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Topic: Countably Infinite Sets
Replies: 7   Last Post: Dec 29, 2012 12:45 PM

 Messages: [ Previous | Next ]
 James Waldby Posts: 545 Registered: 1/27/11
Re: Countably Infinite Sets
Posted: Dec 28, 2012 6:58 PM

On Fri, 28 Dec 2012 14:16:14 -0800, netzweltler wrote:
> On 28 Dez., 21:53, James Waldby wrote:
>> On Fri, 28 Dec 2012 12:51:29 -0600, fasnsto wrote:
>> > "netzweltler" <reinhard_fisc...@arcor.de> wrote ...
>> >> Does the set {{1}, {1,2}, {1,2,3}, ...} contain all natural numbers?
>>
>> > no, it would contain the set of all natural numbers.
>>
>> Let S = {{1}, {1,2}, {1,2,3}, ...}.  The infinite union of the members
>> of S would be a set containing all natural numbers, but no member of S
>> is itself a set containing all natural numbers.  Of course, as noted in
>> some earlier replies, S is isomorphic to the naturals.

> I understand, that the union of the members of S contains all natural
> numbers. But, did I really write the union here?
> {{1}, {1,2}, {1,2,3}, ...}

Here in math.sci, one expects (or at least hopes for) accurate use of
terminology. {{1}, {1,2}, {1,2,3}, ...} is not a union as such. This
set is a set of sets. That is, this set is a set where each member is
a set. This set is not a list of sets or a list of rows because a list
is not a set and a set is not a row.

> Isn't it a list of infinitely many rows like
> 1
> 1,2
> 1,2,3
> ...
>
> Does the list contain all natural numbers? Is this list an union?

A list is not a union. Regarding the list of infinitely many rows
indicated just above, if we call the display of all the rows a
"tableau", we could say that the tableau contains all natural numbers.
We could also say that the list contains the natural number 1, and
for each natural number contains a list that ends with that natural
number. But in my opinion it is wrong to assert that every natural
number appears in this list of lists, because each thing (except 1)
appearing in this list is in turn a list, not a number.

--
jiw

Date Subject Author
12/28/12 James Waldby
12/28/12 netzweltler
12/28/12 Virgil
12/28/12 James Waldby
12/29/12 netzweltler
12/29/12 Scott Berg