Set manipulation

**Showcase_22** · January 25th 2011, 09:56 AM

Prove that:

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

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

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

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

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

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

where x and y are sets.

I can't get anywhere with this! I've tried using $x \subset Px$ 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 $x \subset y$ ). 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?

**Plato** · January 25th 2011, 12:00 PM

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

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

**Showcase_22** · January 25th 2011, 01:52 PM

Ohh, [LaTeX ERROR: Convert failed] .

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

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

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

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: $\mathcal{A}\subseteq\mathcal{P}\bigcup {\mathcal{A}}$ .

**Plato** · January 25th 2011, 02:05 PM

Originally Posted by Showcase_22

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

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

Does that work?

**DrSteve** · January 25th 2011, 02:54 PM

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

**Plato** · January 25th 2011, 03:11 PM

.

**Showcase_22** · January 25th 2011, 11:10 PM

Originally Posted by Plato

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

Does that work?

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

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.

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