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 $U \in {\cal E}\setminus \{\emptyset\}$?



Picking a real for every non-empty open set in $\mathbb{R}$
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
$$
U\backslash\{f(V):V\leq U\}.
$$


