TITLE: Bijection between closed uncountable subset of $\Bbb R$ and $\Bbb R$.
QUESTION [17 upvotes]: This is probably a really stupid question but I'm hoping it's true: can we find a bijection between any uncountable closed subset of $\Bbb R$ and $\Bbb R$ (real number line)?

REPLY [4 votes]: Here is a proof that does not use any transfinite induction at all!

Take any uncountable closed subset $S$ of $\mathbb{R}$.
Let $S_n = S \cap [n,n+1]$ for each integer $n$.
Then $S_n$ is uncountable for some integer $n$, so we can assume that $S \subseteq [a,b]$ for some reals $a,b$.
Let $p = \sup \{ x : x \in [a,b] \land \text{$S \cap [a,x)$ is countable} \}$, which exists because $S \cap [a,a)$ is countable.
Then $S \cap [a,p) = \bigcup_{n=1}^\infty S \cap [a,p-\frac1n)$ is countable.
Similarly let $q = \inf \{ x : x \in [a,b] \land \text{$S \cap (x,b]$ is countable} \}$.
Then $S \cap (q,b]$ is likewise countable.
Clearly $p < q$, otherwise $S$ would be countable.
Let $r = \frac12(p+q)$. Then $S \cap [a,r]$ and $S \cap [r,b]$ are both uncountable, by definition of $p$ and $q$.
Thus $S \cap [p,r]$ and $S \cap [r,q]$ are uncountable closed subsets of $[p,q]$, and $r-p = q-r < \frac12(b-a)$.
Therefore we can iterate this process starting from $[0,1]$. Let $T$ be the binary tree with root $[0,1]$ where each node $[a,b]$ has children $[p,r]$ and $[r,q]$ with length at most half their parent's, such that $S \cap [p,r]$ and $S \cap [r,q]$ are uncountable.
For every downward path $P$ in $T$:
  $P$ is a sequence of nested closed intervals with length decreasing to zero.
  Let $c$ be the unique point in their intersection.
  If $c \notin S$:
    Let $u = \inf \{ x : x \in [0,c] \land S \cap [x,c] = \varnothing \}$.
    Let $v = \sup \{ x : x \in [c,1] \land S \cap [c,x] = \varnothing \}$.
    Then $c \in (u,v)$ and $S \cap (u,v) = \varnothing$.
    But eventually the interval along $P$ will have length less than $\min(c-u,v-c)$.
    Thus some interval $I$ in $P$ is contained in $(u,v)$, contradicting $S \cap P$ being uncountable.
  Therefore $c \in S$.
  Also $c$ is different for any downward path in $T$ that is different from $P$.
Therefore $S$ has at least as many elements as there are countably infinite binary sequences.
Thus the reals inject into $S$, and hence there is a bijection from $S$ to the reals.

As pointed out by Andrés E. Caicedo in the comments, my initial comment was false, because this proof uses countable choice twice, via the theorem that a countable union of countable sets is also countable.