Re: Cantor For Dummies ...

On Mar 28, 12:23 pm, riderofgiraffes <mathforum.org...@xxxxxxxxxxxxxx>
wrote:
Consider the committee C that is made up as
follows. For each delegate Di, if they are
on the committee that convenes in their room,
leave them out of C. If they are not in the
committee that convenes in their room, put
them in C. C is the committee of those
delegates displaced when their room is used.

Ask - where does committee C meet?

Before we answer that, please explicitly list
the delegates in that committee. If you can't,
there must not be such a committee. Nonexistent
committees don't meet and therefore aren't
assigned a room.

I would be delighted to do so. Please provide
for me the assignment of committees to rooms.
The committee C is derived from that assignment,
and the subsequent discussion is valid for any
such assignment.

However, if you want me to list committee C
explicitly, then I will do so once you have
produced for me the assignment.

I can't explicitly do so, since my lifetime
is finite in length. But if I had an unbounded
life, I would gladly supply you with such a list.
(Assuming you pay me by the hour, say $1000/hr,
and I only need to work few hours per week.)
You, however, would be left with no possible
chance at creating such a committee since there
will be no time left after all of eternity.

But surely you could name the first committee
in one hour, then the next in 30 minutes, the
next in 15 minutes, the next in 7.5 minutes,
and so on, each time taking half the previous
time to name the next committee. In that way
you'd be done in two hours.

But really I guess I must be a bit thick, because
I don't follow the point you're making. I've
shown that every time there's an assignment of
committees to rooms, there's a committee left
off. I've even shown how to construct a specific
committee that's been left off. It's an explicit
rule to follow, and seems to me to be perfectly
well defined.

So let me try again. Let's suppose you present
me with an assignment of committees to rooms. In
the first hour of examining this assignment I can
specify whether D1 is in C or not, because I can
check to see if D1 is in the committee that meets
in R1. Then in the next 30 minutes I can move on
to determining whether D2 is on C, and so on. Thus
I have an explicit algorithm for determining C.

Now, would you be so kind as to explain clearly
your objection?

I don't have a single objection, but if we list all
the algorithms for generating committes, and you have
an algorithm for generating a committee, then the
committe that your algorithm generates is in such
a list.

.

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