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I attended a lecture today in which we were given the proof of non-existence of homeomorphisms between $\mathbb R$ and $\mathbb R^2$. I came up with the following bijection between $\mathbb R$ and $\mathbb R^2$ but could not prove why this doesn't qualify as a valid homeomorphism.

Map $(x,y) \rightarrow (\frac 1 \pi (\tan ^{-1} x + \pi/2), \frac 1 \pi (\tan ^{-1} y + \pi/2))$. This map is continuous, bijective and maps $\mathbb R^2$ to the unit square.

Now, for every pair $(x,y)$ with $x = 0.a_1a_2 \cdots, y = 0.b_1b_2 \cdots$ being their decimal expansions, define $$f(x,y) = 0.a_1b_1a_2b_2 \cdots$$ $f(x,y)$ is a continuous bijection between the unit square and the interval $(0,1)$

Now, just map $x \rightarrow \tan (\pi x - \pi/2)$ to get a continuous bijection between $(0,1)$ and $\mathbb R$.

Why isn't the composition of these functions a homeomorphism between $\mathbb R$ and $\mathbb R^2$?

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Why exactly $f$ is continuous? (Actually, it isn't.) – Grigory M Nov 1 '12 at 14:14
I have the same question as Grigory - I can think of a heuristic reason why $f$ might be continuous, but I guess that in fact it isn't. – Matthew Pressland Nov 1 '12 at 14:15
Not related to the actual problem here, but note for future reference that not every continuous bijection is a homeomorphism! One counterexample is that any bijection from $\mathcal P(\mathbb N)$ with the discrete metric to $\mathbb R$ is continuous, but neither of these are homeomorphisms, because their inverses are not continuous. – Henning Makholm Nov 1 '12 at 14:35
@Henning: Another good example of this is the function $f: [0, 1) \rightarrow C$ where $C$ is the unit circle in the complex plane, and $f(z) = e^{2 \pi i z}$. This is a continuous bijection whose inverse is not continuous. – Benjamin Dickman Nov 1 '12 at 15:50
up vote 11 down vote accepted

Your middle function is not continuous. Let $x_0=0.1$, $y_n=0.4\overbrace{9\cdots9}^{n\text{ times }}0\cdots$. Then $(x_0,y_n)\to(0.1, 0.5)$. We have $$ f(0.1,0.5)=0.15,\ \ f(0.1,y_n)=0.14090909\cdots $$ So $$ |f(0.1,0.5)-f(0.1,y_n)|>0.009 $$ for all $n$.

As a general comment, every time your proof includes a "waving hand part", you should suspect that part. A lot. (said from extensive own experience)

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Out of curiosity, what do you mean with "waving hand part"? – Arthur Nov 1 '12 at 14:30
Anything where you've looked at it and said "I'm sure that's true", but haven't bothered to check. – Cameron Buie Nov 1 '12 at 14:42

Your function $f$ is not continuous. If it were, then fixing $y$ to some arbitrary value is would give a strictly increasing continuous function of $x$ from the unit interval to itself, whose image would have to be an interval; however this cannot be the case given the fact that half of the decimal places have unchanging digits throughout the image of the map (this restriction of $f$ very serverly violates the intermediate value theorem). In fact you can check that $f$ is discontinuous in any point at least one of whose two coordinates has a finite decimal representation. For such coordinates you have to choose one of the two possible decimal representations to work with in the definition of $f$; assuming you choose the finite one (rather than the one that ends with all digits $9$), you get a discontinous jump when you lower that coordinate.

You may note that your function $f$ also fails to be surjective, for essentially the some reason (for instance $0.3919491959992969593959\ldots$ is not in the image). As a consolation, it does happen to be continuous almost everywhere.

I may add that the intermediate value theorem consideration shows that for a map $f:\mathbf R^2\to\mathbf R$ (or $(0,1)^2\to(0,1)$), even the weaker requirement of just being both continuous and injective cannot be met. It suffices to consider two distinct points $p_0,p_1\in\mathbf R^2$, where we may assume $f(p_0)<f(p_1)$, and two paths in $\mathbf R^2$ from $p_0$ to $p_1$ that are disjoint except for the end points. The restriction of $f$ to one such path is still continuous, so the image of this restriction must contain all of the interval $[f(p_0),f(p_1)]$ by the intermediate value theorem. But since this holds for both paths, we get a contradiction with injectivity (for every value between $f(p_0)$ and $f(p_1)$).

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Note that $0.4\bar{9}=.5$, but $$f(0.4\bar{9},0.4\bar{9})=0.44\bar{9}=0.45\neq 0.55=f(0.5,0.5),$$ so you need to be a bit more cautious with your interleaving to get a function. Even if you give a convention for how to choose the decimal expansion, this function still won't be continuous. Fix one of the arguments and play with it to see why.

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Thats a good point – Amr Nov 17 '12 at 17:38

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