Martin's axiom produces an uncountable Q-set

Given: Work in ZFC. Assume $\mathrm{M}\mathrm{A}+\mathrm{\neg }\mathrm{C}\mathrm{H}$ and a subset $E\subseteq \mathbb{R}$ with $\mathrm{\mid }E\mathrm{\mid }={\omega }_1$.

$\mathrm{\neg }\mathrm{C}\mathrm{H}$ says that there is a set $A$ with $\mathbb{N}\prec A\prec \mathcal{P}(\mathbb{N})$ (The continuum hypothesis, and what this page does not prove); under choice every uncountable cardinal is at least ${\omega }_1$, so ${\omega }_1\le \mathrm{\mid }A\mathrm{\mid }<2^{{\mathrm{\aleph }}_0}$ and hence ${\omega }_1<2^{{\mathrm{\aleph }}_0}$ (The successor cardinal ${\kappa }^{+}$, the alephs ${\mathrm{\aleph }}_{\alpha }$, the beths ${\mathrm{\beth }}_{\alpha }$, successor and limit cardinals, and the identifications ${\mathrm{\aleph }}_0=\omega$ and ${\mathrm{\aleph }}_1={\omega }_1$, Cardinal (initial ordinal) and cardinality, The Axiom of Choice). Moreover $\mathcal{P}(\mathbb{N})$ injects into $\mathbb{R}$: send a subset to its characteristic binary sequence, then use the stated bijection from binary sequences onto the Cantor subset of $\mathbb{R}$ (The Cantor set is exactly the set of ${\sum }_{k\ge 1}{a}_k{3}^{-k}$ with every $a_k\in \{0,2\}$, and this gives a bijection with $\{0,1{\}}^{\mathbb{N}}$).

$\mathrm{M}\mathrm{A}$ is the scheme $\mathrm{M}\mathrm{A}(\kappa )$ for every infinite $\kappa <\mathfrak{c}$ (Martin's Axiom at a cardinal and Martin's Axiom).

Solovay's almost-disjoint extension lemma: if $\mathcal{B}\subseteq \mathcal{P}(\omega )$ is almost disjoint and $\mathrm{\mid }\mathcal{B}\mathrm{\mid }<\mathfrak{c}$, then every $A\subseteq \mathcal{B}$ all of whose members are infinite is extended by a $d\subseteq \omega$ that is infinite on $A$ and finite on $\mathcal{B}\setminus A$ (Martin's axiom extends families almost disjoint from a subfamily).

Put $W(k,n)=(k\mathrm{/}{2}^n-2^{-n-1},k\mathrm{/}{2}^n+2^{-n-1})$, $k\in \mathbb{Z}$, $n\in \mathbb{N}$. Enumerate pairs without repetition: encode $k\ge 0$ by $2k$ and $k<0$ by $-2k-1$, and use the bijection of $\mathbb{N}\times \mathbb{N}\approx \mathbb{N}$. For each $x,n$, the floor of $2^{n}x$ supplied by Integer part: for every real $x$ there is exactly one integer $m$ with $m\le x<m+1$ shows that either $x$ lies in an interval of scale $n$, or $x=(2k+1)\mathrm{/}{2}^{n+1}$ is exactly a midpoint. In the latter case it is a grid center at every scale $m\ge n+1$ and therefore belongs to an interval at every such scale. If no midpoint case occurs, it belongs at every scale. Thus every $x$ belongs to infinitely many distinct indexed intervals. These intervals form a base: a containing interval of sufficiently fine scale lies in any prescribed neighbourhood of $x$, since its diameter is $2^{-n}$ and these diameters tend to zero. Indeed $2^n\ge n+1$ by induction and For every $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$ supplies arbitrarily small reciprocal bounds. Relative $G_{\delta }$ and Q-set have the meaning of Q-sets and Bing's tangent-disk Moore-space interface.

If $x\mathrm{\ne }y$ both belong to $W(k,n)$, then $\mathrm{\mid }x-y\mathrm{\mid }<2^{-n}$. At each fixed scale the intervals are pairwise disjoint, so at most one contains both points. By the decay proved in [L1], only finitely many scales can satisfy $2^{-n}>\mathrm{\mid }x-y\mathrm{\mid }$. Because the pair enumeration has no repetitions, $\{i:x,y\in W_i\}$ is finite. This uses the triangle inequality on the real line and the explicit interval endpoints.