Is there such an expression as $\displaystyle \forall B \mathcal,{P}(B)$?

It's very hard for me imagine there would be more than one $\displaystyle B$ in $\displaystyle \mathcal{P}(B)$.

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- Nov 4th 2010, 10:21 AMnovicePower set of B
Is there such an expression as $\displaystyle \forall B \mathcal,{P}(B)$?

It's very hard for me imagine there would be more than one $\displaystyle B$ in $\displaystyle \mathcal{P}(B)$. - Nov 4th 2010, 10:26 AMPlato
- Nov 4th 2010, 11:26 AMnovice
- Nov 4th 2010, 11:28 AMMoeBlee
- Nov 4th 2010, 11:34 AMMoeBlee
That's a formula, though it's an odd one. It seems not what you intend.

What you probably mean is

$\displaystyle \forall A \in \mathcal{P}(B)(A \subseteq B)$

No.

$\displaystyle \{Y:Y\subseteq B\}$

is a term; it's a "name"; it names a particular set; it doesn't make an assertion.

$\displaystyle \forall Y \in \mathcal{P}(B)(Y \subseteq B)$

is not a term, but rather it's a formula; it makes an assertion (once we agree on what 'B' stands for, or, alternatively, once we decide it should be taken in the sense of a universal generalization on 'B').

You very much need to get that logic book I recommended, as soon as possible; it will get you on the right track to writing correct symbolizations. - Nov 4th 2010, 11:40 AMnovice
I am supposed to provide an example of a set $\displaystyle A$ that satisfies$\displaystyle \forall B P(B)(A\subseteq B)$.

$\displaystyle A= \varnothing$ seems to fit the expression, but I know there is only one $\displaystyle B$ in $\displaystyle P(B)$

If B={1,2}, P(B)={$\displaystyle \varnothing$ {1},{2},B}. I wondering why $\displaystyle \forall B$ is used. It doesn't make sense. - Nov 4th 2010, 11:54 AMnovice
- Nov 4th 2010, 11:54 AMMoeBlee
I don't recognize that as a formula. Yes, it doesn't make sense. I can't find an A that satisfies it, because it's not even a formula.

The earlier formula you gave was

$\displaystyle \forall B \in \mathcal{P}(B)(A \subseteq B)$

That is equivalent to:

For all B (If B is a subset of B, then A is subset of B).

An A that satisfies that is B itself, or any other A that is a subset of B, depending on what B is.

Though maybe you've not rendered the formula correctly (it is rather odd)? - Nov 4th 2010, 12:03 PMPlato
It surely is a typo of some sort.

I know of several books that use this notation, $\displaystyle P(B:A)$.

It stands for $\displaystyle A\in P(B)$ and $\displaystyle P(B:A)=\{X\in P(B):A\subseteq X\} $.

All subsets of B that have A as a subset.

Is that close to what you think it may mean? - Nov 4th 2010, 12:22 PMnovice
Thank you both. I am going scrape off everything I got so far and study logic first to rebuild it from ground up.

- Nov 4th 2010, 01:00 PMMoeBlee