Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb R^2$?

share|improve this question
6  
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. –  Matt Pressland Nov 1 '12 at 14:15
3  
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

3 Answers 3

up vote 10 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)

share|improve this answer
    
Out of curiosity, what do you mean with "waving hand part"? –  Arthur Nov 1 '12 at 14:30
3  
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 $f(p_1)$).

share|improve this answer

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.

share|improve this answer
    
Thats a good point –  Amr Nov 17 '12 at 17:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.