1) { }=Q , E={x X : g(x)<f(x)} Show that E= [{x X : g(x)< } {x X : f(x)> }] 2) {x X : g(x)+f(x)<a} = {x X : g(x)<a-f(x)}. If f(x) is a countable function show that a-f(x) is countable function too.
Follow Math Help Forum on Facebook and Google+
Originally Posted by SENTINEL4 1) { }=Q , E={x X : g(x)<f(x)} Show that E= [{x X : g(x)< } {x X : f(x)> }] 2) {x X : g(x)+f(x)<a} = {x X : g(x)<a-f(x)}. If f(x) is a countable function show that a-f(x) is countable function too. Before spending time on this please make sure this is what you mean.
Also, what do you mean by "countable function"? I know what "countable sets" are but not "countable function". It may be a translation problem. Do you mean "measurable function".
@Plato Under the is not n. @HallsofIvy It means measurable not countable.
Last edited by Plato; May 12th 2009 at 01:39 PM.
View Tag Cloud