FOM: Re: AC and Tychonov

Matt Insall montez at
Tue Feb 22 19:32:46 EST 2000

Thank you, professor Ross, for clearing this up.  I should have thought of
this myself.  It's interesting, then, that we can essentially show that
Tychonoff's Theorem in the Hausdorff setting follows from BPI.  Although I
have not gone through the details of the converse, it seems that the
Hausdorff Tychonoff Theorem is equivalent to BPI.  I shall check to see if
Rubin and Howard note this fact.  (They probably do.)

Matt Insall
Associate Professor
Mathematics and Statistics Department
University of Missouri - Rolla
insall at
montez at

----- Original Message -----
From: David Ross <ross at>
To: <montez at>
Sent: Tuesday, February 22, 2000 2:25 PM
Subject: AC and Tychonov

> Of course, full AC is used to construct the nonstandard model and
> its properties
> in the first place.  However, your intuition is good: in the case of the
> product of _Hausdorff_ spaces,
> the nonstandard proof is otherwise choice-free, and can almost certainly
> translated into a proof involving an 'ad-hoc' nonstandard model employing
> ultraproducts of the original topological spaces.
> This reflects the fact that Tychonov for Hausdorff spaces only needs BPI.
> the non-Hausdorff
> case, recall that in the nonst'd proof one starts with an sequence  <x_i>
> *\Prod{X_i}, (both indexed by *I) and shows it is nearstandard to
> (indexed by I).  Since we have no standard part for non-Hausdorff spaces,
> instead *choose* a standard y_i with x_i in its monad for each standard i,
> and show that <x_i> is near the standard sequence <y_i>.
> Regards,
> David A. Ross
> ross at

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