Set Theory: Power sets of Unions

[WWCY]

Homework Statement

I'm having issues understanding a mistake that I'm making, any assistance is appreciated! I know a counterexample but my attempt at proving the proposition is what's troubling me.

Prove or disprove
$$P(A \cup B) \subseteq P(A) \cup P(B) $$

Homework Equations

The Attempt at a Solution

Let $C$ be an particularly, arbitrarily picked element of element of $P(A \cup B)$
Then by definition of a powerset
$$C \subseteq A \cup B$$
This means that
$$\forall x \in C, \ x \in A \cup B$$
$$\forall x \in C, \ x \in A \ \text{or} \ x \in B \ \ \ \ ^{**} $$
$$\forall x \in C, \ x \in A \ \text{or} \ \ \ \forall x \in C , x \in B \ \ \ \ ^{**}$$

Then $C \subseteq A$ or $C \subseteq B$ and thus $C \in P(A) \ \text{or} \ C \in P(B)$ and $C \subseteq P(A) \cup P(B)$. The bits I left the asterisks were where I felt I made some error.

But, in a proof of a similar problem: $P(A \cap B) \subseteq P(A) \cap P(B)$, I wrote
$$\forall x \in C, \ x \in A \cap B$$
$$\forall x \in C, \ x \in A \ \text{and} \ \forall x \in C, x \in B $$
Then $C \subseteq A$ and $C \subseteq B$, and thus $C \in P(A) \ \text{and} \ C \in P(B)$, and $C \subseteq P(A) \cap P(B)$. This gave me the right answer.

Why did the first one give me the wrong answer while the second one gave a right one?

PS It would be nice if the explanation isn't too technical as it's my first course on the subject. Thanks!

 

[PeroK]

Homework Statement

I'm having issues understanding a mistake that I'm making, any assistance is appreciated! I know a counterexample but my attempt at proving the proposition is what's troubling me.

Prove or disprove
$$P(A \cup B) \subseteq P(A) \cup P(B) $$

Homework Equations

The Attempt at a Solution

Let $C$ be an particularly, arbitrarily picked element of element of $P(A \cup B)$
Then by definition of a powerset
$$C \subseteq A \cup B$$
This means that
$$\forall x \in C, \ x \in A \cup B$$
$$\forall x \in C, \ x \in A \ \text{or} \ x \in B \ \ \ \ ^{**} $$
$$\forall x \in C, \ x \in A \ \text{or} \ \ \ \forall x \in C , x \in B \ \ \ \ ^{**}$$

Then $C \subseteq A$ or $C \subseteq B$ and thus $C \in P(A) \ \text{or} \ C \in P(B)$ and $C \subseteq P(A) \cup P(B)$. The bits I left the asterisks were where I felt I made some error.

But, in a proof of a similar problem: $P(A \cap B) \subseteq P(A) \cap P(B)$, I wrote
$$\forall x \in C, \ x \in A \cap B$$
$$\forall x \in C, \ x \in A \ \text{and} \ \forall x \in C, x \in B $$
Then $C \subseteq A$ and $C \subseteq B$, and thus $C \in P(A) \ \text{and} \ C \in P(B)$, and $C \subseteq P(A) \cap P(B)$. This gave me the right answer.

Why did the first one give me the wrong answer while the second one gave a right one?

PS It would be nice if the explanation isn't too technical as it's my first course on the subject. Thanks!

You can't start with $C \subseteq A \cup B$ and conclude that $C \subseteq A$ or $C \subseteq B$.

You should be able to find a counterexample to that.

Or, draw a Venn diagram. For example, with disjoint $A$ and $B$.

 

[WWCY]

Thanks for your response!

You can't start with $C \subseteq A \cup B$ and conclude that $C \subseteq A$ or $C \subseteq B$.

You should be able to find a counterexample to that.

Or, draw a Venn diagram. For example, with disjoint $A$ and $B$.

Ah, I see it now. Was my mistake down to the misuse of quantifiers?

 

[PeroK]

Thanks for your response!

Ah, I see it now. Was my mistake down to the misuse of quantifiers?

The mistake is either one of logic or of dissociating the use of quantifiers from basic logic. The false steps should be easy to find if you follow the "proof" through with an example.