Question regarding subsets

[chadman201]

Assume that A ⊆ N(natural numbers) , and that if k ⊆ A then k+1 ⊆ A. Can we conclude that k-1 ⊆ A and therefore conclude that A = N ?

Thanks.

 

[Dr.Peterson]

Please check what you wrote, which as written is meaningless.

If k is a subset of A, then it is a set, not a number, so you can't add 1 to it. Did you mean to say that k is an element of A? But there are still other issues after dealing with that.

Also, please tell us what you think the answer is, and why. That will help clarify your meaning.

 

[chadman201]

Sorry I made a mistake, I meant to say that k ∈ A and that k +1 ∈ A.

i'm wondering if we can assume that k - 1 is also a subset of A and therefore conclude that all numbers above and below k are all elements of A.

Example:
k = 25
so k+1 is 26
and they are both elements of A

i'm asking if we can assign 25 to k+1 and have k equal to 24,
so that if we assign any number to k we can conclude that every natural number greater or less than k is an element of A

 

[Dr.Peterson]

Sorry I made a mistake, I meant to say that k ∈ A and that k +1 ∈ A.

i'm wondering if we can assume that k - 1 is also a subset of A and therefore conclude that all numbers above and below k are all elements of A.

Example:
k = 25
so k+1 is 26
and they are both elements of A

i'm asking if we can assign 25 to k+1 and have k equal to 24,
so that if we assign any number to k we can conclude that every natural number greater or less than k is an element of A

On what grounds would you conclude that?

You are saying that if k is in the set, then k+1 is in the set. How can you then conclude that if k+1 is in the set, then k is in the set? The converse is not equivalent to the original statement.

 

[pka]

Example:
k = 25
so k+1 is 26
and they are both elements of A
i'm asking if we can assign 25 to k+1 and have k equal to 24,
so that if we assign any number to k we can conclude that every natural number greater or less than k is an element of A

The answer is NO! Consider the set \(A=\{k\in\mathbb{N} : 25+k\}\).
We know that \(0\in\mathbb{N}\) so is \(25\in A~?\)
If \(J\in A\) does it follow that \(J+1\in A~?\)
If \(J\in A\) does it follow that \(J-1\in A~?\)
Could that mean that \(A=\mathbb{N}~?\)

 

[chadman201]

so what you're saying is that i cant conclude that the numbers below 25 are in the subset A?
what about all the numbers greater than 25,
can we say that the subset A contains all the natural numbers greater than 25?

 

[Dr.Peterson]

so what you're saying is that i cant conclude that the numbers below 25 are in the subset A?
what about all the numbers greater than 25,
can we say that the subset A contains all the natural numbers greater than 25?

If 25 is in the set, then all natural numbers greater than 25 are in it, by induction.

But there is no way to conclude that 24 is in the set. Do you not see this?

Please tell us what course you are taking, and what you have learned about sets and about logic.

 

[pka]

so what you're saying is that i cant conclude that the numbers below 25 are in the subset A?
what about all the numbers greater than 25, can we say that the subset A contains all the natural numbers greater than 25?

Can you at least answer these questions?
The answer is NO! Consider the set \(A=\{25+k\text{ for all }k\in\mathbb{N}\}\).
We know that \(0\in\mathbb{N}\) so is \(25\in A~?\)
If \(J\in A\) does it follow that \(J+1\in A~?\) Please explain.
If \(J\in A\) does it follow that \(J-1\in A~?\) Please explain.
If \(J<25\) can \(J\in A~?\) If you don't understand any of these please tell what you don't understand.

 

[Steven G]

Assuming that N = {1,2,3,...} then what happens if k=1? That should bother you greatly. After all if k=1, then k-1 = 0 which is NOT in N