Rudin shrinking obstruction

For a coordinate $n$, the set $\{h\in X:h(n)<{\mathrm{\aleph }}_n\}$ is open: it is the intersection with $X$ of the box with factor $[0,{\mathrm{\aleph }}_n)$ at $n$ and whole factors elsewhere, and $[0,{\mathrm{\aleph }}_n)$ is an open initial ray of the ordinal interval. Thus the set with $h(n)={\mathrm{\aleph }}_n$ is closed. Each $F_k$ is the intersection of finitely many such sets, hence closed, and $F_{k+1}\subseteq F_k$ because it imposes all preceding equalities. The empty initial coordinate set gives $F_0=F_1=X$. If $h$ belonged to every $F_k$, taking $k=n$ would give $h(n)={\mathrm{\aleph }}_n$ for every $n\in B$. F3 would make these coordinate cofinalities ${\mathrm{\aleph }}_n$, which have no finite uniform aleph bound: the infinite subset $B$ of the natural numbers is unbounded, since a bounded subset would be finite. This contradicts F1. Hence ${\bigcap }_k{F}_k=\mathrm{\varnothing }$.

For every $k$, F2 gives a nonempty set of functions $z_k\in {\prod }_{n\in B}{\mathrm{\aleph }}_n$ whose tails are contained in $U_k$. Use A1 to choose one for each $k$ and put $b(n)={\sup }_{k<\omega }{z}_k(n)$. Each ${\mathrm{\aleph }}_n$ is uncountable regular by F3, so F4 gives $b(n)<{\mathrm{\aleph }}_n$. Thus $b$ belongs to the product in the statement. If $b<h$ with $h\in X$, then $z_k(n)\le b(n)<h(n)$ at every coordinate and every $k$. The defining property of $z_k$ puts $h$ in every $U_k$.

Define $h(n)=b(n)+{\omega }_1$ using ordinal addition. By F5 this is the order type of the disjoint ordered sum of $b(n)$ and ${\omega }_1$. With $\mu =\max (\mathrm{\mid }b(n)\mathrm{\mid },{\mathrm{\aleph }}_1)$, one has $\mu <{\mathrm{\aleph }}_n$ because $b(n)<{\mathrm{\aleph }}_n$ and $n\ge 2$. F6 bounds that sum's cardinality by $\mu$, hence its order type $h(n)$ is below the initial ordinal ${\mathrm{\aleph }}_n$: an ordinal at least ${\mathrm{\aleph }}_n$ would contain a copy of ${\mathrm{\aleph }}_n$ and have cardinality at least ${\mathrm{\aleph }}_n$. The final copy of ${\omega }_1$ is cofinal in the sum, so $\mathrm{cf}(h(n))\le {\omega }_1$. A shorter cofinal set cannot exist. Its part in the final copy would be bounded there by regularity F3 and F4, and a larger point of that copy would then bound the entire set, including all points in the initial $b(n)$ part. Therefore $\mathrm{cf}(h(n))={\omega }_1$. The final copy is nonempty and has no largest point, so $h(n)>b(n)$, also when $b(n)=0$. All coordinate cofinalities of $h$ are ${\omega }_1<{\mathrm{\aleph }}_2$, so F1 gives $h\in X$.

The point $h$ in step 2.1 belongs to the tail above $b$, making that tail nonempty. Step 1.2 puts the entire tail in ${\bigcap }_k{U}_k$, and step 1.1 gives the decreasing closed sequence with empty intersection. These are all the stated conclusions. QED.