4 ms·
Just a guess, but I'm pretty sure that if you have a totally-ordered set (like in your infinite pixel screen, dictionary order on first coordinate followed by s
by rtpg 12y ago
Just a guess, but I'm pretty sure that if you have a totally-ordered set (like in your infinite pixel screen, dictionary order on first coordinate followed by second), and all subsets S = {x/ a<=x<=b } are countably infinite, then the set is countably infinite.
A constructive proof of this is probably doable by inspiring oneself off of Hilbert's hotel paradox (http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Gran...)
- ontario 12y agoEven though this sounds plausible, it's actually false. The counterxample is the first uncountable ordinal ω₁ [1], viewed as a well-ordered set (and thus a fortiori totally ordered). For every a,b in ω₁, the closed interval [a,b] is countable (by definition), but ω₁ itself is not. [1] https://en.wikipedia.org/wiki/First_uncountable_ordinal https://en.wikipedia.org/wiki/First_uncountable_ordinal
- rtpg 12y agoah, I remember seeing that in a set theory book I was struggling through at one point. Might be worth a second try so that I don't miss this