Free tail ultrafilters and bounded real ultralimit calculus

We first discharge the reciprocal input underlying the real-field/completeness interface. If $a\in \mathcal{C}\setminus \mathcal{N}$, choose rational $\delta >0$ and $N_0\ge 1$ with $\mathrm{\mid }{a}_n\mathrm{\mid }>\delta$ for $n\ge N_0$. Set $b_n=1$ for $n<N_0$ and $b_n=1\mathrm{/}{a}_n$ otherwise; all denominators used are nonzero.

The sets containing a tail of $\mathbb{N}$ form a proper filter: intersections contain the later tail, supersets preserve containment, and no tail is empty. AC and [F8] extend it to an ultrafilter. No finite set belongs to the extension because a disjoint tail belongs; in particular it is not principal.

Least-upper-bound completeness implies that the natural multiples of $1$ are unbounded: if their supremum were $s$, then $s-1$ would not be an upper bound, giving a natural $k>s-1$ and $k+1>s$. Consequently $(B-A)2^{-j}\to 0$ for $A\le B$, since $2^j\ge j+1$.

For a sequence $u_n\in [A,B]$, repeatedly bisect the current interval $I_j=[a_j,b_j]$, retaining its left half if that half's preimage under $u$ is large, otherwise its right half. The right half is large in the latter case: intersect the large preimage of $I_j$ with the large complement of the left-half preimage. Thus each $I_j$ has large preimage, the intervals nest, and $b_j-a_j=(B-A)2^{-j}$. This is a deterministic recursion.

For rational $\epsilon >0$, take $N\ge N_0$ also beyond a Cauchy index of $a$ at tolerance $\epsilon {\delta }^2$. For $m,n\ge N$, $\mathrm{\mid }{b}_m-b_n\mathrm{\mid }=\mathrm{\mid }{a}_m-a_n\mathrm{\mid }\mathrm{/}(\mathrm{\mid }{a}_m\mathrm{\mid }\mathrm{\mid }{a}_n\mathrm{\mid })\le \mathrm{\mid }{a}_m-a_n\mathrm{\mid }\mathrm{/}{\delta }^2<\epsilon$. The nonstrict comparison includes $a_m=a_n$. Thus $b\in \mathcal{C}$.

Set $u={\sup }_j{a}_j$. All $a_j\le u\le b_j$: for fixed $j$, every later left endpoint lies below $b_j$, and earlier ones lie below $a_j$. For any $\epsilon >0$, a sufficiently late $I_j$ has length less than $\epsilon$, so its large preimage lies in $\{n:\mathrm{\mid }{u}_n-u\mathrm{\mid }<\epsilon \}$. Thus $u$ is an ultralimit, also when $A=B$.

The sequence $ab-1$ is eventually zero, hence null. The ideal $\mathcal{N}$ is proper since the constant $1$ fails the null test with tolerance $1\mathrm{/}2$. Every ideal containing $\mathcal{N}$ and $a$ contains $ab-(ab-1)=1$, hence all of $\mathcal{C}$. This proves its maximality and the reciprocal interface directly, without the erroneous strict intermediate comparison in the older maximality proof. The real completeness conclusion of [F7] is used with this corrected input.

If $u\mathrm{\ne }{u}^{\mathrm{\prime }}$, their neighbourhoods of radius $\mathrm{\mid }u-u^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}3$ are disjoint by the triangle inequality. Their preimages cannot both belong to a proper filter. Hence the limit is unique. Equality of two sequences on a large set lets every neighbourhood test for one pass to the other by intersection; their limits therefore agree. A free ultrafilter contains no finite set: if a finite set were large but none of its singletons were large, intersecting their large complements from [F9] would contradict its largeness. A large singleton would make the ultrafilter principal by upward closure and the filter intersection axiom. Hence the complement of every finite set is large by [F9], so an ordinary convergent bounded sequence satisfies the same tests.

Intersect the two large error sets for $u_n,u$ and $v_n,v$, each at tolerance $\epsilon \mathrm{/}2$. There $\mathrm{\mid }(u_n+v_n)-(u+v)\mathrm{\mid }\le \mathrm{\mid }{u}_n-u\mathrm{\mid }+\mathrm{\mid }{v}_n-v\mathrm{\mid }<\epsilon$. For $c\mathrm{\ne }0$ use tolerance $\epsilon \mathrm{/}\mathrm{\mid }c\mathrm{\mid }$ to obtain $\mathrm{\mid }c{u}_n-cu\mathrm{\mid }<\epsilon$; for $c=0$ the sequence is constant zero. Uniqueness identifies the stated limits.

If $\mathrm{\mid }{u}_n\mathrm{\mid }\le H$, intersect the large sets where both errors are less than $\epsilon \mathrm{/}(H+\mathrm{\mid }v\mathrm{\mid }+1)$. Then $\mathrm{\mid }{u}_n{v}_n-uv\mathrm{\mid }\le H\mathrm{\mid }{v}_n-v\mathrm{\mid }+\mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }{u}_n-u\mathrm{\mid }<\epsilon$. Also $\mid \mathrm{\mid }{u}_n\mathrm{\mid }-\mathrm{\mid }u\mathrm{\mid }\mid \le \mathrm{\mid }{u}_n-u\mathrm{\mid }$, so absolute values converge to $\mathrm{\mid }u\mathrm{\mid }$. These arguments apply to zero and unit constants as well.

If $u_n\le v_n$ on a large set but $u>v$, intersect that set with the two error sets of radius $(u-v)\mathrm{/}3$. It would give $u_n>v_n$, impossible. Thus $u\le v$. All calculus constructions used only deterministic interval bisection and finite intersections once the ultrafilter was supplied; AC was spent only in the free-filter existence step.