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Suppose $f:[0,1)\to (0,1)$ is bijective. Prove that $f$ is not continuous.


I know that $(0,1)$ and $[0,1)$ are not homeomorphic spaces because of the connected property. Can we then conclude that $f$ is not continuous because of that?

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  • $\begingroup$ There have been many questions asked already related to this $\endgroup$
    – TripleA
    Commented Dec 28, 2016 at 1:10
  • $\begingroup$ If $f$ is continuous, then the preimages of open sets are open. Under this $f$, is the preimage of (0,1) an open set? $\endgroup$
    – JuliusL33t
    Commented Dec 28, 2016 at 1:11
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    $\begingroup$ @JuliusL33t: the preimage of $(0,1)$ is open in $[0,1)$. $\endgroup$
    – user9464
    Commented Dec 28, 2016 at 1:14
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    $\begingroup$ @Jack: wow, I'm almost a little ashamed, of course you're right, my bad. $\endgroup$
    – JuliusL33t
    Commented Dec 28, 2016 at 1:28
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    $\begingroup$ Simply, $(0,1)$ gets disconnected by the removal of $f(0)$, while $[0,1)$ is still connected if we remove $0$. Continuous maps must send connected sets into connected sets. $\endgroup$ Commented Dec 28, 2016 at 1:52

6 Answers 6

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Assume $f(0) = a \in (0,1).$ Since $f$ takes values both less than and greater than $a$ on the interval $(0,1)$ (by surjectivity), the intermediate value theorem implies that it takes the value $a$ again on the interval $(0,1),$ contradicting injectivity.

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Suppose otherwise such map exists. Let $f:[0,1)\to(0,1)$ be a continuous bijection. Then we have an induced bijection $$ g:(0,1)\to(0,1)\setminus\{f(0)\} $$ with $g=f|_{(0,1)}$.

Note that $g$ is also continuous. But this is impossible since continuous map preserve connectedness while $(0,1)\setminus\{f(0)\}$ is not connected.

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Here is yet another way.

Suppose $f$ is a continuous function, not necessarily bijective, from $[0,1)$ to $(0,1)$. We can show that $f$ can't be bijective.

Since the continuous image of a connected set is connected and the only connected subsets of $(0,1)$ are intervals, the image of $(0,1)$ must be some interval contained in $(0,1)$. If it were the entirety of $(0,1)$, we'd have nowhere to put $f(0)$, so $f$ wouldn't be injective. If the image of $(0,1)$ were some strict subinterval $I$ of $(0,1)$, $f(0)$ couldn't possibly hit every point in the complement of $I$ in $(0,1)$, so $f$ wouldn't be surjective. Either way, $f$ can't be bijective.

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The result below shows that $f$ is strictly monotonic. Suppose that $f$ is strictly increasing, $f(0)\in (0,1)$ implies there exists $y<f(0)$, $y=f(x)$ and $x>0$ contradiction.

If $f$ is strictly decreasing, you have $f(0)<y=f(x)$, $x>0$ contradiction.

https://proofwiki.org/wiki/Continuous_Injection_of_Interval_is_Strictly_Monotone

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continuous bijections between intervals of real numbers are monotonic, and so their inverses are also continuous.

So if $f$ was a continuous bijection it would be a homeomorphism, but $[0,1)$ and $(0,1)$ are not homeomorphic, as the first interval has a point which does not disconnect the set upon removal.

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Let $f(0)=a$, and let $p=f^{-1}(a/2)$ and $q=f^{-1}(1-a/2)$. Suppose WLOG that $0<p<q<1$. Then $f|_{[p,q]}\colon[p,q] \to (0,1)$ is continuous. Given that $f(p)<a$ and $f(q)>a$, there exists an element $z\in[p,q]$ such that $f(z)=a$ by the intermediate value theorem. This contradicts the fact that $f$ is a bijection and $f(0)=a$.


I should add that you don't need all the power of the intermediate value theorem here, if you don't want. You can make do with the fact that the continuous image of a connected set is connected, and then note that $f|_{(0,1)}$ has a disconnected image by bijectivity and the fact that $(0,a)\cup (a,1)$ is not connected.

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