Tychonoff’s theorem implies AC
In this entry, we prove that Tychonoff’s theorem implies that product of non-empty set of non-empty sets is non-empty, which is equivalent to the axiom of choice (AC). This fact, together with the fact that AC implies Tychonoff’s theorem, shows that Tychonoff’s theorem is equivalent to AC (under ZF). The proof was first discovered by John Kelley in 1950, and is now an exercise in axiomatic set theory.
First, some notations: for each , set , , the generalized cartesian product of all the ’s, and the projection from onto .
We break down the proof into several steps:
is equipollent to .
An element of is a function , such that for each . In other words, either , or . An element of is a function such that for each . Finally, iff .
Next, we topologize each in such a way that is compact.
To show that is compact under , let be an open cover for . We want to show that there is a finite subset of covering . If , then we are done. Otherwise, pick a non-empty element in , so that , and is finite. By assumption, each element in belongs to some open set in . So to cover , only a finite number of open sets in are needed. These open sets, together with , cover . Hence is compact.
Finally, we prove that , and therefore , is non-empty.
To show that is non-empty, we employ a characterization of compact space: is compact iff every collection of closed sets in having FIP has non-empty intersection (http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty). Let us look at the collection . Given , pick an element , since by assumption. Note that this is possible, since there are only a finite number of sets. Define as follows:
Since , for each . Therefore,
Since are arbitrarily picked from , the collection has finite intersection property, and since is compact, must be non-empty.
This completes the proof. ∎
Remark. In the proof, we see that the trick is to adjoin the set to each set . Instead of , we could have picked some arbitrary, but fixed singleton , as long as for each , and the proof follows essentially the same way.
- 1 T. J. Jech, The Axiom of Choice. North-Holland Pub. Co., Amsterdam, 1973.
- 2 J. L. Kelley, The Tychonoff’s product theorem implies the axiom of choice. Fund. Math. 37, pp. 75-76, 1950.
|Title||Tychonoff’s theorem implies AC|
|Date of creation||2013-03-22 18:45:38|
|Last modified on||2013-03-22 18:45:38|
|Last modified by||CWoo (3771)|