Re: infinity

Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> writes:

> If you want to think of it that way, I suppose you can, even though
> there is no point at which you are taking away more balls than you
> put in. While set theorists claim that order is irrelevant within a
> set,

For determining the cardinality of a set the order is irrelevant. It
turns out that the cardinality of the set of balls put into the vase
and the cardinality of the set of balls taken out again are the same.
What does this mean? That there is _one_ way of mapping the balls
taken in and taken out 1:1. But since this cardinality happens to be
infinite, and infinite sets can be mapped to a proper subset of
themselves, this does not mean that there might not also be a mapping
1:10. So the question "do balls remain, and how many?" can't be
answered by looking just at the set sizes, because the sets are
infinite. Instead one has to take a look at every ball individually
and has to check whether it will be put in, but not removed. Because
we are talking about infinite sets, and the pigeon hole principle is
no longer valid.

> here they are arguing that there is a difference between whether the
> one ball removed is labeled one way or another. If we add 10x
> through 10x+9 (for x=0 to N) and then remove 10x, then 10x+1 through
> 10x+9 are NEVER removed. However, if you say you are adding 10x
> through 10x+9, and taking away x+1, then you claim that you remove
> all elements. There is a serious inconsistency here between claiming
> that order doesn't matter for sets in general, and then claiming
> that here it does. What difference does the label on the removed
> ball really make?

Without "labels" or other identifying features, you can't decide
whether a given ball, once put in, is taken out again at some time.
Since it is feasible to remove just a subset (we are talking about
infinite sets here), there is no way except checking the balls. I
could equally well forget ball 0 and start the removal action with
ball 1. In that case, at the end ball 0 will be left. I could remove
just the odd-numbered balls, in which case the even-numbered balls
will be left. And so on.

> The solution to the problem lies in forgetting about your
> bijections, and noting that at each of an infinite number of steps
> you are adding a net 9 balls to the vase,

Oh, but counting to 9 is all about bijections.

> and the overall number never decreases from one step to the next.
> You see, it's not that I am missing anything, but simply noting that
> your system produces contradictory results, none of which make very
> good sense.

Once you realize that an infinite set can be mapped 1:1 to a proper
subset of itself, it becomes clear that at the "end of the process",
we can no longer rely on the pigeon hole principle: the number of
things put in and taken out alone does not suffice for determining
anything (well, it is sufficient for determining that the remaining
number of balls must be countable, but that is not really impressive).

--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
.

Relevant Pages

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  • Re: infinity
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