Re: Infinities
Sun, 16 Nov 1997 20:34:50 +0000

> Date: Sun, 16 Nov 1997 08:13:56 -0800 (PST)
> From: John K Clark <>
> Infinite numbers do not obey the same laws of arithmetic that finite numbers
> do, however, you do count them, that is determine how big they are, in
> exactly the same way, by putting them in a one relationship with something
> else. I know that if I can put each of the fingers on my right hand in a one
> to one correspondence with some apples and have no apples or fingers left
> over, then there must be 5 apples. In the same way I can put the odd integers
> in a one to relationship with all the integers, both the odd AND the even,
> so there must be an equal number of both.
> 1 -1
> 3- 2
> 5- 3
> 7- 4
> 9- 5
> .
> .
> Or you can prove that all lines are composed of the same number of points
> regardless of length. Draw 2 parallel lines, a short one and a long one below
> it, pick a point midway along the short line but above it.

> /\
> / \
> /________\
> / \
> /________________\

> Draw a line from that point to any place on the short line, then continue it
> until you hit the long line. You've made a one to one correspondence between
> all the points in the short line and all the points in the long line, so they
> must have an equal number of points.

How did you get this? It seems obvious that because we admitted that
the longer line *was* longer, after you have used all of the
(infinity of) points on the shorter line, there will be points on the
longer line in between the lines you drew across.

> But not all infinities are equal. Let's try to put the integers in a one to
> one correspondence with all the points in the line from 0 to 1 expressed as a
> decimal.
> 1 - 0.a1,a2,a3,a4,a5 ...
> 2 - 0.b1,b2,b3,b4,b5 ...
> 3 - 0.c1,c2,c3,c4,c5 ...
> 4 - 0.d1,d2,d3,d4.d5 ...
> .
> .
> The trouble is it doesn't work, there are decimals not included, for example,
> the point 0.A1,B2,C3,D4,E5 ... where A1 is any digit except a1, B2 is any
> digit except b2, C3 is any digit except c3 etc. This point differs in at
> least one decimal place with any point in our one to one scheme, we've used
> all the integers but there are still points remaining, so there must be more
> points on a line than integers.

But doesn't this apply to the odd number example above? If we use
what appears to be this same logic, since we have used all of the
integers, by pairing them up with only half of the *same* integers,
it follows that the infinity of the integers is equal to half of
itself. Therefore the whole argument must be wrong.