# Power set of B

• Nov 4th 2010, 11:21 AM
novice
Power set of B
Is there such an expression as $\forall B \mathcal,{P}(B)$?

It's very hard for me imagine there would be more than one $B$ in $\mathcal{P}(B)$.
• Nov 4th 2010, 11:26 AM
Plato
Quote:

Originally Posted by novice
Is there such an expression as $\forall B \mathcal,{P}(B)$?
It's very hard for me imagine there would be more than one $B$ in $\mathcal{P}(B)$.

Can you possibly fill out that question with more details?

Is it just $\mathcal{P}(B)=\{Y:Y\subseteq B\}~?$
If $B$ has n elements the $\mathcal{P}(B)$ has $2^n$ elements.
• Nov 4th 2010, 12:26 PM
novice
Quote:

Originally Posted by Plato
Can you possibly fill out that question with more details?

Is it just $\mathcal{P}(B)=\{Y:Y\subseteq B\}~?$
If $B$ has n elements the $\mathcal{P}(B)$ has $2^n$ elements.

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

Could it be a typo?

Does $\{Y:Y\subseteq B\}$ mean $\forall Y \in \mathcal{P}(B)(Y \subseteq B)$
• Nov 4th 2010, 12:28 PM
MoeBlee
Quote:

Originally Posted by novice
Is there such an expression as $\forall B \mathcal,{P}(B)$?

I don't recognize that as a meaningful formula.

What is it that you're trying to say there?
• Nov 4th 2010, 12:34 PM
MoeBlee
Quote:

Originally Posted by novice
$\forall B \in \mathcal{P}(B)(A \subseteq B)$

That's a formula, though it's an odd one. It seems not what you intend.

What you probably mean is

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

Quote:

Originally Posted by novice
Does $\{Y:Y\subseteq B\}$ mean $\forall Y \in \mathcal{P}(B)(Y \subseteq B)$

No.

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

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

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

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, 12:40 PM
novice
Quote:

Originally Posted by MoeBlee
I don't recognize that as a meaningful formula.

What is it that you're trying to say there?

I am supposed to provide an example of a set $A$ that satisfies $\forall B P(B)(A\subseteq B)$.

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

If B={1,2}, P(B)={ $\varnothing$ {1},{2},B}. I wondering why $\forall B$ is used. It doesn't make sense.
• Nov 4th 2010, 12:54 PM
novice
Quote:

Originally Posted by MoeBlee
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.

It's coming my way.
• Nov 4th 2010, 12:54 PM
MoeBlee
Quote:

Originally Posted by novice
$A$ that satisfies $\forall B P(B)(A\subseteq B)$.

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

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

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, 01:03 PM
Plato
It surely is a typo of some sort.
I know of several books that use this notation, $P(B:A)$.
It stands for $A\in P(B)$ and $P(B:A)=\{X\in P(B):A\subseteq X\}$.
All subsets of B that have A as a subset.

Is that close to what you think it may mean?
• Nov 4th 2010, 01:22 PM
novice
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, 02:00 PM
MoeBlee
Quote:

Originally Posted by novice
It's coming my way.

You'll enjoy it. You're doing the right thing.