The scale subspace is closed

For any $z,w\in X$, choose indices witnessing F1. If their indices are equal, $z{=}^{\ast }w$; if one index is smaller, scale strictness and the two finite exceptional sets give the corresponding strict eventual inequality. Thus exactly one of $z{<}^{\ast }w$, $w{<}^{\ast }z$, or $z{=}^{\ast }w$ holds, since $B$ is infinite. Fix $z\in X$ with $z\le t$, and let $E(z,t)=\{n:z(n)=t(n)\}$. Set $a(n)=0$ on $E(z,t)$ and $a(n)=z(n)$ elsewhere. Point-coordinates of $t$ are positive; outside $E(z,t)$ we have $z(n)<t(n)$. Hence $a<t$. By closure and F3 there is $w\in X\cap (a,t{]}_R$. If $z{<}^{\ast }w$, the equality coordinates of $z,t$ must lie in the finite exception set, because at such a coordinate $w(n)\le t(n)=z(n)$. Thus $E(z,t)$ is finite. In the other two cases $w{\le }^{\ast }z$. But at every coordinate outside $E(z,t)$, membership in the box gives $w(n)>a(n)=z(n)$. Hence that complement is finite, so $E(z,t)$ is cofinite.

Put $A=\{z\in X:z\le t\}$, and for each finite or cofinite $E\subseteq B$ put $A_E=\{z\in A:E(z,t)=E\}$. Every local box at $t$ meets $X$ and contains only points at most $t$, so $t\in {\bar{A}}^R$ by F3. Step 1.1 partitions $A$ into the $A_E$. There are countably many possible $E$: finite subsets of $B$ inject into the natural numbers by $S\mapsto {\sum }_{n\in S}{2}^n$ (the largest differing exponent exceeds the sum of all smaller powers), and complementation identifies cofinite subsets with finite ones. If $t$ avoided every ${\overline{A_E}}^R$, the countable intersection of their open complements would be an open neighborhood of $t$ by F3 and would miss $A$, contradicting $t\in \bar{A}$. Choose an $E$ with $t\in \overline{A_E}$. If $E$ is cofinite, $A_E$ is nonempty; take a member $z$. Then $t{=}^{\ast }z$, and $t\in R$ makes this an admissible finite modification, so $t\in X$ by F1. It remains to consider finite $E$.

Since $t\in R$, some finite $q$ bounds all its coordinate cofinalities strictly below ${\mathrm{\aleph }}_q$ and above $\omega$. By F4 the possible cofinalities are among ${\mathrm{\aleph }}_1,\dots ,{\mathrm{\aleph }}_{q-1}$. Put $D_i=\{n\in B:\mathrm{cf}(t(n))={\mathrm{\aleph }}_i\}$. At least one $D_i$ is infinite, since these finitely many sets cover the infinite $B$. Let $m\ge 1$ be the least such index. The union of the $D_i$ for $i<m$ is finite. Choose a natural number $k\ge m$ at least all members of this union and of the finite $E$; if the union is empty, use $k=m$. Consequently every tail coordinate $n>k$ has cofinality at least $\kappa ={\mathrm{\aleph }}_m$, and no such coordinate is in $E$. On the infinite set $D_m\cap (k,\omega )$ choose increasing cofinal functions ${\gamma }_n:\kappa ⟶t(n)$. F4 and A1 supply them: enumerate a size-$\kappa$ cofinal subset and recursively take increasing bounds, with every proper initial segment bounded because its size is less than the cofinality.

Construct $z_{\xi }\in A_E$ for $\xi <\kappa$ as follows. At stage $\xi$, for $n>k$ put $r_{\xi }(n)={\sup }_{\eta <\xi }{z}_{\eta }(n)$, with empty supremum zero. Every previous value is below $t(n)$, because its equality set is $E$ and $n\mathrm{\notin }E$. Since $\mathrm{\mid }\xi \mathrm{\mid }<\kappa \le \mathrm{cf}(t(n))$, F4 gives $r_{\xi }(n)<t(n)$. Define a lower function $a_{\xi }<t$ by setting it to zero on $n\le k$, to $\max (r_{\xi }(n),{\gamma }_n(\xi ))$ on $D_m\cap (k,\omega )$, and to $r_{\xi }(n)$ on other tail coordinates. All these inequalities are strict by step 3.1. As $t\in \overline{A_E}$, the open box $(a_{\xi },t{]}_R$ meets $A_E$ by F3. Fix by A1 a choice function for all such nonempty box intersections, a set-indexed family, before applying F6 to the displayed rule. The chosen $z_{\xi }$ then satisfies $z_{\xi }\le t$, and at every $n>k$ it is strictly above all earlier values and strictly below $t(n)$. Also $z_{\xi }(n)>{\gamma }_n(\xi )$ on $D_m\cap (k,\omega )$. These verifications inductively justify every stage of the recursion.

Apply F2 to the sequence of step 4.1 with this $1\le m\le k$. It gives $h\in X$ agreeing with $g(n)={\sup }_{\xi <\kappa }{z}_{\xi }(n)$ on the tail $n>k$. Replace $h(n)$ by $t(n)$ on the finite prefix $n\le k$, obtaining $h^{\mathrm{\prime }}$. It is still a Rudin point: use a finite aleph bound larger than the two uniform bounds for $h$ and $t$. Hence F1 gives $h^{\mathrm{\prime }}\in X$. On the tail $g(n)\le t(n)$, since every term is at most $t(n)$; on the prefix equality holds by construction, so $h^{\mathrm{\prime }}\le t$. Moreover $g(n)=t(n)$ for every $n\in D_m\cap (k,\omega )$, since the sequence dominates the cofinal ${\gamma }_n$ there and is bounded by $t(n)$. This infinite set is contained in $E(h^{\mathrm{\prime }},t)$. Step 1.1 therefore forces $E(h^{\mathrm{\prime }},t)$ to be cofinite. Thus $t{=}^{\ast }{h}^{\mathrm{\prime }}$, and $t\in R$ implies $t\in X$ by F1.

Every point of ${\bar{X}}^R$ belongs to $X$, by the cofinite case of step 2.1 or the finite case of step 5.1. Therefore $X$ is closed in $R$. QED.