1. ## identity and inverse

Here is another part of the question:

For part (c), this is what I've done so far:

suppose both e and é are identities of $\displaystyle \mathcal{P}(M)$ and suppose $\displaystyle x \in \mathcal{P}(M)$. Therefore

xe=x .....(1)
xé=x .....(2)

Since $\displaystyle e,é \in \mathcal{P}(M)$, we can set x=é in equation (1) and set x=e in equation (2)

ée=é
eé=e

Thus e and é are equal and the identity is uniqe.

Is this correct? I think the problem here is that I've shown that the identity of $\displaystyle (\mathcal{P}(M), \star)$ is unique but I have not shown that it must have an identity. Can anyone help?

For part (d), I did this:

Let $\displaystyle a,b,c \in \mathcal{P}(M_n)$.
Suppose b and c are both inverses of a. Then $\displaystyle ab=e$ also $\displaystyle ac=e$. Hence $\displaystyle ab=bc \iff b=c$.

Again, I think my approach just shows the uniqueness, and not the existence of an inverse... I appreciate any help with this.

2. You need to exhibit the identity. What does the identity do, in this case? Well, given any set $\displaystyle x \in \mathcal{P}(M),$ you have that $\displaystyle x\star e=e\star x=x,$ right? Now, $\displaystyle x$ and $\displaystyle e$ must be the same kind of object; that is, members of $\displaystyle \mathcal{P}(M).$ Which member of $\displaystyle \mathcal{P}(M)$ will do what you need?

3. Originally Posted by Ackbeet
You need to exhibit the identity. What does the identity do, in this case? Well, given any set $\displaystyle x \in \mathcal{P}(M),$ you have that $\displaystyle x\star e=e\star x=x,$ right? Now, $\displaystyle x$ and $\displaystyle e$ must be the same kind of object; that is, members of $\displaystyle \mathcal{P}(M).$ Which member of $\displaystyle \mathcal{P}(M)$ will do what you need?
I'm still confused... I think $\displaystyle \mathcal{P}(M) = \{ \emptyset , \{1 \} \}$, so

$\displaystyle 1 \star \emptyset = (\emptyset \cup 1) \backslash (\emptyset \cap 1)$. Then what?

4. Originally Posted by demode
I'm still confused... I think $\displaystyle \mathcal{P}(M) = \{ \emptyset , \{1 \} \}$, so

$\displaystyle 1 \star \emptyset = (\emptyset \cup 1) \backslash (\emptyset \cap 1)$. Then what?
$\displaystyle A \star \emptyset = \left( {A\backslash \emptyset } \right) \cup \left( {\emptyset \backslash A} \right) = A \cup \emptyset = A$

5. So, Plato has told you that the identity is the empty set. You now need to prove this this identity is unique, and you need to find the inverse of an element.

To prove the identity is unique you do not need to use this example - any identity of a set is unique. This is because if you have two identities $\displaystyle e_1$ and $\displaystyle e_2$ then clearly $\displaystyle e_1e_2 = e_1$ but also $\displaystyle e_1e_2=e_2$. Thus they are equal.

To find the inverse, you want to find a set $\displaystyle B$ such that $\displaystyle A \cup B = A \cap B$. Note that $\displaystyle |A \cup B| \geq \max\{|A|, |B|\}$ and $\displaystyle |A \cap B| \leq \min\{|A|, |B|\}$. What does this tell you? (Remember you want $\displaystyle A \cap B = A \cup B$).

6. ## Thanks...

Originally Posted by Swlabr
To find the inverse, you want to find a set $\displaystyle B$ such that $\displaystyle A \cup B = A \cap B$. Note that $\displaystyle |A \cup B| \geq \max\{|A|, |B|\}$ and $\displaystyle |A \cap B| \leq \min\{|A|, |B|\}$. What does this tell you? (Remember you want $\displaystyle A \cap B = A \cup B$).
I think $\displaystyle A \cup B = A \cap B$ iff $\displaystyle A=B$, which means every element is its own inverse.

To prove that the inverse for each element is unique I let $\displaystyle A, B, C \in \mathcal{P}(M_n)$, where C and B are inverses of A. Then

$\displaystyle AB=e$ and $\displaystyle AC=e$. Therefore $\displaystyle AB=AC \iff B=C$.

Is it correct?

7. Originally Posted by demode
I think $\displaystyle A \cup B = A \cap B$ iff $\displaystyle A=B$, which means every element is its own inverse.

To prove that the inverse for each element is unique I let $\displaystyle A, B, C \in \mathcal{P}(M_n)$, where C and B are inverses of A. Then

$\displaystyle AB=e$ and $\displaystyle AC=e$. Therefore $\displaystyle AB=AC \iff B=C$.

Is it correct?
Yes, that is correct.