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.