Given $I$ a set of indexes and $X_i$ a set of topological spaces, define

The Cartesian product: $\prod_{i \in I}X_i = \{ f:I \rightarrow \bigcup X_i | f(i) \in X_i \}$

I have read that we need the axiom of choice in order to show that the cartesian product of non-empty collection of non-empty sets is not empty.

My question is, why do we need the axiom of choice for that?

If all the spaces are non-empty, pick arbitrarly from each $X_i$ a point $x_i$, and the product is not empty.

I know that this what the axiom of choice is all about. But, The axiom states that we will have a well defined choice function. I don't see where we need a well defined choice function in the case of showing the non-emptyness of an infinite product..

Any help?

Thank you!

  • 2
    $\begingroup$ "pick arbitrarly" <- That's choice. In fact, the axiom of choice is precisely the assertion that a product of nonempty sets always is nonempty. $\endgroup$ – Daniel Fischer Jul 20 '14 at 13:18

Picking an arbitrary $x_i$ is exactly where the axiom of choice gets into the picture.

Otherwise, how can you justify the existence of such a function which picks this arbitrary $x_i$?

Recall that the axiom of choice is needed when the product is infinite. When the product is finite, then indeed by induction we can choose an arbitrary $x_i$ and it's fine. But this induction is not transfinite.

  • $\begingroup$ Also, the axiom of countable choice is much weaker than the general axiom of choice. Actually, there's an axiom of choice for each cardinality, but we usually only care about the countable one or the general one. $\endgroup$ – Arthur Jul 20 '14 at 13:25
  • 1
    $\begingroup$ This is more complicated than just "each cardinality", we can have several parameters in "axiom of choice for ..." sort of axioms. But I disagree that "we usually only care about the countable one or the general one". $\endgroup$ – Asaf Karagila Jul 20 '14 at 13:28
  • $\begingroup$ Didn't Hilbert use the pairs of socks vs pairs of shoes analogy? $\endgroup$ – Henno Brandsma Jul 20 '14 at 13:28
  • $\begingroup$ @Henno: Russell. Not Hilbert. $\endgroup$ – Asaf Karagila Jul 20 '14 at 13:30
  • $\begingroup$ @AsafKaragila Thx. I found it illuminating at the time. $\endgroup$ – Henno Brandsma Jul 20 '14 at 13:41

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