Timeline for How to construct a continuous finite additive measure on the natural numbers

12 events

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Dec 1, 2020 at 5:26	comment	added	Alex Ravsky		The range of $f$ is $[0,1]$, not $\{0,1\}$. Thus any set $A\subset\Bbb N$ such that $0<f(A)<1$ intersects any set of $f^{-1}(\{1\})$, but $A\not\in f^{-1}(\{1\})$.
Feb 19, 2020 at 16:13	comment	added	YCor		Conversely for any such $f$, $f^{-1}(\{1\})$ is a non-principal ultrafilter. So it's just the same as asking about the "construction" of a non-principal ultrafilter on $\mathbf{N}$.
S May 9, 2012 at 14:28	vote	accept	Jialiang He
S May 9, 2012 at 14:27	vote	accept	Jialiang He
S May 9, 2012 at 14:28
May 9, 2012 at 11:12	answer	added	KP Hart		timeline score: 7
May 5, 2012 at 14:08	vote	accept	Jialiang He
S May 9, 2012 at 14:27
May 4, 2012 at 9:42	history	edited	Neil Strickland	CC BY-SA 3.0	Spelling, grammar, language
May 4, 2012 at 6:15	comment	added	Yemon Choi		@David: I think this means the set of natural numbers
May 4, 2012 at 5:50	answer	added	Stefan Geschke		timeline score: 16
May 4, 2012 at 5:47	comment	added	David Roberts♦		What do you mean by 'nature'?
May 4, 2012 at 5:03	comment	added	user5810		How do you define whether or not $f$ is continuous? $\:$
May 4, 2012 at 4:41	history	asked	Jialiang He	CC BY-SA 3.0