GRE 0568 #65

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This has been asked about here: Why are there no continuous one-to-one functions from (0, 1) onto [0, 1]?

There are proofs by:

None of the proofs are like mine, so I guess my proof is somehow wrong:

Such $f$ in $(III)$ would be a homeomorphism and thus $f^{-1}$ is a map that is a homeomorphism too. In particular, $f^{-1}$ is a map and is continuous. I believe now the same argument that disproves $(II)$, also disproves $(III)$:

In $(II)$, we must have $g(\text{compact}) \ \text{is compact}$ ($g$ is the supposed function in $(II)$). Now, we must have $f^{-1}(\text{compact}) \ \text{is compact}$.

Where did I go wrong ($\emptyset$ is an answer), and why/why not?

UPDATE: $f$ isn't necessarily an open map.


That $f$ is continuous and bijective, I believe implies that $f^{-1}$ is a closed map. This is a contradiction because the image of $[0,1]$ under $f^{-1}$ should be closed, but we instead get $(0,1)$.

Is this still wrong?


$(III)$ can be understood in trying to construct the continuous function in $(I)$: It will never be injective. (I didn't see at first $(III)$ was just $(I)$ with injective.)

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    $\begingroup$ What makes you think that all continuous bijections are homeomorphisms? $\endgroup$ – Kavi Rama Murthy Oct 25 '18 at 9:54
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    $\begingroup$ Continuous bijections are not homeomorphisms. Continuous bijections with continuous inverses are homeomorphisms. This makes the first assertion of your proof, namely that $f$ in $III$ is a homeomorphism, incorrect. $\endgroup$ – астон вілла олоф мэллбэрг Oct 25 '18 at 9:57
  • $\begingroup$ @KaviRamaMurthy OOHHHH LOL OK THANKS!!! $\endgroup$ – BCLC Oct 25 '18 at 9:59
  • $\begingroup$ @астонвіллаолофмэллбэрг OOHHHH LOL OK THANKS!!! $\endgroup$ – BCLC Oct 25 '18 at 9:59
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    $\begingroup$ No, that doesn't prove that $f$ is a homeomorphism. $\endgroup$ – Kavi Rama Murthy Oct 25 '18 at 10:16

I think it is worth giving a counterexample to the assertion : "continuous bijections are homeomorphisms".

The example is a rather simple one : the identity map is always a bijection. If you take two topologies on a set, one which is (strictly) coarser than the other, then the identity map will be continuous precisely in one direction : from the topology with more open sets to the one with less open sets. The other way, if you take a pullback of an open set which is not in the smaller topology, it won't be open in the smaller topology, hence continuity is not possible.

For example, take $\mathbb R$ with the usual topology, and say $\mathbb R$ with indiscrete/discrete topology with the identity map.

This is an example of a continuous bijection which is NOT a homeomorphism. There are conditions under which a continuous bijection is a homeomorphism (compact to Hausdorff), but it is not true in general.

It is true that $f^{-1}$ is closed : in fact, the fact that $f^{-1}$ carries closed sets to closed sets is equivalent to the continuity of $f$. Therefore, we do not expect this fact to be of help to us in the question. I will expand.

When we look at $[0,1]$ and $(0,1)$ as topological spaces, it is with the subspace topology derived from $\mathbb R$. That is, a description of "open" or "closed" in each of these topologies, is given by the intersection of the set with a set that is open or closed in $\mathbb R$ respectively.

For example :

  • $[0,1]$ is closed and open in the $[0,1]$ topology, because $[0,1] = [0,1] \cap \mathbb R$, and so it is the intersection of $[0,1]$ with a set both open and closed in $\mathbb R$.

  • Similarly, $(0,1)$ is both open and closed in the subspace topology derived from $\mathbb R$ (in the subspace topology, not the one on $\mathbb R$).

With this in mind, since the map is from $(0,1)$ to $[0,1]$, the fact that $f^{-1}([0,1]) = (0,1)$ is closed is true, and not a contradiction : this is because $f$, as defined as a topological map, is operating with the subspace topologies, not those directly derived from $\mathbb R$. That is the problem with your logic.

  • $\begingroup$ Follow up: Wait but $f^{-1}$ is a closed map, and that could be argued...? $\endgroup$ – BCLC Oct 25 '18 at 10:10
  • $\begingroup$ Note: I updated my post. $\endgroup$ – BCLC Oct 25 '18 at 10:17
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    $\begingroup$ I have also updated my post. You can clarify things which you do not understand, since I believe this takes you into slightly alien territory. $\endgroup$ – астон вілла олоф мэллбэрг Oct 25 '18 at 10:26
  • $\begingroup$ Lol had a feeling it was something like that hehehe THANKS! $\endgroup$ – BCLC Oct 25 '18 at 10:31
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    $\begingroup$ You are welcome! $\endgroup$ – астон вілла олоф мэллбэрг Oct 25 '18 at 10:32

Overkill solution: by the invariance of domain theorem (1-dimensional version), a 1-1 continuous function $f$ on an open subset $U$ of $\mathbb{R}$ into $\mathbb{R}$ is open, and in particular $f[U]$ is open too.

$[0,1]$ is not open, so no such map can exist onto $[0,1]$.

Other more elementary solution: Suppose $f: (0,1) \to [0,1]$ is 1-1, continuous and onto. Let $0<t_0,t_1 < 1$ be such that $f(t_0) = 0, f(t_1)=1$. If e.g. $t_0 < t_1$, then $f[t_0,t_1] = [0,1]$ and otherwise ($t_1 < t_0$), $f[t_1,t_0]=[0,1]$ by the intermediate value theorem. Now in the first case any $x \in (0,t_0)$ will assume an image already assumed by a different point from $[t_0,t_1]$, contradicting 1-1-ness of $f$. Otherwise the same holds for an $x \in (0,t_1)$ and $[t_1,t_0]$ resp. In either case we see that any continuous map $f$ from $(0,1)$ to $[0,1]$ that assumes the values $0$ and $1$ cannot be 1-1.

  • $\begingroup$ Thanks Henno Brandsma! 1. Why wouldn't астон вілла олоф мэллбэрг's answer about the closed map contradict your answer? 2. Why doesn't Wikipedia imply $f$ is indeed a homeomorphism as I (incorrectly) deduced? 'f is a homeomorphism between U and V.' $\endgroup$ – BCLC Oct 25 '18 at 14:28

A continuous 1-1 map from (0, 1) to {\mathbb R} is order preserving or order reversing. Assume order preserving. If c maps to 0 then any d in (0, 1) less than c would map to a point less than 0 in [0,1]. There are no such.


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