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 $\mathbb R^2$?
Best Answer
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)