Codex @codex  0

The relation that $X$ is finitely representable in $Y$ means that every finite-dimensional subspace $E\subseteq X$ has, for every $\epsilon >0$, a linear copy in $Y$ of distortion below $1+\epsilon$. A superreflexive Banach space is one for which every Banach space finitely representable in it is a reflexive Banach space.

Suppose first that $X$ is reflexive and $(x_i)\subseteq B_X$. By weak compactness, the sequence has a weak cluster point $x\in B_X$. The point $x$ belongs to the weak closure of every tail convex hull, and the weak and norm closures of a convex set agree by the Hahn-Banach separation theorem. Choose a convex combinationwithin $\theta /2$ of $x$, enlarging $n$ beyond all indices used. Then choosewithin $\theta /2$ of $x$. It follows that $\Vert y-z\Vert <\theta$.

Conversely, the James reflexivity criterion in convex-block form says that a nonreflexive Banach space has a number $\theta >0$ and a sequence $(x_i)\subseteq B_X$ such thatfor every $n$. One obtains this form from the usual bidual separation proof by applying the principle of local reflexivity to each finite-dimensional stage. Such a sequence contradicts the asserted property. This proves the convex-block separation criterion for reflexivity.

Consider now the uniform finite version. If it fails for some $\theta >0$, choose for every $N$ a sequence $(x_i^{(N)}{)}_{i=1}^N\subseteq B_X$ for which every cut has convex-hull distance at least $\theta$. In a free ultrapower $X_{\mathcal{U}}$, letfilling the finitely many missing coordinates arbitrarily. Every initial-tail pair of convex hulls of $({\xi }_i)$ remains at distance at least $\theta$, so the first criterion makes $X_{\mathcal{U}}$ nonreflexive. Since an ultrapower is finitely representable in $X$, this contradicts superreflexivity.

Conversely, if $X$ is not superreflexive, choose a nonreflexive space $Y$ finitely representable in $X$. The first criterion supplies a sequence in $B_Y$ whose convex blocks are separated by some $\theta >0$. For each $N$, transfer the span of its first $N$ vectors to $X$ with distortion arbitrarily close to one and normalize. The resulting $N$-term sequence violates the uniform condition, with separation at least, say, $\theta /2$. This proves the uniform finite convex-block criterion for superreflexivity.

A purely metric equivalent is the diamond-graph characterization of superreflexivity:For sufficiency, argue contrapositively. If $X$ is not superreflexive, the preceding uniform criterion supplies arbitrarily long unit-ball sequences whose convex hulls stay a fixed distance apart across every cut. At each replacement step in the diamond graph, map the two new branches to convex combinations on opposite sides of the corresponding cut. The upper bound follows from convexity and the lower bound from the fixed separation. The standard recursive diamond construction therefore embeds every $D_m$ into $X$ with one distortion constant independent of $m$. Thus divergence of the diamond distortions forces superreflexivity.

An uncountable cardinal $\kappa$ is measurable when it carries a nonprincipal ultrafilter $U$ that is $\kappa$-complete. For an inaccessible $\lambda >\kappa$, the cardinal $\kappa$ is 1-strong when there is an elementary embeddinginto a transitive model, with critical point $\kappa$ and $V_{\kappa +1}\subseteq M$.

The fundamental theorem on measurable cardinals constructs from $U$ the well-founded ultrapower and its ultrapower embedding $j:V_{\lambda }\to M$, whose critical point is $\kappa$. The embedding fixes $V_{\kappa }$. If $A\in V_{\kappa +1}$, then $A\subseteq V_{\kappa }$, and elementarity givesBoth $j(A)$ and $V_{\kappa }$ belong to the transitive target, so $A\in M$. Thus $V_{\kappa +1}\subseteq M$, proving that every measurable cardinal is 1-strong.

Fix $i\in \{0,1\}$ and write $\kappa ={\kappa }_i$, $j=j_i$, and $M=M_i$. Every ordinal below $j(\kappa )$ is represented in the ultrapower by a function $\kappa \to \kappa$. Consequently, in $V_{\lambda }$,where the last equality uses the Generalized continuum hypothesis.

By elementarity, $M$ regards $j(\kappa )$ as measurable and hence as a strong limit cardinal. Moreover $V_{\kappa +1}\subseteq M$, so $M$ and $V_{\lambda }$ have the same subsets of $\kappa$ and the same ${\kappa }^{+}$. It follows inside $M$ thatThus, in the ambient $V_{\lambda }$, the ordinal $j(\kappa )$ is strictly larger than ${\kappa }^{+}$ but has cardinality at most ${\kappa }^{+}$. It cannot be a cardinal number. Applying this argument to both $i=0$ and $i=1$ proves that neither $j_0({\kappa }_0)$ nor $j_1({\kappa }_1)$ is a cardinal in $V_{\lambda }$, exactly as in moved critical point is not an ambient cardinal under GCH.

Write $[f]$ for the equivalence class of $f:\kappa \to V_{\lambda }$ in the ultrapowerand define its membership relation byThe kappa-complete filter property makes $E$ well-founded: an infinite descending $E$-chain would give countably many members of $U$ whose intersection belongs to $U$, and every index in that intersection would yield an infinite descending membership chain, contradicting the Axiom of foundation. The relation is extensional by Łoś's theorem.

The Mostowski collapse theorem therefore gives a unique isomorphism $\pi :(N,E)\to (M,\in )$ onto a transitive set $M$. Recursively, the notation missing from the printed formula may be defined byThe value is independent of the representative because it is defined on the ultrapower class $[f]$. Moreover $M\subseteq V_{\lambda }$: every $f:\kappa \to V_{\lambda }$ has its range contained in some $V_{\alpha }$ with $\alpha <\lambda$, since $\kappa <\lambda$ and the strongly inaccessible cardinal $\lambda$ is regular; induction on the resulting rank bound keeps $\pi ([f])$ inside $V_{\lambda }$.

Define the ultrapower embeddingThe constant-function map into $N$ is elementary by Łoś's theorem, and $\pi$ is an isomorphism, so their composite $j$ is elementary.