Re: infinity

Tony Orlow says...

>When I am applying the fact that the largest element in an
>initial segment of the naturals starting at 1 is always the
>same number as the set size

As has been pointed out to you many times, it is only a "fact"
for sets of the form A_n = { 1, 2, ..., n }. It is not a fact
for sets with no largest element. If there is no largest element,
then of course the size is not equal to the largest element.

>> On the contrary, aleph_0 is very precisely defined:
>>
>> aleph_0 is the smallest infinite ordinal
>
>Do you honestly think that is a precise definition?

Yes.

>First of all, what makes you think there IS a smallest infinite
>ordinal?

It's true by definition of "ordinal". By definition, the ordinals
are well-ordered, which means that any nonempty collection of ordinals
has a smallest element. It immediately follows that there is a smallest
infinite ordinal.

The simplest representation of the ordinals is to identify
them with the Von Neumann numerals. These numerals have the
nice property that for any ordinal alpha:

alpha = the set of all ordinals less than alpha

So 0 = {} (the empty set, since there are no ordinals less than 0)
1 = { 0 } (since 0 is the only ordinal less than 1).
2 = { 0, 1 }
etc.

Using this representation of the ordinals, for any two ordinals
alpha and beta,

alpha < beta
<->
alpha is an element of beta

Then we can define aleph_0 to be the set of all finite ordinals.

aleph_0 then is clearly *not* finite (since it contains all finite
ordinals, and no ordinal can be an element of itself). There can
be no infinite ordinal less than aleph_0, because:

x < aleph_0
<-> x is an element of aleph_0
<-> x is a finite ordinal

>If you subtract 1 from an infinite, do you get a finite number?

The operation of subtraction is not defined for all ordinals. By
definition,

x - 1 = that number y such that y+1 = x.

However, if x is a limit ordinal, there *is* no such y. So x-1
is undefined. That's the definition of a limit ordinal.

--
Daryl McCullough
Ithaca, NY

.

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