Editing Net (mathematics) (section)

=== Tychonoff's theorem and relation to the axiom of choice ===

If no <math>L \in X</math> is given but for every <math>i \in I,</math> there exists some <math>L_i \in X_i</math> such that <math>\pi_i\left(f_\bull\right) \to L_i</math> in <math>X_i</math> then the tuple defined by <math>L = \left(L_i\right)_{i \in I}</math> will be a limit of <math>f_\bull</math> in <math>X.</math> 
However, the [[axiom of choice]] might be need to be assumed to conclude that this tuple <math>L</math> exists; the axiom of choice is not needed in some situations, such as when <math>I</math> is finite or when every <math>L_i \in X_i</math> is the {{em|unique}} limit of the net <math>\pi_i\left(f_\bull\right)</math> (because then there is nothing to choose between), which happens for example, when every <math>X_i</math> is a [[Hausdorff space]]. If <math>I</math> is infinite and <math>{\textstyle\prod} X_\bull = {\textstyle\prod\limits_{j \in I}} X_j</math> is not empty, then the axiom of choice would (in general) still be needed to conclude that the projections <math>\pi_i : {\textstyle\prod} X_\bull \to X_i</math> are [[surjective map]]s.

The axiom of choice is equivalent to [[Tychonoff's theorem]], which states that the product of any collection of compact topological spaces is compact. 
But if every compact space is also Hausdorff, then the so called "Tychonoff's theorem for compact Hausdorff spaces" can be used instead, which is equivalent to the [[ultrafilter lemma]] and so strictly weaker than the [[axiom of choice]]. 
Nets can be used to give short proofs of both version of Tychonoff's theorem by using the characterization of net convergence given above together with the fact that a space is compact if and only if every net has a convergent [[Subnet (mathematics)|subnet]].