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Comment by seanhunter | Hacker News Reader

“So there must be an equal number of them” is not correct.

Q is indeed dense in R, but firstly it’s very clear that there isn’t an equal number of them because rational numbers are a subset of the real numbers and there exists at least one irrational number (I pick “e”) that is in R but not in Q. So R must be at least bigger than Q.

Additionally you can’t say that between any two rationals there must be a real number because all rational numbers are also real numbers. You can say that Q is dense in R, but if you try to say R is dense in Q what you’re trying to say is “the bits of R that are not in Q are dense in the rest of R” which ends up with a bit of set logic to just be the same statement as the first one.

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> Q is indeed dense in R, but firstly it’s very clear that there isn’t an equal number of them because rational numbers are a subset of the real numbers and there exists at least one irrational number (I pick “e”) that is in R but not in Q. So R must be at least bigger than Q.

This isn't a correct explanation, because I can use the same explanation to show that there are more integers than that there are even integers.

"it’s very clear that there isn’t an equal number of them because even numbers (let's call it E) are a subset of the natural numbers (let's call that N) and there exists at least one odd number (I pick 1) that is in N but not in E. So N must be at least bigger than E."

But as I explain in another thread, that doesn’t apply because E and N are in 1-to-1 correspondence which is not the case with Q and R

That’s what makes this statement incorrect:

> firstly it’s very clear that there isn’t an equal number of them because rational numbers are a subset of the real numbers and there exists at least one irrational number (I pick “e”) that is in R but not in Q

There are N not in E, but E and N have the same cardinality.

You have a second technical mistake as well:

> Additionally you can’t say that between any two rationals there must be a real number because all rational numbers are also real numbers.

They’re obviously referring to Q as a subset of R, and for any two elements of subset Q there is indeed a member of R not in Q.

Why does this work? Just because there is 1 number which is in R but not in Q doesn’t prove anything, does it?

If it did, why are Even N and (Even + Odd N) the same size?

Even N and (even + odd) N have the same size because there is a 1 to 1 correspondence between the sets. For every even number x in N I can pair it with x/2 which is also in N and that gives me all of N so N must be at least as large as even N. Likewise for x in N I can pair x with 2x and that gives me all the even N, so even N must be at least as large as N. So N and even N must have the same size.

However, while I can pair every rational number up with a real number, if I try to go the other way I find there is no rational number to pair with some (actually infinitely many) real numbers. I picked one, e, to show this. So the real numbers must be strictly larger than the rationals. Because the rationals are a strict subset of the reals it’s simpler than the Even/odd natural numbers because you don’t need the correspondence- every x in Q is in R but there exists at least one x in R not in Q, and no amount of correspondence shenanigans can get around that so |R|>|Q|.

> every x in Q is in R but there exists at least one x in R not in Q, and no amount of correspondence shenanigans can get around that so |R|>|Q|.

This also applies to evens and naturals: evens are a strict subset of naturals and there exists naturals which are not evens.

A counter example in the reals is (0,1) and R have the same cardinality, despite there being real numbers not in the interval (0,1).

You have the fact incorrect: for infinite sets, all B being a strict subset of A shows is that |A| >= |B|.

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