Construct a bijection from $\mathbb{R}$ to $\mathbb{R}\setminus S$, where $S$ is countable

Question

Two questions:

1. 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.

2. $S$ 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 1

An explicit bijection $(0,1) \to [0,1]$ for part $1$ is given by:

$$f\left(\frac{1}{2}\right) = 0,\quad f\left(\frac{1}{3}\right) = 1,\quad f\left(\frac{1}{n}\right) = \frac{1}{n-2}\ \textrm{for}\ n > 3,$$

$$f(x) = x\ \textrm{for}\ x\ \textrm{not equal to a reciprocal of an integer}$$

For a bijection $\mathbb{R}$ to $\mathbb{R}\setminus S$, we can number the elements of $S$ (because $S$ is countable) as $s_1, s_2, s_3, \dots$ Choose an infinite sequence $t_1, t_2, \dots$ of distinct elements of $\mathbb{R}$, none of which are in $S$. Then define:

$$g(s_i) = t_{2i},\quad g(t_i) = t_{2i+1},\quad g(x) = x\ (\textrm{otherwise}).$$

This is a bijection from $\mathbb{R}$ to $\mathbb{R}\setminus S$.

Answer 2

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.

Answer 3

I'm going to give a constructive procedure that hints at generalizations of the cut-and-paste/split function approaches of Akhil/Simon. The top-level idea is that given a countable $S \subseteq \mathbb{R}$ to be deleted, we will find a countably infinite subset $T \subseteq \mathbb{R} \setminus S$. We will then split $T$ into two disjoint sets $A$ and $B$ having size $|S|$ and $|T|$, respectively and identify the deleted $S$ from $\mathbb{R}$ with $A$ and $T$ with $B$. In particular, this will fix all points in $\mathbb{R} \setminus (S \cup T)$ and will send the elements from $S \cup T$ to $A \cup B = T$. The point is that we work with well-orderable subsets of $\mathbb{R}$ which admit nice bijections. (e.g., a well-order order isomporphic to $\mathbb{N}$ can be put into bijective correspondence with $\mathbb{N} \times \{0, 1\}$ by sending the evens $2n$ to $(n, 0)$ and the odds $2n+1$ to $(n, 1)$.) Our desired bijections are thus formed from a cut/paste method where we cut out well-ordered sets, work with bijections of them, and then paste the identity mapping with the induced maps from these bijections as Akhil suggests in his answer and as described below.

(1) First consider any injection $j: \mathbb{N} \rightarrow [0, 1]$ including $0$ and $1$ in the range (e.g., $n \mapsto \frac{1}{n}$ for $n > 0$ and $0 \mapsto 0$). Then we have a $1:1$ correspondence between $\mathbb{N}$ and $C = \text{range}(j)$. Because $C$ includes $0$ and $1$, we have $[0, 1] \setminus C = (0, 1) \setminus C$ so given a bijection $b$ between $C$ and $(0, 1) \cap C$, we can paste together the identity map and $b$ on these respective domains to get a bijection between $[0, 1]$ and $(0, 1)$. Now note that $j$ maps exactly two elements $m, n \in \mathbb{N}$ to the $\text{range}(j)$ that are not in $(0, 1)$, mainly the ones being mapped to $0$ and $1$. Consequently, letting $g: \mathbb{N} \rightarrow \mathbb{N} \setminus \{m, n\}$ be a bijection, say one simply listing the Natural numbers in order that skips $m$ and $n$, you can verify that $j \circ g \circ j^{-1}: C \rightarrow (0, 1) \cap C$ is a well-defined bijection.

(2) For this part, extend your countable bijection $s: \omega \rightarrow S$ to a countable injection $k: \omega + \omega \rightarrow \mathbb{R}$ ($\omega = \mathbb{N}$ and $\omega + \omega$ has an ordering isomorphic to one on $\mathbb{N} \times \{0, 1\}$ given by $(m, i) \leq (n, i)$ when $m \leq n$ and $(m, 0) \leq (n, 1)$ always). The reason such an injection exists is because given a set of Reals that resulted from the subtraction of countably many elements, we can always find a countable number that haven't been removed by iteratively diagonalizing against all the ones in our list, adding each new one obtained by this method to the top of the list so that it too gets diagonalized against in the construction of the next Real not in the list.

Now we have a $1:1$ correspondence between $\omega + \omega$ and $D = \text{range}(k)$. Because $D$ includes all elements in $S$ by construction, we have $\mathbb{R} \setminus D = (\mathbb{R} \setminus S) \setminus D$ so given a bijection $b^*$ between $D$ and $(\mathbb{R} \setminus S) \cap D$, we can paste together the identity map and $b^*$ on these respective domains to get a bijection between $\mathbb{R}$ and $\mathbb{R} \setminus S$. Now note that all elements listed by $k$ at and beyond $\omega$ are in $\mathbb{R} \setminus S$ by the fact that $k$ only lists each element once and already listed all elements in $S$ before $\omega$. Consequently letting $h: \omega + \omega \rightarrow (\omega + \omega) \setminus \omega$ be a bijection, you can verify that $k \circ h \circ k^{-1}: D \rightarrow (\mathbb{R} \setminus S) \cap D$ is a well-defined bijection.

[I realize that this can be technical without knowing about ordinals but just think of $\omega + \omega$ as being the set of Natural numbers with a copy of the Natural numbers on top and $(\omega + \omega) \setminus \omega$ as the top set of Natural numbers. Establishing a bijection between the two sets is similar to establishing a bijection between the set of evens and the set of Natural numbers.]