Randomly selecting from an infinite number of choices

I made a mistake... I was only thinking of reals between zero and one, so all numbers chosen were to the right of the decimal point.

Having infinite digits to the left of the decimal point poses a problem, in that the number selected will be infinite... and I think that we need to guarantee that the selected number will be finite.

So it looks like selecting a random real from a uniform distribution of all reals might indeed be a hypertask... which is puzzling to me if randomly selecting from a subset of the reals in a given range is only a supertask.

So, to the mathematicians out there - what substantial difference is there between the set of reals that can be selected by randomly choosing infinite digits, and the set of all reals? And why does this seem to make a difference in the number of tasks required to select a random element from the set? (Does it make a difference?)

Do the two sets have the same cardinality?
I think the first set is bounded... does that make a difference?

 

Log in or Sign up to hide all adverts.

Ah, okay... although... doesn't the real segment [0,1] actually have the same cardinality as the entire real line?

Right, but that's again the same problem with drawing from the set of naturals, right? Put it this way: given some (randomly selected) natural number, we could then produce a randomly selected real number by generating a (countably) infinite number of digits to the right of the decimal.

Yeah, I think it's actually a hypertask to choose a real from, say, [0,1]. But I'm definitely on thin ice here, so...

 

Log in or Sign up to hide all adverts.

Yes, I think it does. Which is why I'm puzzled. The set of reals in [0,infinity) seems to be a lot like the set in [0,1].

Right. Which is why I now agree that choosing randomly from a uniform distribution of all reals might be a hypertask.

Why?
Choosing countably infinite random digits to place after the decimal point is a supertask, right?
And the pool of possible results is the same as the reals in [0,1], right?

 

Log in or Sign up to hide all adverts.

Another mistake... I meant closed, not bounded.
But it doesn't make a difference, because we can make the first set open by simply removing the largest element - the one containing all nines.

I'm still not 100% sure on cardinality... does the set [0,1) have the same cardinality as [0,infinity)?
I suspect that it does, but how do you demonstrate it?

 

Yeah, there is a bijection between them (i.e., f(x) = 1/x... well, that only gets you [1,infinity), but clearly that has the same cardinality as [0,infinity) )

I believe so, yes.

Yeah, that's where I get stuck too. I think the answer must have to do with the issue that all natural numbers must eventually end up with an infinite string of zeros, whereas the reals do not have this constraint. But I'm not sure how to make that concrete...

Yeah, see above. There are various ways to demonstrate that kind of thing, but finding a bijection between the two sets is probably the most direct way.

Note that you can also show that R^n has the same cardinality as R.

 

You guys are stuck because the underlying idea is nonsense. The concept of a uniform distribution over an infinite range (or a uniform distribution over a countably infinite set) does not make any sense. There are a plethora of perfectly valid distributions over an infinite range (e.g. guassian, exponential, ...) and over a countably infinite set, but a uniform distribution is not one of them.

 

It's all nonsense, in fact. Even having an idea that it's all nonsense, is complete nonsense.

 

The selected set is countable - same size as the integers.

You could, for example, time-stamp each choice.

 

After a bit of thought, it seems to me that the relevant measure here is not cardinality, but order type.

Also, an overview of related topics.