Cardinal numbers proof

[tashe]

How can I prove these lines:

Let N be a set of natural numbers and |N|=x[sub:15ih65jb]o[/sub:15ih65jb].
How can i prove that
x[sub:15ih65jb]0[/sub:15ih65jb] + k = x[sub:15ih65jb]0[/sub:15ih65jb]
or
x[sub:15ih65jb]0[/sub:15ih65jb]*x[sub:15ih65jb]0[/sub:15ih65jb]=x[sub:15ih65jb]0[/sub:15ih65jb]

Help anyone?

 

[pka]

Here is some bad news for you: Unless you can tell exactly what your text material is, then I doubt that we can help you. I have about 20 books in my library that contain discussions of these problems. There are no two of which use exactly the same approach. In fact, several of them use widely different notations and definitions. If I were to post help using a wildly different approach than the one in you textbook, that would only confuse you badly.

 

[tashe]

The lessons is about cardinal numbers.
X[sub:949tr7ty]o[/sub:949tr7ty] is the cardinal number of the set N
and k is some constant number.

I found the first example solved in the book this way minutes ago:
Prove that:
X[sub:949tr7ty]o[/sub:949tr7ty] * X[sub:949tr7ty]o[/sub:949tr7ty] = X[sub:949tr7ty]o[/sub:949tr7ty]

Let S=NxN i.e. a set of all the pairs of the natural numbers. S[sub:949tr7ty]k[/sub:949tr7ty] is the set
[(1,k),(2,k),...,(n,k),...]. Then S=US[sub:949tr7ty]k[/sub:949tr7ty] (which means that S is a union of all the S[sub:949tr7ty]k[/sub:949tr7ty] ).
From the above seen we can see that S is a countable union of countable sets, which by definition makes S a countable set.

 

[stapel]

The lessons is about cardinal numbers.

Yes, we're aware of that. What you seem not aware of is that this material is not presented in exactly the same way in any two texts. Unless we were in your class or happened to have taught from your particular text, or unless you can provide us with all of the theorems, definitions, properties, axioms, and such at your disposal, we almost certainly cannot assist.

Sorry!

Eliz.

 

[pka]

Prove that: X[sub:fcgxxmot]o[/sub:fcgxxmot] * X[sub:fcgxxmot]o[/sub:fcgxxmot] = X[sub:fcgxxmot]o[/sub:fcgxxmot]
Let S=NxN i.e. a set of all the pairs of the natural numbers. S[sub:fcgxxmot]k[/sub:fcgxxmot] is the set
[(1,k),(2,k),...,(n,k),...]. Then S=US[sub:fcgxxmot]k[/sub:fcgxxmot] (which means that S is a union of all the S[sub:fcgxxmot]k[/sub:fcgxxmot] ).
From the above seen we can see that S is a countable union of countable sets, which by definition makes S a countable set.

Well I now have some vague idea of the approach that your text uses.
But note that it is not by definition that the countable union of pair-wire disjoint sets is a countable set. That is a theorem and must be proven.

 

[tashe]

don't bother man
i got an answer for the question elsewhere

http://www.mathhelpforum.com/math-help/ ... proof.html

thanks for the effort

 

[stapel]

i got an answer for the question elsewhere

You got an answer. Depending upon your particular text, the answer might or might not be acceptable to your grader. :shock:

Eliz.

 

[PAULK]

Re:

i got an answer for the question elsewhere

You got an answer. Depending upon your particular text, the answer might or might not be acceptable to your grader. :shock:

Eliz.

I agree. So you might want to quote THIS theorem:

The union of a countable collection of finite sets is countable.

NOW your X0 * X0 set, which means the set of ordered pairs of integers, can be arranged this way:

Let Ak = { (x,y), s.t. x + y = k }

Clearly Ak is a finite set, and:
1. the union of all the Ak's is X0*X0
2. { Ak } is a countable collection.