Baire category in $\mathbb{R}$, by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets

Fix $x_0\in \mathbb{R}$ and a real ${\epsilon }_0>0$; by [L1] it suffices to produce a point of ${\bigcap }_n{U}_n$ lying in $N_{{\epsilon }_0}(x_0)$, since $x_0$ and ${\epsilon }_0$ are then arbitrary.

By [L4] fix a bijection $\beta :\mathbb{N}\to \mathbb{Q}$ and put $e:=\iota \circ \beta$, where $\iota (q)=\widehat{q}$, so that $e$ is a bijection from $\mathbb{N}$ onto ${\mathbb{Q}}_{\mathbb{R}}$.

Recall the terminology of the Given: a pair $(p,q)$ of elements of ${\mathbb{Q}}_{\mathbb{R}}$ is good when $p<q$, and $G$ is the set of good pairs.

Refinement claim. For every good $(p,q)$ and every $n\in \mathbb{N}$ there is a good $(p^{\mathrm{\prime }},q^{\mathrm{\prime }})$ with $[p^{\mathrm{\prime }},q^{\mathrm{\prime }}]\subseteq (p,q)\cap U_n$. To see it, note first that $(p,q)$ is nonempty, since [L4] supplies an element of ${\mathbb{Q}}_{\mathbb{R}}$ strictly between $p$ and $q$, and that $(p,q)$ is open by [L2]; fix $y_1\in (p,q)$ and, by [L2], a real ${\rho }_1>0$ with $N_{{\rho }_1}(y_1)\subseteq (p,q)$. Since $U_n$ is dense, [A1] and [L1] give $y\in N_{{\rho }_1}(y_1)\cap U_n$, so $y\in (p,q)\cap U_n$, and that set is open by [A1], [L2] and [L3], so there is a real $\rho >0$ with $N_{\rho }(y)\subseteq (p,q)\cap U_n$. By [L4] fix $p^{\mathrm{\prime }},q^{\mathrm{\prime }}\in {\mathbb{Q}}_{\mathbb{R}}$ with $y-\rho <p^{\mathrm{\prime }}<y<q^{\mathrm{\prime }}<y+\rho$. Then $p^{\mathrm{\prime }}<q^{\mathrm{\prime }}$, so $(p^{\mathrm{\prime }},q^{\mathrm{\prime }})$ is good, and every $t\in [p^{\mathrm{\prime }},q^{\mathrm{\prime }}]$ satisfies $y-\rho <p^{\mathrm{\prime }}\le t\le q^{\mathrm{\prime }}<y+\rho$, whence $\mathrm{\mid }t-y\mathrm{\mid }<\rho$ and $t\in N_{\rho }(y)$; thus $[p^{\mathrm{\prime }},q^{\mathrm{\prime }}]\subseteq N_{\rho }(y)\subseteq (p,q)\cap U_n$.

Successor rule. For $(k,(p,q))\in \mathbb{N}\times G$ let $m$ be the least natural for which some natural $j$ makes $(e(m),e(j))$ good with $[e(m),e(j)]\subseteq (p,q)\cap U_k$, and let $j$ be the least natural with that property for that $m$; put $T(k,(p,q)):=(\sigma (k),(e(m),e(j)))$. The set of eligible $m$ is nonempty by step 2.1 applied with $n=k$, since $e$ is onto ${\mathbb{Q}}_{\mathbb{R}}$ by step 1.2, so both minima exist by [L5] and $T:\mathbb{N}\times G\to \mathbb{N}\times G$ is a total function defined without any selection.

The recursion. By [L4] fix $p_0,q_0\in {\mathbb{Q}}_{\mathbb{R}}$ with $x_0-{\epsilon }_0<p_0<x_0<q_0<x_0+{\epsilon }_0$; then $(p_0,q_0)$ is good and, as in step 2.1, $[p_0,q_0]\subseteq N_{{\epsilon }_0}(x_0)$ by [L2]. Apply [L6] with $Y=\mathbb{N}\times G$, seed $(0,(p_0,q_0))$ and map $T$ to get $h:\mathbb{N}\to \mathbb{N}\times G$ with $h(0)=(0,(p_0,q_0))$ and $h(\sigma (k))=T(h(k))$; an induction on $k$ shows that the first coordinate of $h(k)$ is $k$, so write $h(k)=(k,(p_k,q_k))$, every $(p_k,q_k)$ being good.

Write $I_k:=[p_k,q_k]$, a nonempty closed bounded interval by [L2]. The rule of step 3.1 gives, for every $k\in \mathbb{N}$, that $I_{k+1}\subseteq (p_k,q_k)\cap U_k\subseteq I_k$; in particular the family $(I_k)$ is nested and $I_{k+1}\subseteq U_k$.

By [L7] applied to the nested family $(I_k)$ of nonempty closed bounded intervals, ${\bigcap }_k{I}_k\mathrm{\ne }\mathrm{\varnothing }$; fix $x$ in it.

For every $n\in \mathbb{N}$ one has $x\in I_{n+1}\subseteq U_n$ by steps 5.1 and 6.1, so $x\in {\bigcap }_n{U}_n$; and $x\in I_0\subseteq N_{{\epsilon }_0}(x_0)$ by steps 4.1 and 6.1. So $N_{{\epsilon }_0}(x_0)$ meets ${\bigcap }_n{U}_n$.

Since $x_0\in \mathbb{R}$ and the real ${\epsilon }_0>0$ were arbitrary, every neighbourhood of every point of $\mathbb{R}$ meets ${\bigcap }_n{U}_n$, so that set is dense by [L1].

For the consequence, let $(A_n)$ be a sequence of nowhere dense sets and put $U_n:=\mathbb{R}\setminus \overline{A_n}$, which is open by [L3] and [L8] and dense by [L8]; by step 8.1 the set ${\bigcap }_n{U}_n$ is dense, hence nonempty, and any $x$ in it lies outside every $\overline{A_n}$ and so outside every $A_n$, giving $x\mathrm{\notin }{\bigcup }_n{A}_n$ and therefore ${\bigcup }_n{A}_n\mathrm{\ne }\mathbb{R}$. By [L9] the same conclusion covers a union of an at most countable family of nowhere dense sets, so no meager set is all of $\mathbb{R}$.