# GRE 0568 #65: Disprove 'There is a continuous one-to-one function from $(0,1)$ onto $[0,1].$'

GRE 0568 #65

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?

## NEW PROOF:

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?

## UPDATE:

$$(III)$$ 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.)

• What makes you think that all continuous bijections are homeomorphisms? – Kavi Rama Murthy Oct 25 '18 at 9:54
• 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. – астон вілла олоф мэллбэрг Oct 25 '18 at 9:57
• @KaviRamaMurthy OOHHHH LOL OK THANKS!!! – BCLC Oct 25 '18 at 9:59
• @астонвіллаолофмэллбэрг OOHHHH LOL OK THANKS!!! – BCLC Oct 25 '18 at 9:59
• No, that doesn't prove that $f$ is a homeomorphism. – 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.

• Follow up: Wait but $f^{-1}$ is a closed map, and that could be argued...? – BCLC Oct 25 '18 at 10:10
• Note: I updated my post. – BCLC Oct 25 '18 at 10:17
• 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. – астон вілла олоф мэллбэрг Oct 25 '18 at 10:26
• Lol had a feeling it was something like that hehehe THANKS! – BCLC Oct 25 '18 at 10:31
• You are welcome! – 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 $$0 be such that $$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.

• 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.' – 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.