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I’m learning proofs on my own and I’m on the section about cases. I have to prove the following but don’t how to do this. Can someone tell me how you would prove this? (where $P()$ represents the power set): $P(A) \cup P(B) = P(A \cup B) \Rightarrow (A \subseteq B) \lor (B \subseteq A)$

Your help is greatly appreciated! :)

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Try to prove this with contraposition (meaning $(A \Rightarrow B) \equiv (\neg B \Rightarrow \neg A)$):

Assume neither $A \subseteq B$ nor $B \subseteq A$ holds, hence you get elementes $a \in A \setminus B$ and $b \in B \setminus A$. Now consider the subset $\{a,b\} \in P(A \cup B)$. Is this in $P(A) \cup P(B)$ (hence in $P(A)$ or $P(B)$)?

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    $\begingroup$ This is very cool. I will have to look at it more. The chapter however is all about using cases in the proof for any OR statements and it seems there are just too many different cases such as x ∈ P(A), x ∈ P(B), P(x ∈ A U B). Just getting confused on how to actually do this. This is actually an exercise question in the book "how to prove it" for Chapter 3.5 but there is no answer and I've been thinking about it for hours. My head is hurting ;). I do appreciate your help :) $\endgroup$ – maybedave Sep 16 '16 at 16:41
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A similar way without the need for the contrapositive: get an arbitrary $x$ such that $x\in A$ and an arbitrary $y$ such that $y\in B$, then the set $\{x,y\}\in\mathcal P(A\cup B)$, by definition of the power set.

But now it must hold that $\{x,y\}\in\mathcal P(A)\cup\mathcal P(B)$, this mean that $\{x,y\}\in\mathcal P(A)$ or $\{x,y\}\in\mathcal P(B)$.

If $\{x,y\}\in\mathcal P(A)$ then $y\in B$, and because $y$ was arbitrary we have that $B\subseteq A$. In an analogous way if $\{x,y\}\in\mathcal P(B)$ we have that $A \subseteq B$.

EDIT: well, we need to formalize this proof a bit more to be a valid proof. Suppose an arbitrary $X\subseteq\mathcal P(A\cup B)$. If $A=B$ we are done. If $A\neq B$ then we divide four cases for $X$:

  1. $X$ is empty
  2. $X$ is composed of points exclusively of $A$
  3. $X$ is composed of points exclusively of $B$
  4. $X$ is composed of points of $A$ and points of $B$

Now we can check that when the four cases hold then the statement of the question hold.

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