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Herkfixer t1_j2963q7 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

Yes, for real numbers. My point is that the OP not specifying the two classes of numbers makes it untrue.

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M8dude t1_j29dx0r wrote

i think it's fair to assume that OP is talking about the real numbers.

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Haven_Stranger t1_j29vkgd wrote

Two classes? What two classes? OP specified "numbers between 0 and 1" and "[numbers] from 0 to infinity". The class involved (entailed, even) is "numbers on the number line". In other words, reals. That's one class. That's the basis of the comparison.

Both specified ranges are uncountably infinite.

Here's an easier comparison: There are as many numbers between 0 and 1 as there are between 1 and infinity. It's an easier comparison because now both sets are strictly between their bounds, the two sets don't overlap, and the bijection formula is simpler.

If A is a real between 0 and 1, and B is the matching real greater than 1, then the mapping is:

A -> 1 / B

Also

B -> 1 / A

That's it. For every real number larger than one, there exists exactly one matching number between zero and one, and vice-versa. No exceptions, no excuses, nothing left unaccounted.

The size of the two sets are exactly the same, even though the extents of the two sets are wildly different.

Also also, that's the comparison OP meant to express. Even so, the comparison posted still holds true. It just that, instead of mapping A to the inverse of B, we map A to the inverse of one more than B.

So, no, we don't have to compare reals on one side and rationals on the other, or anything else where we'd have to specify two classes. Those comparisons can be made, of course, but they're not relevant to the post.

It takes different real numbers to make it all the way out to infinity, but it doesn't take more of them. There are exactly as many real numbers greater than 1 as there are between 0 and 1.

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