Variables in Set Theory

[IDontEvenHaveABrain]

I am doing an exercise in the book how to prove it by velleman and I have problems understanding the logical expression in the picture I attached.
The exercise was: Analyze the logical form of the following statement. My answer was correct, but I don't really understand it.

The following expressions should all be equal, right?
[MATH] \mathscr{P}(A) \subseteq \mathscr{P}(B) \\ \forall x(x \in \mathscr{P}(A) \rightarrow x \in \mathscr{P}(B))\\ \forall x(x \subseteq A \rightarrow x \subseteq B) \\ \forall x[ (\forall y(y \in x \rightarrow y \in A)) \rightarrow (\forall y(y \in x \rightarrow y \in B))] [/MATH]
I understand that for every x, when x is a subsset of A, then it is also a subset of B

[MATH] \forall x(x \subseteq A \rightarrow x \subseteq B) \\ [/MATH]
But the following expression:

[MATH] \forall x[ (\forall y(y \in x \rightarrow y \in A)) \rightarrow (\forall y(y \in x \rightarrow y \in B))] [/MATH]
Doesn't that mean that for every x and every y, when y is in x, then y is also in A.
That doesn't make sense in my opionen. Let's take:

[MATH] x = \{ 1, 2, 3 \} \\ y = 3 [/MATH]
There could be the case, that x is not a subset of A, but y is in x. So the last two statements of the 4 I stated in the beginning should not be equal, right?
I am obviously making a mistake somewhere. I hope someone could clear things up.

Have a good day.

 

[IDontEvenHaveABrain]

Just some clarification:

The exercise:
Analyze the logical forms of the following statements.
[MATH] 1.)x \epsilon \mathscr{P}(A) \\ \\ 2.) \mathscr{P}(A) \subseteq \mathscr{P}(B) \\ [/MATH]
My answer to the first excerise was:

[MATH] x \in \mathscr{P}(A) \\ = x \subseteq A \\ = \forall y( y \in x \rightarrow y \in A) \\ [/MATH]
My answer to the second exercise:

[MATH] \mathscr{P}(A) \subseteq \mathscr{P}(B)$ \\ = \forall x(x \in \mathscr{P}(A) \rightarrow x \in \mathscr{P}(B)) \\ = \forall x(x \subseteq A \rightarrow x \subseteq B) \\ = \forall x[ (\forall y(y \in x \rightarrow y \in A)) \rightarrow (\forall y(y \in x \rightarrow y \in B))] \\ [/MATH]
Now my problem is, do I first have to assume that x is a subset of A:
Assume [MATH] x \subseteq A [/MATH][MATH] \forall x[ (\forall y(y \in x \rightarrow y \in A)) \rightarrow (\forall y(y \in x \rightarrow y \in B))] \\ [/MATH]
Or would

[MATH] \forall x[ (\forall y(y \in x \rightarrow y \in A)) \rightarrow (\forall y(y \in x \rightarrow y \in B))] \\ [/MATH]
be enough. Because the [MATH] \forall x [/MATH] seems to me, like there is no constrain on x, so it wouldn't need to be a subset of A. But then again this expression [MATH] \forall y(y \in x \rightarrow y \in A [/MATH] would make no sense for me.

The answer from the book is attached as a picture. Sorry for posting again, but I could not edit my post and I want to be sure, that there is no misunderstanding.

 

[pka]

I am doing an exercise in the book how to prove it by velleman and I have problems understanding the logical expression in the picture I attached.
The exercise was: Analyze the logical form of the following statement. My answer was correct, but I don't really understand it.

The following expressions should all be equal, right?
[MATH] 0)~\mathscr{P}(A) \subseteq \mathscr{P}(B) \\ i)~\forall x(x \in \mathscr{P}(A) \rightarrow x \in \mathscr{P}(B))\\ ii)~\forall x(x \subseteq A \rightarrow x \subseteq B) \\ iii)~\forall x[ (\forall y(y \in x \rightarrow y \in A)) \rightarrow (\forall y(y \in x \rightarrow y \in B))][/MATH]
But the following expression:
[MATH]\forall x[ (\forall y(y \in x \rightarrow y \in A)) \rightarrow (\forall y(y \in x \rightarrow y \in B))][/MATH]

You are correct, iii) does not follow from the three above it. For one the scope of the \(x\) is not clear as to what it is.
I cannot find my copy of Velleman. Maybe if you would post the exact wording of the question someone may know the question or even have that book.

 

[lex]

[MATH] \forall x[ (\forall y(y \in x \rightarrow y \in A)) \rightarrow (\forall y(y \in x \rightarrow y \in B))] [/MATH]
Doesn't that mean that for every x and every y, when y is in x, then y is also in A.
That doesn't make sense in my opinion.

for all x (IF it is the case that for all y, y being a member of x implies that y is a member of A, THEN it is also the case that for all y, y being a member of x implies y is a member of B)

 

[Dr.Peterson]

But the following expression:

[MATH] \forall x[ (\forall y(y \in x \rightarrow y \in A)) \rightarrow (\forall y(y \in x \rightarrow y \in B))] [/MATH]
Doesn't that mean that for every x and every y, when y is in x, then y is also in A.

You are putting [MATH]\forall y[/MATH] in the wrong place, which may be your difficulty.

Observe that we get from

[MATH]\forall x(x \subseteq A \rightarrow x \subseteq B)[/MATH]​

to

[MATH]\forall x[ (\forall y(y \in x \rightarrow y \in A)) \rightarrow (\forall y(y \in x \rightarrow y \in B))][/MATH]​

by replacing [MATH]x \subseteq A[/MATH] with [MATH](\forall y(y \in x \rightarrow y \in A))[/MATH]. Is it not clear that the latter is the definition of the former?

Let's take:

[MATH] x = \{ 1, 2, 3 \} \\ y = 3 [/MATH]
There could be the case, that x is not a subset of A, but y is in x. So the last two statements of the 4 I stated in the beginning should not be equal, right?

If x is not a subset of A, then there will be y in x that are not in A, so the condition will not apply, and the statement is irrelevant.

 

[IDontEvenHaveABrain]

Thank you all for your replies. I see the problem in my reasoning now. I am really happy, that it all makes sense now. Now I can finally go to bed.