Editing Tychonoff's theorem (section)

== Proof of the axiom of choice from Tychonoff's theorem ==

To prove that Tychonoff's theorem in its general version implies the axiom of choice, we establish that every infinite [[cartesian product]] of non-empty sets is nonempty. The trickiest part of the proof is introducing the right topology. The right topology, as it turns out, is the [[cofinite topology]] with a small twist. It turns out that every set given this topology automatically becomes a compact space. Once we have this fact, Tychonoff's theorem can be applied; we then use the [[finite intersection property]] (FIP) definition of compactness. The proof itself (due to [[J. L. Kelley]]) follows:

Let {''A<sub>i</sub>''} be an indexed family of nonempty sets, for ''i'' ranging in ''I'' (where ''I'' is an arbitrary indexing set). We wish to show that the cartesian product of these sets is nonempty. Now, for each ''i'', take ''X<sub>i</sub>'' to be ''A<sub>i</sub>'' with the index ''i'' itself tacked on (renaming the indices using the [[disjoint union]] if necessary, we may assume that ''i'' is not a member of ''A<sub>i</sub>'', so simply take ''X<sub>i</sub>'' = ''A<sub>i</sub>'' ∪ {''i''}).

Now define the cartesian product
<math display=block>X = \prod_{i \in I} X_i</math>
along with the natural projection maps ''π<sub>i</sub>'' which take a member of ''X'' to its ''i''th term.

We give each ''X<sub>j</sub>'' the topology whose open sets are: the empty set, the singleton {''i''}, the set ''X<sub>i</sub>''. This makes ''X<sub>i</sub>'' compact, and by Tychonoff's theorem, ''X'' is also compact (in the product topology). The projection maps are continuous; all the ''A<sub>i</sub>'''s are closed, being complements of the [[singleton (mathematics)|singleton]] open set {''i''} in ''X<sub>i</sub>''. So the inverse images π<sub>''i''</sub><sup>−1</sup>(''A<sub>i</sub>'') are closed subsets of ''X''. We note that
<math display=block>\prod_{i \in I} A_i = \bigcap_{i \in I} \pi_i^{-1}(A_i) </math>
and prove that these inverse images have the FIP. Let ''i<sub>1</sub>'', ..., ''i<sub>N</sub>'' be a finite collection of indices in ''I''. Then the ''finite'' product ''A<sub>i<sub>1</sub></sub>'' × ... × ''A<sub>i<sub>N</sub></sub>''
is non-empty (only finitely many choices here, so AC is not needed); it merely consists of ''N''-tuples. Let  ''a'' = (''a''<sub>1</sub>, ..., ''a<sub>N</sub>'') be such an ''N''-tuple. We extend ''a'' to the whole index set: take ''a'' to the function ''f'' defined by ''f''(''j'') = ''a<sub>k</sub>'' if ''j'' = ''i<sub>k</sub>'', and ''f''(''j'') = ''j'' otherwise. ''This step is where the addition of the extra point to each space is crucial'', for it allows us to define ''f'' for everything outside of the ''N''-tuple in a precise way without choices (we can already choose, by construction, ''j'' from ''X<sub>j</sub>'' ). π''<sub>i<sub>k</sub></sub>''(''f'') = ''a<sub>k</sub>'' is obviously an element of each ''A<sub>i<sub>k</sub></sub>'' so that ''f'' is in each inverse image; thus we have
<math display=block>\bigcap_{k = 1}^N \pi_{i_k}^{-1}(A_{i_k}) \neq \varnothing.</math>

By the FIP definition of compactness, the entire intersection over ''I'' must be nonempty, and the proof is complete.