proof of Tychonoff’s theorem in finite case
(The finite case of Tychonoff’s Theorem is of course a subset of the infinite
case,
but the proof is substantially easier, so that is why it is presented here.)
To prove that is compact
if the are compact, it suffices (by induction
) to prove that is compact
when and are. It also suffices to prove that
a finite subcover can be extracted from every open cover of
by only the basis sets of the form , where is open in and is open in .
Proof.
The proof is by the straightforward strategy of composing a finite subcover from a lower-dimensional subcover. Let the open cover of by basis sets be given.
The set is compact, because it is the image of a
continuous embedding
of the compact set .
Hence has a finite subcover in : label the subcover
by .
Do this for each .
To get the desired subcover of , we need to pick a finite number
of . Consider .
This is a finite intersection of open sets, so is open in .
The collection
is an open covering of , so pick
a finite subcover .
Then is a finite subcover
of .
∎
Title | proof of Tychonoff’s theorem in finite case |
---|---|
Canonical name | ProofOfTychonoffsTheoremInFiniteCase |
Date of creation | 2013-03-22 15:26:27 |
Last modified on | 2013-03-22 15:26:27 |
Owner | stevecheng (10074) |
Last modified by | stevecheng (10074) |
Numerical id | 4 |
Author | stevecheng (10074) |
Entry type | Proof |
Classification | msc 54D30 |