MHB Why Prove \( x = y \) in \( \mathcal{P} \cup A \)?

[evinda]

Hi! (Smile)

I want to prove that for each set $A$:
$$A \subset \mathcal P \cup A$$

According to my notes, we prove it like that:

Let $x \in A$. We want to show that $x \in \mathcal P \cup A$, so, that: $\exists y \in \mathcal P \cup A$, such that $x=y$.
It suffices to show that if $z \in x$, then $z \in \cup A$, that stands, since $x \in A$.

Could you explain me why we want to show that :
$$\exists y \in \mathcal P \cup A, \text{ so that } x=y$$
?
At the beginning, we suppose that $x \in A$ and we want to prove that $x \in \mathcal P \cup A$, so we want to prove that $x \subset \cup A$.
I haven't understood what we could do to prove this.. (Worried)

 

[Evgeny.Makarov]

I want to prove that for each set $A$:
$$A \subset \mathcal P \cup A$$

Do you mean $\mathcal{P}\left(\bigcup A\right)$? In LaTeX, \cup is a binary operator, while \bigcup is a prefix unary operator.

Could you explain me why we want to show that :
$$\exists y \in \mathcal P \cup A, \text{ so that } x=y$$?

This statement is equivalent to $x\in\mathcal P( \bigcup A)$, and I don't see how it simplifies it.

At the beginning, we suppose that $x \in A$ and we want to prove that $x \in \mathcal P \cup A$, so we want to prove that $x \subset \cup A$.

Yes. The latter statement means $\forall z\;z\in x\to z\in\bigcup A$. But this follows from the definition of $\bigcup A$.

 

[evinda]

Do you mean $\mathcal{P}\left(\bigcup A\right)$? In LaTeX, \cup is a binary operator, while \bigcup is a prefix unary operator.

Yes, that's what I meant. I will take it into consideration! (Smile)

Yes. The latter statement means $\forall z\;z\in x\to z\in\bigcup A$. But this follows from the definition of $\bigcup A$.

A ok! I got it! (Nod)

And does the equality $A= \mathcal{P}\left(\bigcup A\right) $ also stand? (Thinking)

 

[Evgeny.Makarov]

And does the equality $A= \mathcal{P}\left(\bigcup A\right) $ also stand?

No. For example, when $A=\{\{1,2\},\{3,4\}\}$, then $\bigcup A=\{1,2,3,4\}$ and $\mathcal{P}(\bigcup A)$ contains all 16 subsets of $\{1,2,3,4\}$. Of those 16, $A$ contains only two.

 

[evinda]

No. For example, when $A=\{\{1,2\},\{3,4\}\}$, then $\bigcup A=\{1,2,3,4\}$ and $\mathcal{P}(\bigcup A)$ contains all 16 subsets of $\{1,2,3,4\}$. Of those 16, $A$ contains only two.

I understand! (Nod) Does the equality never hold or are there cases, where it stands that $A=\mathcal{P}(\bigcup A)$ ? (Thinking)

 

[Evgeny.Makarov]

Does the equality never hold or are there cases, where it stands that $A=\mathcal{P}(\bigcup A)$ ?

This happens iff $A=\mathcal{P}(B)$ for some $B$.

 

[evinda]

This happens iff $A=\mathcal{P}(B)$ for some $B$.

How did you conclude that it happens iff $A=\mathcal{P}(B)$ for some $B$?

And how could we prove this? (Thinking)