Math Forum Discussions

David R Tribble

Posts: 3,426
Registered: 7/21/05

Re: Cantor's first proof,
Posted: Nov 16, 2012 3:11 PM

Plain Text

WM wrote:
> I think I have posted this argument already, but it
> is easily repeated:
>
> Consider the set of all positive rational numbers as existing.
> Fill into the vase all numbers of the interval (0, 1].
> Then enumerate one of them by 1 and take it off the vase.
> Fill into the vase all numbers of the interval (1, 2].
> Then enumerate one of them by 2 and take it off the vase.
> Fill into the vase all numbers of the interval (2, 3].
> Then enumerate one of them by 3 and take it off the vase.
> Continue until all rational numbers have been enumerated.
> Then all have been taken off the vase.
> The remaining set of not enumerated rationals is empty.

Here it is again, but with minor corrections for clarity:

> Consider the set of all positive rational numbers.
> Put into the vase all rationals in the interval (0, 1].
> Then denote one of them by '1' and remove it from the vase.
> Put into the vase all rationals in the interval (2, 3].
> Then denote one of them by '2' and remove it from the vase.
> Continue until all rational numbers have been added.
> Then all rationals have been denoted and taken out of the vase.
> The remaining set of non-denoted rationals is empty.

Okay, so at step 1 (assuming we start with an empty vase),
we add all the rationals in (0,1] to the vase. We then denote
one of them, let's say 1/2, by '1', and remove it from the vase.
So we're left with all the rationals in (0,1] \ 1/2.

Step 2, we add all the rationals in (1,2]. Then we denote one
of them, let's say 3/2, by '2', and remove it from the vase.
So we're left with all the rationals in (0,2] \ {1/2, 3/2}.

Step 3, we add all the rationals in (2,3]. Then we denote one
of them, let's say 5/2, by '3', and remove it from the vase.
So we're left with all the rationals in (0,3] \ {1/2, 3/2, 5/2}.

And so on.

So what's left in the vase is all the rationals in (0,oo) except
for {1/2, 3/2, 5/2, 7/2, ...} (i.e., all odd multiples of 1/2).
Obviously, a (countably) infinite number of rationals remain
in the vase.

Since this is doubtless not the result what you expected, you
should see that the problem is that your conditions are not stated
precisely enough.

But even if we modify your procedure, we still don't get the results
you expect. Let's say that at each step we only denote and remove
a rational in the interval (0,1) from the vase. So in step 1 we
remove, say, 1/2. At step 2, we remove 1/3. At step 3, we remove
1/4. And so on. But even with this modification, we still end
up with a vase full of a countably infinite set of rationals,
specifically, Q+ \ {1/2, 1/3, 1/4, 1/5, ...}.

You're trying to add a countably infinite number of rationals to
the vase at each step and remove a single one. The problem is that
you are not precise enough at specifying _which_ rational should
be removed at each step.

You would like, it seems, for all of the rationals to be removed
by the end. That is indeed one possible outcome, _provided_ that
you specify with more precision _which_ specific rational gets
removed at each step. Without that, most of the possible outcomes
are all pretty much the same: an infinite number of rationals remain
in the vase.

-drt