Set theory. Is the converse true?

**omoplata** · June 10th 2011, 09:12 AM

1. The problem statement, all variables and given/known data

Prove that $\cup_{x \in C} 2^{x} \subseteq 2^{\cup C}$

2. Relevant equations

$\cup_{x \in C} 2^{x} = \{ A | \exists x \in C, A \subseteq 2^{x} \}$

$2^{x}$ is the powerset of $x$ . i.e. $2^{x} = \{ y | y \subseteq x \}$

3. The attempt at a solution

Suppose $A \in \cup_{x \in C} 2^{x}$ . Then,

$\exists x \in C, A \in 2^{x}$

$\exists x \in C, A \subseteq x$

$A \subseteq ( \cup C )$

$A \in 2^{\cup C}$

Therefore, $A \in \cup_{x \in C} 2^{x} \Rightarrow A \in 2^{\cup C}$

Therefore, $\cup_{x \in C} 2^{x} \subseteq 2^{\cup C}$

But I think there might be something wrong with my proof. Because why can't I start assuming $A \in 2^{\cup C}$ and go to $A \in \cup_{x \in C} 2^{x}$ . That means $A \in 2^{\cup C} \Rightarrow A \in \cup_{x \in C} 2^{x}$ and therefore $2^{\cup C} \subseteq \cup_{x \in C} 2^{x}$ also. That means $\cup_{x \in C} 2^{x} = 2^{\cup C}$

Is there something wrong with this proof?

**MoeBlee** · June 10th 2011, 09:29 AM

To me that seems like odd use of the '{ }' notation.

Ordinarily, $\cup_{x \in C} \{ 2^{x} \}$ = {A | Ex(x in C & A in {Px})} = {A | Ex(x in C & A = Px)}.

**MoeBlee** · June 10th 2011, 09:41 AM

But I glean that the actual problem is this:

Show {A | Ex(x in C & A in Px)} subset PUC

So, let x in C and A subset x. Show A subset UC.

Suppose z in A. So z in x in C. So z in UC.

Done.

**omoplata** · June 10th 2011, 09:48 AM

Originally Posted by MoeBlee

To me that seems like odd use of the '{ }' notation.

Ordinarily, $\cup_{x \in C} \{ 2^{x} \}$ = {A | Ex(x in C & A in {Px})} = {A | Ex(x in C & A = Px)}.

Sorry, it should be $\cup_{x \in C} 2^{x}$ , not $\cup_{x \in C} {2^{x}}$ . I edited the original post and corrected it.

**omoplata** · June 10th 2011, 09:54 AM

Originally Posted by MoeBlee

But I glean that the actual problem is this:

Show {A | Ex(x in C & A in Px)} subset PUC

So, let x in C and A subset x. Show A subset UC.

Suppose z in A. So z in x in C. So z in UC.

Done.

Thanks. Now I understand why I can't go backwards. z in UC does not imply z in x.

**MoeBlee** · June 10th 2011, 10:22 AM

So I take it you also wanted to check whether

PUC subset {A | Ex(x in C & A in Px)} is a theorem.

But counterexample:

Let C = {{0} {1}}. Let A = {0 1}.

There, A in PUC but there is no x in C such that A in Px.

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