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JoePoe247 t1_j2ayejy wrote

Reply to comment by M8dude in There’s just as many numbers between 0 and 1 as there is from 0 to infinity. by Resinate1

Ok thanks, didnt realize that with the different corresponding function. My logic would be that y=2*(x+1/2) can be used to define every number in both sets (where y is 1-3 and x is 0-1). But there are numbers outside of the set 1-3, say 15. So if I made a different set, inclusive of 1-3 and 15, then there are more numbers in this new set.

I guess I understand that I'm wrong since mathematicians smarter than me have come up with theories/proofs to what you're saying, but I think there's logic in my argument.

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M8dude t1_j2bv9x8 wrote

yeah, there's plenty logic in your argument, there are many different so-called 'measures' to quantify sets of numbers, example the 'distance measure' (i think) of an interval [a, b] is just denoted by b - a.

This is makes the sets have different measure (and more useful ones than just "infinite"), even though they have the same number of numbers.

the measure we would have used before is called the 'counting measure', telling us we'd have to count to infinity for both sets, but that doesn't mean they have the same number of elements (see cantor's second diagonal argument, or yours with the 15), so it has to be shown using a so-called 'bijective function' (our correspondence function), which thank god is pretty easy to construct for any two intervals.

But anyway, good thinking and yep you are right about the example set including [1, 3] and 15, for the counting measure.

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