
@StevenLu: of course! There is always a non-continous extension. – Andrea Mori Dec 28 '20 at 12:58
@AndreaMori I think if $f$ has a continuous extension to $[0,1]$, then $f^{-1}$ is continuous since $[0,1]$ is compact. If $f$ has no continuous extension, then I think $\lim \limits_{x\to 1}f(x)= \infty \text{or} -\infty$. – Steven Lu Dec 28 '20 at 13:24
In classifying the possibilities, it might help to realize that if $f$ is continuous and $1$-$1$, then so is $\tan^{-1}\circ f$. – Paul Sinclair Dec 28 '20 at 20:07

How to show a injective continuous function $f$ from $[0,1)$ to $\Bbb R$ has a continuous inverse?

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