0

Suppose that $X$ is a (edit: separable) complete metric space and $\{X_j:j\ge 1\}$ is a partition of $X$ (each $X_j\ne \emptyset$). Given a (continuous) function $f:X\to \mathbb{R}$, I construct $$ h(x)=\sum_{j\ge 1}f(x_j)1_{X_j}(x), $$ where $x_j$ is an arbitrary point of $X_j$. My question is whether the ACC is implicitly involved in this construction?

Robert W.
  • 767

1 Answers1

1

If $f$ is constant on each $X_j$, then there is no need to use the Axiom of Choice here, since we are not really choosing $x_j$, but rather its value, which is fixed.

If, however, $f$ is not constant on infinitely many $X_j$s, then there is no reason to pick one point over the other, and the value will now depend on that choice. So, in that case, you're absolutely relying on the Axiom of Choice.

So, if, for example $\Bbb R$ has a countable family of subsets without a choice function, say $A_j$, and by happenstance, $f(X_j)\subseteq A_j$ for all $j$, then we have effectively constructed a choice function.

You might be able to subvert the use of the Axiom of Choice if you're assuming that your partition has some topological properties which allows you to ensure that $f(X_j)$ is a closed set or an interval, for example.

Asaf Karagila
  • 405,794
  • That's exactly what I thought about assuming some properties for $X_j$'s. For example, if these are intervals of $\mathbb{R}$, one can simply pick the midpoint of each interval. – Robert W. Jun 18 '24 at 10:54
  • BTW, I've just seen this post, which seems to imply that every separable complete metric space is Loeb (as defined here). In this case, I just need to choose from $\operatorname{cl}(X_j)$ to avoid ACC. Is this a known result? – Robert W. Jun 18 '24 at 14:23
  • Note that you're choosing a value from $\operatorname{cl}(f(X_j))$, and continuity is not enough to guarantee that $f(\operatorname{cl}(X_j))$ is a closed set. – Asaf Karagila Jun 18 '24 at 14:49
  • I must be missing something. However, doesn't it suffice to choose a value from $\operatorname{cl}(X_j)$, say $x_j$; then $f(x_j)$ is given? – Robert W. Jun 18 '24 at 14:58
  • Well, first of all, $X_j$ is not a set of reals. Choosing from closed sets in arbitrary complete metric space might not be possible. It is choice-free in $\Bbb R$, but not in general. – Asaf Karagila Jun 18 '24 at 15:00