Picking a real for every non-empty open set in $\mathbb{R}$

Question

Let ${\cal E}$ denote the collection of open sets of $\mathbb{R}$ with respect to the Euclidean topology. It is well known that $|{\cal E}| = 2^{\aleph_0}$. Is there an injective map $f:{\cal E}\setminus \{\emptyset\} \to \mathbb{R}$ such that $f(U)\in U$ for all $U \in {\cal E}\setminus \{\emptyset\}$?

Answer 1

Choose a well-ordering of $\cal{E}\backslash\{\emptyset\}$ with order type the initial ordinal of size $2^{\aleph_0}$. Consider injective functions $f$ from initial segments of $\cal{E}\backslash\{\emptyset\}$ to $\mathbb{R}$ satisfying $f(U)\in U$. The set of such functions is partially ordered by extension, and Zorn's lemma implies there is a maximal such function. The domain of the maximal function $f$ must by all of $\cal{E}\backslash\{\emptyset\}$, for if it weren't, there would be a smallest open set $U$ not in the domain. Now $U$ has cardinality $2^{\aleph_0}$ and the set $\{V\in\cal{E}:V\leq U\}$ has cardinality less than $2^{\aleph_0}$, so we can define $f(U)$ to be any element of $$ U\backslash\{f(V):V\leq U\}. $$

Answer 2

Fix an injective map $h:\mathcal{E}\to\mathbf{R}/\mathbf{Q}$. Lift it to a map $g:\mathcal{E}\to\mathbf{R}$. Then for every $U\in\mathcal{E}$ there exists $r(U)\in\mathbf{Q}$ such that $f(U)=g(U)-r(U)\in U$. Then $f$ is injective, hence is a function as desired.

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Addendum: this can be made without choice. To simplify I'll replace $\mathbf{Q}$ with $\mathbf{Z}[1/2]$. First, there is an explicit injection from $\mathcal{E}$ to $2^{\mathbf{Q}}$ [see edit below]. Then we use a bijection $\mathbf{Q}\to\mathbf{N}$ to get an injection $\mathcal{E}\to 2^{\mathbf{N}}$. Third, we find (below) an explicit map $g:2^{\mathbf{N}}\to\mathbf{R}$ such that the composite map $h:\mathcal{E}\to\mathbf{R}/(\mathbf{Z}[1/2])$ is injective. Then using an enumeration of $\mathbf{Z}[1/2]$, the map $f$ as above can be constructed explicitly.

Here is a construction of $g$ (using the ugly convention $\mathbf{N}=\{1,2,\dots\}$). Define $g(I)=\sum_{n\in I}3^{-n^2}$. Let us show that $h$ is injective. If $I\neq J$, then $t:=g(I)-g(J)=\sum_{n\ge 1}u_n3^{-n^2}$, where $(u_n)$ is not identically zero and takes values in $\{-1,0,1\}$. We have $0<|t|<1/2$. If $(u_n)$ is finitely supported then $t\in\mathbf{Z}[1/3]\smallsetminus\mathbf{Z}$ and hence $t\notin\mathbf{Z}[1/2]$. Otherwise, $t$ has a lacunary ternary expansion and hence is irrational. So in all cases $t\notin\mathbf{Z}[1/2]$ and hence $h$ is injective.

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Edit (May '20): as user Arno told me, the map I proposed, mapping $U$ to $U\cap\mathbf{Q}$, is not injective (just take $U=\mathbf{R}$ vs the complement of an irrational singleton). Let me fix this: define $U_n=\{x\in U:d(x,U^c)> 1/n\}$. Define $\theta:\mathcal{E}\to (2^{\mathbf{Q}})^\mathbf{N}\simeq 2^{\mathbf{Q}\times\mathbf{N}}$ by $\theta(U)=(U_n\cap\mathbf{Q})_{n\in\mathbf{N}}$. I claim that $\theta$ is injective. Indeed, suppose that $U\neq V$, say there exists $x\in U\smallsetminus V$. There exists $n_0$ such that $[x-2/n_0,x+2/n_0]\subset U$. For $n\ge n_0$, there exists a rational $y$ such that $|y-x|< 1/n$. So $y\notin V_n$, while $y\in U_n$. Hence $(U_n)_{n\in\mathbf{N}}\neq (V_n)_{n\in\mathbf{N}}$.

Answer 3

We can get a rather simple such function $f$. If if we measure this in terms of descriptive set theory, we get a Baire class 1 function (we need DST for Quasi-Polish spaces here, as $\mathcal{E} = \mathcal{O}(\mathbb{R})$ is not metric). If we want to do this in a weak axiom system, then $\mathrm{ACA}_0$ would do.

Let us start by fixing some enumeration $(I_n)_{n \in \mathbb{N}}$ of the non-trivial closed intervals with rational endpoints. Now consider $\chi : \mathcal{O}(\mathbb{R}) \to 2^\mathbb{N}$ defined as $\chi(U) = \{n \in \mathbb{N} \mid I_n \subseteq U\}$.

The function $\chi$ is injective and Baire class 1 respectively definable in $\mathrm{ACA}_0$, and this is where all the complexity lies. If instead we would have chosen the type of $\chi$ as being $\chi : \mathcal{O}(\mathbb{R}) \to \mathcal{O}(\mathbb{N})$, then $\chi$ would even have been continuous.

For each interval $I_n$, we pick a finite prefix $w_n$ of a ternary expansion of a real number such that any extension of $w_n$ belongs to $I_n$. As the intervals are non-trivial, such thing always exists.

We can now built $f$ as $f(U) = (w_{\min \chi(U)}2\chi(U))_3$, ie by building the reals via their ternary expansions. This is well-defined whenever $U \neq \emptyset$. By construction of the $w_n$, we have that $f(U) \in U$. We can recover $\chi(U)$ from $f(U)$ ("look after the last 2 in the ternary expansion"), so $f$ inherits being injective from $\chi$. This part of the construction is continuous, so $f$ does not gain any complexity over $\chi$.

Answer 4

With the axiom of choice:
if $\kappa$ is an infinite cardinal, then any collection of (at most) $\kappa$ sets, each of cardinality (at least) $\kappa$, has an injective choice function. In this case $\kappa=2^{\aleph_0}$, and not only the collection of all nonempty open subsets of $\mathbb R$ but even the collection of all uncountable Borel subsets of $\mathbb R$ has an injective choice function.

Without the axiom of choice:
Construct a set $S\subset\mathbb R$ of cardinality $2^{\aleph_0}$ such that $x,y\in S,\ x\ne y\implies x-y\notin\mathbb Q$, and define an injection $h:\mathcal E\to S$. (This can all be done without choice.) Let $\mathbb Q=\{r_n:n\in\mathbb N\}$. Given $U\in\mathcal E\setminus\{\emptyset\}$, find the least $n$ such that $r_n+h(U)\in U$, and define $f(U)=r_n+h(U)$. Then $f(U)=f(V)\implies h(U)-h(V)\in\mathbb Q\implies h(U)=h(V)\implies U=V$.

P.S. Here is a simple way to construct such a set $S$. Enumerate the positive rational numbers as $\{d_n:n\in\mathbb N\}$. For $X\subseteq\mathbb R$ define $D(X)=\{x-y:x,y\in X,\ x\gt y\}$. The following construction can easily be made choice-free by using intervals with rational endpoints.

Construct a closed interval $I$ with $d_0\notin D(I)$; then construct disjoint closed intervals $I_0,I_1\subset I$ with $d_1\notin D(I_0\cup I_1)$; then construct disjoint closed intervals $I_{00},I_{01}\subset I_0$ and disjoint closed intervals $I_{10},I_{11}\subset I_1$ with $d_2\notin D(I_{00}\cup I_{01}\cup I_{10}\cup I_{11})$; and so on. The limiting set $$S=I\cap(I_0\cup I_1)\cap(I_{00}\cup I_{01}\cup I_{10}\cup I_{11})\cap\cdots$$

is a Cantor set with $D(S)\cap\mathbb Q=\emptyset$.

In a comment FrançoisG.Dorais pointed to a more general construction in an answer to the question explicit big linearly independent sets.