From: israel@math.ubc.ca (Robert Israel)
Subject: Re: sequences in infinite product topology
Date: 15 Mar 1999 22:10:27 GMT
Newsgroups: sci.math
In article , "Dr. Michael Albert" writes:
|> Consider the set of all functions from the unit interval
|> to the reals. Definite a set as L-open iff for each
|> function f in the set, for any sequence f_n which converges
|> to f pointwise, f_n is eventually in the set. This defines
|> a topology which is at least as fine as the usual product
|> topology (that is, any set which is open in the usual product
|> topology is open in this topology). Is this equivalent
|> to the usual product topology?
No. Let U be the set of (Lebesgue) nonmeasurable functions.
Since the limit of a pointwise convergent sequence of measurable
functions is measurable, U is L-open. But U is not open in the
product topology: every nonempty open set in the product topology
contains measurable functions.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
==============================================================================
From: Fred Galvin
Subject: Re: sequences in infinite product topology
Date: Mon, 15 Mar 1999 12:39:42 -0600
Newsgroups: sci.math
On Mon, 15 Mar 1999, Dr. Michael Albert wrote:
> Consider the set of all functions from the unit interval
> to the reals. Definite a set as L-open iff for each
> function f in the set, for any sequence f_n which converges
> to f pointwise, f_n is eventually in the set. This defines
> a topology which is at least as fine as the usual product
> topology (that is, any set which is open in the usual product
> topology is open in this topology). Is this equivalent
> to the usual product topology?
No; {f:{x:f(x)>0} is uncountable} is L-open but is not open in the usual
product topology.