Two questions:
Find a bijective function from $(0,1)$ to $[0,1]$. I haven't found the solution to this since I saw it a few days ago. It strikes me as odd--mapping a open set into a closed set.
$S$ is countable. It's trivial to find a bijective function $f:\mathbb{N}\to\mathbb{N}\setminus S$ when $|\mathbb{N}| = |\mathbb{N}\setminus S|$; let $f(n)$ equal the $n^{\text{th}}$ smallest number in $\mathbb{N}\setminus S$. Are there any analogous trivial solutions to $f:\mathbb{R}\to\mathbb{R}\setminus S$?
Answer
The proof of the Schroeder-Bernstein theorem allows you to get a bijection for 1, since we have an injection $(0,1) \to [0,1]$ and a bijection $f: [0,1] \to [1/4, 3/4] \subset (0,1)$ (say $x \to x/2 +1/4$). The function's definition will be somewhat messy (basically, it depends on how many times you can lift a point under these to injections already defined, and specifically the parity of the number of times), but it'll do it.
For 2, iterate this construction to get a bijection $R \to R - N$. Then given any countable set $S$, define the map of $R$ that interchanges $N$ and $S$ and leaves every other point fixed. Then the composition of the first bijection with this second map is your bijection.
Continuity considerations imply that the map can't be continuous: in 1, for instance, we'd otherwise have that $(0,1)$ is compact, which it's not.
No comments:
Post a Comment