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by Nevermark 1760 days ago
> ... intuitive constructions like "almost all real numbers are outside the closed unit interval 0 <= x <= 1".

But that would be wrong wouldn't it? I can produce all real numbers by pairing a finite number of (in this case 3) real numbers outside the set with each real number inside the set.

For each real x, in 0 <= x <= 1, we also have:

1/x (covers all real x, 1 <= x <= +infinity

-x (covers all real x, -1 <= x <= 0

-1/x (covers all real x, -infinity <= x <= -1)

The cardinality of all those reals outside of 0 <= x <= 1 is therefore 3x the cardinality of those inside 0 <= x <= 1, in this construction. But for infinite cardinalities the 3 can be discarded.

So there are exactly as many real numbers in 0 <= x <= -1 as outside it.

2 comments

I don't disagree that these sets are the same cardinality. But cardinality isn't the only way to describe the "size" of a set.

I suppose the typical measure-theoretic definition of "almost all" / "almost everywhere" insists on "everywhere but a zero-measure set", and you can't define a measure that satisfies sigma-additivity that treats intervals of finite Lebesgue measure as such, while ascribing nonzero measure to sets of infinite measure.

But even so, the Lebesgue measure of R is infinite, while the same measure of the unit interval is 1.

I think so?

If I do it the "normal" way (repeating the [0, 1] interval infinitely many times), there are infinite times as many reals outside the [0, 1] interval as there are in it.

But that "infinite times" is a countable infinity - the number of integers. How does "the number of reals in [0, 1] times the number of integers" compare to "the number of reals in [0, 1]"? Are they the "same" infinity?

What if we use rationals instead of reals? We can do the same x 3 thing, right? But the number of rationals is countably infinite, and "3 times countably infinite" is the same as "countably infinite times countably infinite", isn't it?