Sequential Uniform Boundedness with Countable Choice: Examples

We first prove scalar completeness from the real supremum property. A real Cauchy sequence $(a_m)$ is bounded: take $M$ with $\mathrm{\mid }{a}_m-a_M\mathrm{\mid }<1$ for $m\ge M$ and successively take the larger of $\mathrm{\mid }{a}_M\mathrm{\mid }+1$ and each of $\mathrm{\mid }{a}_0\mathrm{\mid },\dots ,\mathrm{\mid }{a}_{M-1}\mathrm{\mid }$. This finite process produces a bound $B$. Each tail has a unique infimum $b_N=\inf \{a_m:m\ge N\}$ in $[-B,B]$, and $L=\sup \{b_N:N\in \mathbb{N}\}$ exists. Given $\eta >0$, take $M$ with $\mathrm{\mid }{a}_l-a_m\mathrm{\mid }<\eta$ for $l,m\ge M$. For fixed $m\ge M$ and every $N\ge M$, all terms in the $N$-tail are in $[a_m-\eta ,a_m+\eta ]$, so its nonempty tail has infimum in that same interval. For $N<M$, $b_N\le b_M\le a_m+\eta$. Thus $a_m-\eta \le b_M\le L\le a_m+\eta$. Using $\eta =\epsilon \mathrm{/}2$ proves $a_m\to L$. If two scalar limits differed by $d>0$, sufficiently late terms within $d\mathrm{/}3$ of both would give $d<2d\mathrm{/}3$ by triangle inequality. Hence the limit is unique.

Each $e_j$ is a null sequence of norm one. Computing each coordinate in the finite sum gives $(P_{N}x{)}_i=x_i$ for $i\le N$ and $(P_{N}x{)}_i=0$ for $i>N$: when $i\le N$ the sole nonzero contribution is $x_i(e_i{)}_i=x_i$, and otherwise every contribution is zero. This calculation follows directly from appending the summands in the recursive finite sum. Thus $P_{N}x$ is eventually zero and lies in $c_0$. At each coordinate, $P_N(ax+by{)}_i=a(P_{N}x{)}_i+b(P_{N}y{)}_i$, so $P_N$ is linear. Also $\mathrm{\mid }(P_{N}x{)}_i\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$ for all $i$, hence $\mathrm{\parallel }{P}_{N}x{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $P_N$ is bounded.

For $z=a+ib$, the modulus formula yields $\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$: the first inequalities follow from $a^2,b^2\le a^2+b^2$, and the last from $(\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }{)}^2=a^2+b^2+2\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }b\mathrm{\mid }\ge a^2+b^2$, comparing nonnegative squares. Therefore the real and imaginary parts of a complex Cauchy sequence are real Cauchy sequences. Their unique limits $a,b$ from step 1.1 give complex limit $a+ib$, since the modulus of the difference is bounded by the sum of the two coordinate errors. The scalar triangle inequality again proves uniqueness.

The estimate in step 1.2 bounds the unit-ball image by $1$. Since $P_N{e}_0=e_0$ and $\mathrm{\parallel }{e}_0{\mathrm{\parallel }}_{\mathrm{\infty }}=1$, this image contains $1$ as a norm value. Its supremum is therefore exactly $\mathrm{\parallel }{P}_N\mathrm{\parallel }=1$ for every $N\ge 0$, including the one-coordinate projection $P_0$. Also $P_{N}0=0$.

The coordinate formula gives $\mathrm{\parallel }{P}_{N}x-x{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{j>N}\mathrm{\mid }{x}_j\mathrm{\mid }$. Given $\epsilon >0$, take $J$ with $\mathrm{\mid }{x}_j\mathrm{\mid }<\epsilon \mathrm{/}2$ for $j\ge J$. For $N\ge J$ the tail supremum is at most $\epsilon \mathrm{/}2<\epsilon$, proving $P_{N}x\to x$. If $x$ already vanishes beyond $J$, the error is exactly zero for $N\ge J$.

Let $(x^{(m)})$ be any norm-Cauchy sequence in $c_0$. For each fixed $j$, $\mathrm{\mid }{x}_j^{(m)}-x_j^{(l)}\mathrm{\mid }\le \mathrm{\parallel }{x}^{(m)}-x^{(l)}{\mathrm{\parallel }}_{\mathrm{\infty }}$, so the coordinate sequence is scalar Cauchy and has a unique limit $x_j$. Replacement applied on $\mathbb{N}$ to the unique pair $(j,x_j)$ produces a set of these pairs, the graph of a function $x:\mathbb{N}\to \mathbb{K}$. This uses unique existence, not a choice of witnesses from non-singleton sets.

If $\mathrm{\parallel }{x}^{(m)}-x^{(l)}{\mathrm{\parallel }}_{\mathrm{\infty }}<\eta$ for $m,l\ge M$, then for fixed $m\ge M$ and $j$, the scalar triangle inequality gives $\mathrm{\mid }{x}_j^{(m)}-x_j\mathrm{\mid }\le \eta +\mathrm{\mid }{x}_j^{(l)}-x_j\mathrm{\mid }$. Since the last term tends to zero, $\mathrm{\mid }{x}_j^{(m)}-x_j\mathrm{\mid }\le \eta$; a positive excess would be contradicted by making that term smaller than half the excess. Taking $\eta =1$ and one such $m$ gives $\mathrm{\mid }{x}_j\mathrm{\mid }\le \mathrm{\parallel }{x}^{(m)}{\mathrm{\parallel }}_{\mathrm{\infty }}+1$ for every $j$, so $x$ is bounded. All subsequent suprema are legitimate by the real supremum property. Thus for general $\eta$ the uniform coordinate bound implies $\mathrm{\parallel }{x}^{(m)}-x{\mathrm{\parallel }}_{\mathrm{\infty }}\le \eta$ for $m\ge M$. Taking $\eta =\epsilon \mathrm{/}2$ proves norm convergence in the bounded-sequence space.

Given $\epsilon >0$, take one $m$ with $\mathrm{\parallel }x-x^{(m)}{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon \mathrm{/}2$, and then $J$ with $\mathrm{\mid }{x}_j^{(m)}\mathrm{\mid }<\epsilon \mathrm{/}2$ for $j\ge J$, since $x^{(m)}\in c_0$. For such $j$, $\mathrm{\mid }{x}_j\mathrm{\mid }\le \mathrm{\mid }{x}_j-x_j^{(m)}\mathrm{\mid }+\mathrm{\mid }{x}_j^{(m)}\mathrm{\mid }<\epsilon$. Hence $x\in c_0$, and the arbitrary Cauchy sequence converges in $c_0$. This proves $c_0$ is Banach over both fields. The choices here concern a single given $\epsilon$; no simultaneous selection is used.

Under ${\mathrm{A}\mathrm{C}}_{\omega }$, apply the sequential theorem with domain and codomain $c_0$: step 5.1 proves the domain Banach, step 1.2 supplies bounded linear maps and the pointwise bound $M_x=\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$. Its conclusion is a finite common bound for the $\mathrm{\parallel }{P}_N\mathrm{\parallel }$. Step 2.2 has already computed that bound as $1$ without any choice principle, and step 2.3 proves the promised convergence.

Define $c_{00}=\{x\in c_0:\mathrm{\exists }M\in \mathbb{N}\mathrm{\forall }j\ge M,x_j=0\}$. Its zero vector has cutoff $M=0$. If $x,y$ have cutoffs $M,N$ and $a,b\in \mathbb{K}$, then $ax+by$ vanishes for $j\ge M+N$, since $M+N\ge M,N$. It is therefore a vector subspace of $c_0$, normed by the inherited supremum norm. The same argument is valid over $\mathbb{C}$, using the modulus convention.

This is exactly the finite-support space. If $x$ has cutoff $M$, its support $\{j:x_j\mathrm{\ne }0\}$ is a subset of the finite natural $M$, hence finite. Conversely, every finite support $F$ has a bijection $f:m\to F$. We prove the range of every map $f:m\to \mathbb{N}$ has a natural upper bound by induction on $m$. At $m=0$, use bound $0$. At $m=r+1$, the restriction to $r$ has a bound $L$ by the induction hypothesis; choose the larger of $L$ and $f(r)$ by trichotomy. This bounds the old values and the one new value at $r$, hence the whole range. Induction gives a bound $L$ for $F$, so $x_j=0$ for $j\ge L+1$. A scalar sequence with finite support is bounded as well: successively taking the maximum of $0$ and the finitely many $\mathrm{\mid }{x}_{f(i)}\mathrm{\mid }$ gives a real bound, and the cutoff makes it null. Thus this converse applies to arbitrary finite-support scalar sequences, not only those already presented in $c_0$.

For each $n\ge 0$ define $T_n:c_{00}\to \mathbb{K}$ by $T_{n}x=n{x}_n$. For scalars $a,b$, $T_n(ax+by)=n(a{x}_n+b{y}_n)=a{T}_{n}x+b{T}_{n}y$. Also $\mathrm{\mid }{T}_{n}x\mathrm{\mid }=n\mathrm{\mid }{x}_n\mathrm{\mid }\le n\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$, so $T_n$ is bounded and its unit-ball supremum is at most $n$. For $n\ge 1$, let $e_n$ have coordinate $1$ at $n$ and zero elsewhere. It belongs to $c_{00}$, has norm $1$, and $\mathrm{\mid }{T}_n{e}_n\mathrm{\mid }=n$, giving $\mathrm{\parallel }{T}_n\mathrm{\parallel }=n$. At $n=0$, $T_0=0$ and its norm is zero; at $n=1$, $T_1{e}_1=1$.

For $x$ with cutoff $M$, $T_{n}x=0$ when $n\ge M$. For $n<M$, $\mathrm{\mid }{T}_{n}x\mathrm{\mid }\le n\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\le M\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$. Thus $M_x=M\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$ bounds the entire orbit, also when $M=0$ and $x=0$. On the other hand, given any real $C$, Archimedean cofinality gives $n>C$, and $\mathrm{\parallel }{T}_n\mathrm{\parallel }=n>C$. The sequence is pointwise bounded but its operator norms are unbounded.

For $N\ge 0$, let $z_j^{(N)}=1\mathrm{/}(j+1)$ if $j\le N$ and $z_j^{(N)}=0$ otherwise. Its support is $\{0,\dots ,N\}$, so it lies in $c_{00}$. For $M>N$, the difference has nonzero coordinates precisely $N<j\le M$, where its magnitude is $1\mathrm{/}(j+1)\le 1\mathrm{/}(N+2)$. Equality is attained at $j=N+1$. Hence $\mathrm{\parallel }{z}^{(M)}-z^{(N)}{\mathrm{\parallel }}_{\mathrm{\infty }}=1\mathrm{/}(N+2)$; for $M=N$ it is zero. Given $\epsilon >0$, take $K\ge 1$ with $1\mathrm{/}K<\epsilon$ using the reciprocal property in the real field. For any $M,N\ge K$, symmetry and the displayed formula bound the distance by $1\mathrm{/}(K+2)<\epsilon$. Thus this sequence is norm-Cauchy.

If it had norm limit $z\in c_{00}$, then for each fixed $j$ and every $N\ge j$, $\mathrm{\mid }{z}_j-1\mathrm{/}(j+1)\mathrm{\mid }=\mathrm{\mid }{z}_j-z_j^{(N)}\mathrm{\mid }\le \mathrm{\parallel }z-z^{(N)}{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. A constant nonnegative real bounded by a null sequence is zero: if positive, eventually that upper bound is smaller than half the constant. Therefore $z_j=1\mathrm{/}(j+1)\mathrm{\ne }0$ for every $j$. This contradicts the cutoff of $z$ established in step 1.1. The Cauchy sequence has no limit in $c_{00}$, so the domain is not Banach. Together with step 3.1 this exhibits exactly the failure when completeness of the domain is omitted.