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Pietro Majer
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Definitely, the one I like the most is the proof via ultrafilters. You only have to state the compactness of a topological space in terms of ultrafilters, which is easily obtained by the definition via open coverings (warning: the equivalence of the definitions is where one uses AC)

X is compact if and only if every ultrafilter is convergent.

Then one observes that

  1. any continuous image image of an ultrafilter is an ultrafilter (in particular, any projection from a product space)

  2. any filter in the product space converges if and only if all its projections converge .

You really only need a few definitions and few natural properties. My test about how nice is a proof is: can I teach it to somebody just while standing in the queue at the canteen, on into subway car?

Definitely, the one I like the most is the proof via ultrafilters. You only have to state the compactness of a topological space in terms of ultrafilters, which is easily obtained by the definition via open coverings (warning: the equivalence of the definitions is where one uses AC)

X is compact if and only if every ultrafilter is convergent.

Then one observes that

  1. any continuous image of an ultrafilter is an ultrafilter (in particular, any projection from a product space)

  2. any filter in the product space converges if and only if all its projections converge .

You really only need a few definitions and few natural properties. My test about how nice is a proof is: can I teach it to somebody just while standing in the queue at the canteen, on into subway car?

Definitely, the one I like the most is the proof via ultrafilters. You only have to state the compactness of a topological space in terms of ultrafilters, which is easily obtained by the definition via open coverings (warning: the equivalence of the definitions is where one uses AC)

X is compact if and only if every ultrafilter is convergent.

Then one observes that

  1. any image of an ultrafilter is an ultrafilter (in particular, any projection from a product space)

  2. any filter in the product space converges if and only if all its projections converge .

You really only need a few definitions and few natural properties. My test about how nice is a proof is: can I teach it to somebody just while standing in the queue at the canteen, on into subway car?

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Pietro Majer
  • 60.5k
  • 4
  • 122
  • 269

Definitely, the one I like the most is the proof via ultrafilters. You only have to state the compactness of a topological space in terms of ultrafilters, which is easily obtained by the devinitiondefinition via open coverings (warning: the equivalence of the definitions is where one uses AC)

X is compact if and only if every ultrafilter is convergent.

Then one observes that

  1. any projection of an ultrafilter in the product to a factor is an ultrafilter;

    any continuous image of an ultrafilter is an ultrafilter (in particular, any projection from a product space)

  2. any filter in the product space converges if and only if all its projections converge .

    any filter in the product space converges if and only if all its projections converge .

You really only need a few definitions and few natural properties. My test about how nice is a proof is: can I teach it to somebody just while standing in the queue at the canteen, on into subway car?

Definitely, the one I like the most is the proof via ultrafilters. You only have to state the compactness of a topological space in terms of ultrafilters, which is easily obtained by the devinition via open coverings:

X is compact if and only if every ultrafilter is convergent.

Then one observes that

  1. any projection of an ultrafilter in the product to a factor is an ultrafilter;
  2. any filter in the product space converges if and only if all its projections converge .

You really only need a few definitions and few natural properties. My test about how nice is a proof is: can I teach it to somebody just while standing in the queue at the canteen, on into subway car?

Definitely, the one I like the most is the proof via ultrafilters. You only have to state the compactness of a topological space in terms of ultrafilters, which is easily obtained by the definition via open coverings (warning: the equivalence of the definitions is where one uses AC)

X is compact if and only if every ultrafilter is convergent.

Then one observes that

  1. any continuous image of an ultrafilter is an ultrafilter (in particular, any projection from a product space)

  2. any filter in the product space converges if and only if all its projections converge .

You really only need a few definitions and few natural properties. My test about how nice is a proof is: can I teach it to somebody just while standing in the queue at the canteen, on into subway car?

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Pietro Majer
  • 60.5k
  • 4
  • 122
  • 269

Definitely, the one I like the most is the proof via ultrafilters. You only have to state the compactness of a topological space in terms of ultrafilters, which is easily obtained by the devinition via open coverings:

X is compact if and only if every ultrafilter is convergent.

Then one observes that

  1. any projection of an ultrafilter in the product to a factor is an ultrafilter;
  2. any filter in the product space converges if and only if all its projections converge .

You really only need a few definitions and few natural properties. My test about how nice is a proof is: can I teach it to somebody just while standing in the queue at the canteen, on into subway car?