Question related to Cantor's Theory

[amkamk13]

Hi All,

I'm learning set theory, and I have to answer (true or false to) the following question:

If A={2n|n∈ℕ} then P(A) is equivalent to ℕ.

I understand that according to Cantor's theory, P(X) is always larger than X, and therefore P(X) can't be equivalent to X.
Can we take that a step further to say that P(any infinite set) is larger than all other infinite sets?

TIA!

PS: Please forgive me if this is in the wrong forum, I'm new here.

 

[lookagain]

Hi All,

I'm learning set theory, and I have to answer (true or false to) the following question:

If A={2n|n∈ℕ} then P(A) is equivalent to ℕ.

I understand that according to Cantor's theory, P(X) is always larger than X, and therefore P(X) can't be equivalent to X.

If your "N" stands for the positive integers, then set A is the set of positive even integers.

Sets A and N have positive integers as their elements.

P(A), the powerset of A, has sets as its elements. For P(A) to be equivalent to N, there has to be a one-to-one
correspondence between its elements (sets) and the elements of N (positive integers).

Subsets of P(A) are either finite or infinite in the number of elements.

For example, look at this pairing between part of P(A), where the subsets are two elements at most in number, versus those of the integers of N:

{2} <---> 1
{2, 4} <---> 2
{4} <---> 3
{4, 6} <---> 4
{6} <---> 5
{6, 8} <---> 6
.
.
.

This will exhaust the elements of N, but there will be nothing to match the empty set in P(A) to,
or an infinitude of other remaining subsets of the powerset of A to.

 

[Steven G]

If your "N" stands for the positive integers, then set A is the set of positive even integers.

Sets A and N have positive integers as their elements.

P(A), the powerset of A, has sets as its elements. For P(A) to be equivalent to N, there has to be a one-to-one
correspondence between its elements (sets) and the elements of N (positive integers).

Subsets of P(A) are either finite or infinite in the number of elements.

For example, look at this pairing between part of P(A), where the subsets are two elements at most in number, versus those of the integers of N:

{2} <---> 1
{2, 4} <---> 2
{4} <---> 3
{4, 6} <---> 4
{6} <---> 5
{6, 8} <---> 6
.
.
.

This will exhaust the elements of N, but there will be nothing to match the empty set in P(A) to,
or an infinitude of other remaining subsets of the powerset of A to.

Lookagain, I not necessarily saying you're wrong but I am confused by what you say? For example you could have paired the empty set with 1 and continued the way you did. I just don't see that your pairing is correct to make your final statement. For example (and I know you know this example) you could say that the pairing 2N and N is as follows: 2-->1, 6-->2, 10-->3, ...So this will exhaust all elts of N yet there are still many elements of 2N left over, namely all of 4N.

I just get the feeling that you could have chosen a different pairing to include subsets of size 3 and size 4 ...

Please explain what I am missing. Thanks!

 

[lookagain]

Lookagain, I not necessarily saying your [sic] wrong...

I'll use a different set-up (read: ordering), possibly as you suggested.

Every {2n}, a set which is an element of the powerset of A, gets paired to 2n, an even integer which
is an element of the set N, starting with n = 1.

And every {2n, 2n + 2}, set which is an element of the powerset of A, gets paired to 2n - 1, an odd integer
which is an element of the set N, starting with n = 1.

{2, 4} <---> 1
{2} <---> 2
{4, 6} <---> 3
{4} <---> 4
{6, 8} <---> 5
{6} <---> 6
.
.
.

__________________


All of N "gets used up."

But it just takes one element of the powerset of A, such as the empty set, that fails to match up with
any of the elements of N, to show that there is no one-to-one correspondence. I never need to
bring up all of the rest/any of the other finite length or infinite length elements of the powerset of A,
for instance.

 

[Steven G]

I'll use a different set-up (read: ordering), possibly as you suggested.

Every {2n}, a set which is an element of the powerset of A, gets paired to 2n, an even integer which
is an element of the set N, starting with n = 1.

And every {2n, 2n + 2}, set which is an element of the powerset of A, gets paired to 2n - 1, an odd integer
which is an element of the set N, starting with n = 1.

{2, 4} <---> 1
{2} <---> 2
{4, 6} <---> 3
{4} <---> 4
{6, 8} <---> 5
{6} <---> 6
.
.
.

__________________


All of N "gets used up."

But it just takes one element of the powerset of A, such as the empty set, that fails to match up with
any of the elements of N, to show that there is no one-to-one correspondence. I never need to
bring up all of the rest/any of the other finite length or infinite length elements of the powerset of A,
for instance.

I am sorry but I am still confused. I will give you an example of what I think you are saying.

Let W = whole numbers and N = thenatural numbers.

Now consider this pairing: 1-->1, 2-->2, ... ,n-->n, .... Now it seems (to me!) that you would conclude that there is no 1 to 1 pairing between the two sets because 0 is not included

Yet, this pairing is 1 to 1: 0-->1, 1-->2, ... , n-->n+1, ...

What am I missing?

 

[lookagain]

What am I missing?

I don't think you're missing anything. My idea may work for finite sets, but I have one or more flaws with my demonstration
with these infinite sets. It does not work here.

 

[Steven G]

I don't think you're missing anything. My idea may work for finite sets, but I have one or more flaws with my demonstration
with these infinite sets. It does not work here.

OK, thank you.