Let me recall the notation: $(A_x)$ be a family of sets indexed by a set $X$, $A=\prod A_x$ is the product of the sets $A_x$, $\cal U$ be a nonprincipal ultrafilter on a set $X$, and $A^*$ is the ultraproduct of the $A_x$ with respect to $\cal U$.

I then recall the notion of convergence with respect to the ultrafilter $\cal U$. Let $V$ be a topological space and let $(v_x)$ be a family of points of $V$.

- Say $(v_x)$ $\cal U$-converges to $v$ if for any neighborhood $W$ of $v$, the set of $x\in X$ such that $v_x\in W$ belongs to $\cal U$.
- If $V$ is Hausdorff, then $(v_x)$ has at most one limit. In particular, if $(w_x)$ is another family of points of $V$ indexed by $X$ which converges to a point $w\neq v$, the set of $x\in X$ such that $v_x\neq w_x$ belongs to $\cal U$.
- Moreover, if $V$ is compact, then any family $(v_x)$ has a $\cal U$-limit in $V$. This is essentially the definition of compactness in terms of ultrafilters, but it does no harm recalling the proof. If, by contradiction, $(v_x)$ has no $\cal U$-limit, any point $v\in V$ has a neighborhood $W_v$ such that the set $X_v$ $x\in X$ with $x\not\in W_v$ belongs to $\cal U$. Since $V$ is compact, there exists a finite set $T$ in $V$ such that the $W_v$, for $v\in T$, cover $V$. This implies that the intersection of the $X_v$, for $v\in T$, is empty. However, this is a finite intersection of elements of $\cal U$, hence belongs to $\cal U$, hence is non-empty.

Last reminding. The set of nonnegative integers $\mathbf N$ is dense in the Hausdorff and compact set $\mathbf Z_2$ of $2$-adic integers endowed with the $2$-adic distance.

For every $x\in X$, choose a surjection $f_x$ from $A_x$ to the interval of integers $[0,\mathrm{Card}(A_x))$, viewed as a subset of $\mathbf Z_2$. For any family $a=(a_x)$ where $a_x\in A_x$ for every $x\in X$, the family $(f_x(a_x))$ has a unique $\cal U$-limit in $\mathbf Z_2$, denoted $f(a)$. This defines a map $f\colon A\to \mathbf Z_2$.

Let $a=(a_x)$ and $b=(b_x)$ be two such families and assume that $f(a)\neq f(b)$. It follows from the fact that $\mathbf Z_2$ is Hausdorff that the set of $x\in X$ such that $a_x\neq b_x$ belongs to $\cal U$. Said the other way round, if $a$ and $b$ are $\cal U$-equivalent, then $f(a)=f(b)$, so that $f$ defines a map $f^*\colon A^*\to\mathbf Z_2$.

We now show that if the cardinalities $\mathrm{Card}(A_x)$ are not bounded (with respect to $\cal U$), this map $f^*$ is surjective. Let $v\in\mathbf Z_2$; for any $x\in X$, let $v_x$ be the integer in $[0,\mathrm{Card}(A_x))$ which is the closest to $v$ with respect to the $2$-adic distance and let $a_x\in A_x$ be such that $f_x(a_x)=v_x$. I claim that $d(v,v_x)$ $\cal U$-converges to $0$; indeed, for any $k$, the set $X_k$ of $x\in X$ such that $\mathrm{Card}(A_x)\geq 2^k$ belongs to $\cal U$ (this is what it means to be unbounded with respect to $\cal U$) and for $x\in X_k$, $d(v,v_x)\leq 2^{-k}$. By construction, $(f_x(a_x))=(v_x)$, hence $f((a_x))=v$; if $a^*$ is the class in $A^*$ of the family $(a_x)$, we thus have $f^*(a^*)=v$.

Finally, $\mathrm{Card}(A^*) \geq \mathrm{Card}(\mathbf Z_2)=2^{\mathbf N}$, as was to be shown.

(This proof is longer, but only because I needed to recall—I mean, to recall to myself—the topological apparatus of convergence for ultrafilters.)

## No comments :

## Post a Comment