Proving P(A) Union P(B) is a Subset of P(A Union B)

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Homework Statement



Prove [tex]P(A) \cup P(B) \subseteq P(A \cup B)[/tex]


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The Attempt at a Solution



I started out by assuming that [tex]A = \left\{a\right\}[/tex] and [tex]B=\left\{b\right\}[/tex].

So then [tex]P(A) \cup P(B) = \left\{\left\{a\right\},\left\{b\right\},null\right\}[/tex] and [tex]P(A \cup B) = \left\{\left\{a\right\},\left\{b\right\},\left\{a,b\right\},null\right\}[/tex]

So I can conclude that [tex]P(A) \cup P(B) \subseteq P(A \cup B)[/tex]

How does that sound?
 
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When you assume that A={a} and B={b}, you are assuming that A and B are both singleton sets. You want to prove the relation for any sets A and B.

When proving one set is a subset of another, say X is a subset of Y, then you let x be in X and show x is in Y. So let [itex]x\in P(A)\cup P(B)[/itex], and then show [itex]x\in P(A\cup B)[/itex].
 
n!kofeyn said:
When you assume that A={a} and B={b}, you are assuming that A and B are both singleton sets. You want to prove the relation for any sets A and B.

When proving one set is a subset of another, say X is a subset of Y, then you let x be in X and show x is in Y. So let [itex]x\in P(A)\cup P(B)[/itex], and then show [itex]x\in P(A\cup B)[/itex].

Okay, so let [tex]x\in P(A)\cup P(B)[/tex].
Then [tex]x\in P(A)[/tex] or [tex]x\in P(B)[/tex]... which means [tex]x\subseteq A[/tex] or [tex]x\subseteq B[/tex]?

So then [tex]x\subseteq (A\cup B)[/tex]. Am I going in the right direction?