Thread: Set manipulation

Prove that:

$\displaystyle \bigcup Px=x \subset P \bigcup x= P \bigcup P \bigcup x$

where x is a set and $\displaystyle Px$ is the power set of x.

This is the last part of the question, the former parts required me to prove:

$\displaystyle x \subset y \Rightarrow \bigcup x \subset \bigcup y$

$\displaystyle x \subset y \Leftrightarrow Px \subset Py$

$\displaystyle x \subset Py \Leftrightarrow \bigcup x \subset y $

where x and y are sets.

I can't get anywhere with this! I've tried using $\displaystyle x \subset Px$ but I can't get it to work out. My main problem is that the first three parts of the questions gave me something to work with (ie. condition 1 allows me to use $\displaystyle x \subset y$). The last part doesn't give me any properties to use so i'm having some trouble trying to use the things I already know.

Does anyone have any ideas?

I have a hard time with your notation.
Have you shown that $\displaystyle \bigcup {\mathcal{P}\mathcal{A}=\mathcal{A}}~? $

Have you shown that $\displaystyle \mathcal{A}\subseteq\mathcal{P}\bigcup {\mathcal{A}}~? $.

Ohh, $\displaystyle \bigcup {\mathcal{P}\mathcal{A}=\{ t~|~(\exists X \in \mathcal{P} \mathcal{A})[t \in X] \}$.

and we also know that $\displaystyle \mathcal{P} \mathcal{A}=\{X~|~Set(X) ~\&~ X \subset \mathcal{A} \}$

so my $\displaystyle \mathcal{P} \mathcal{A}$ contains only elements from $\displaystyle \mathcal{A}$ since $\displaystyle X \subset \mathcal{A}$. We know we have all the elements since $\displaystyle \mathcal{A} \in \mathcal{P} \mathcal{A}$ and all the other subsets contain only some of these elements.

Hence the unionset axiom gives all the elements in $\displaystyle \mathcal{A}$ giving that $\displaystyle \bigcup \mathcal{P} \mathcal {A}=\mathcal{A}$.

Is that right? The only way I could see of doing it was with analysing the definitions.

My main problem is the subset part. I can't seem to introduce anything that helps.

So yes, this is the part i'm having trouble showing: $\displaystyle \mathcal{A}\subseteq\mathcal{P}\bigcup {\mathcal{A}}$.

Originally Posted by Showcase_22

So yes, this is the part i'm having trouble showing: $\displaystyle \mathcal{A}\subseteq\mathcal{P}\bigcup {\mathcal{A}}$.

If $\displaystyle t\in \mathcal{A}$ then $\displaystyle \{t\}\in \mathcal{P}\mathcal{A}$ which implies that $\displaystyle t\in \bigcup {\mathcal{P}\mathcal{A}}$.

Does that work?

Aha! I think by $\displaystyle \subset$ you mean "implies". Is that correct?

.

Originally Posted by Plato

If $\displaystyle t\in \mathcal{A}$ then $\displaystyle \{t\}\in \mathcal{P}\mathcal{A}$ which implies that $\displaystyle t\in \bigcup {\mathcal{P}\mathcal{A}}$.

Does that work?

But doesn't that mean $\displaystyle t\in \bigcup {\mathcal{P}\mathcal{A}}=\mathcal{A}$ so we're back to $\displaystyle t \in \mathcal{A} $?

I'll ask my lecturer today about the subset being an implies. What i've written is exactly what's written on the example sheet.