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Source: *How to Prove It, 2nd Ed* by Velleman. $\mathcal{P}(…) =$ power set of … & $A, B$ are any sets:

Ex 2.3.10: $\qquad \qquad \qquad \qquad \qquad \qquad \mathcal{P}(A \cap B) = \mathcal{P}(A) \cap \mathcal{P}(B)$

Ex 2.3.11 = Chartrand Ex 7.55: $ \qquad \qquad \color{ #0070FF}{\mathcal{P}(A \cup B)} \neq \color{green}{\mathcal{P}(A) \cup \mathcal{P}(B)}$

What are the intuitions? Where did your intuitions originate from? No proofs please.

- Set theory practice problems?
- Inverse of a Function exists iff Function is bijective
- Element of a set?
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- A intersection (A union B)
- $(A\cap B)\cup C = A \cap (B\cup C)$ if and only if $C \subset A$

I compassed to grok this by writing everything out explicitly.

For instance, for Ex 2.3.11, because $\color{ #0070FF}{\mathcal{P}(A \cup B) = \{\color{green}{subsets}, A \cup B\}}$,

but $\color{green}{\mathcal{P}(A) \cup \mathcal{P}(B) = \{subsets\}}$,

thus (UNintuitively) $\color{ #0070FF}{\mathcal{P}(A \cup B)} = \color{green}{\mathcal{P}(A) \cup \mathcal{P}(B)} \cup (A \cup B)$.

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Intuitively $A\cap B$ is a subset of both $A$ and $B$. Hence the power set of $A\cap B$ must be in the power set of both $A$ and $B$, as the power set contains all subsets of a given set. If a set $X$ is in the intersection of the power set of $A$ and the power set of $B$ then all of the elements in $X$ are in $A$ and in $B$, i.e. in $A \cap B$. Without giving a mathematical proof (as requested) this is a good description as to why $\mathcal{P}(A\cap B) = \mathcal{P}(B) \cap \mathcal{P}(B)$.

To show that $\mathcal{P}(A\cup B) \neq \mathcal{P}(A) \cup \mathcal{P}(B)$ you can consider the case $A \not\subseteq B$ then $A \cup B$ will have more elements than $A$ and more elements than $B$. Hence the power set of $A \cup B$ will contain a set that has more elements than $A$ or $B$, whereas the power set of $A$ and $B$ individually cannot contain sets larger than themselves.

In your update you write

(UNintuitively) $\color{#0070FF}{\mathcal{P}(A \cup B)} = \color{green}{\mathcal{P}(A) \cup \mathcal{P}(B)} \cup (A \cup B)$.

which is not true. Consider the following counter example

$A = \{1\},\quad B = \{2,3\} \\

\mathcal{P}(A) = \{\emptyset, \{1\}\} \\

\mathcal{P}(B) = \{\emptyset, \{2\}, \{3\},\{2,3\}\} \\

\mathcal{P}(A) \cup \mathcal{P}(B) = \{\emptyset, \{1\},\{2\},\{3\},\{2,3\}\} \\

\mathcal{P}(A \cup B) = \{\emptyset, \{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$

which shows

$\mathcal{P}(A \cup B) \neq \mathcal{P}(A) \cup \mathcal{P}(B) \cup (A \cup B)$

The power set of $A \cup B$ is the power set of $A$ plus the power set of $B$ plus all the subsets of $A \cup B$ which are neither in $A$ nor $B$.

$\newcommand{\P}[1]{\mathcal P ( #1 )}$Like Arthur Fisher commented, my intuition comes from proving these.

And when trying to prove the first one I quickly see, just by expanding the definitions of $\;\P \cdot\;$ and $\;\cap\;$, that this comes down to

$$

X \subseteq A \cap B \;\equiv\; X \subseteq A \:\land\: X \subseteq B

$$

for any $\;X\;$. And that, using the definitions of $\;\subseteq\;$ and $\;\cap\;$, comes down to the fact that $\;\forall\;$ distributes over $\;\land\;$.

Now the second one similarly comes down to the fact that $\;\forall\;$ does *not* in general distribute over $\;\lor\;$ (which is the essence of the definition of $\;\cup\;$).

Alternatively, you can of course get an intuition about these statements by drawing Venn diagrams: draw some elements of $\;\P{A \cup B}\;$, i.e., some subsets of $\;A \cup B\;$, and see how they relate to elements of $\;\P A\;$ and $\;\P B\;$, i.e., subsets of $\;A\;$ and $\;B\;$. That will quickly give you an example for the second statement (the inequality).

The statement $$P(A \cap B)=P(A) \cap P(B)$$

just says that the sets that are included by $A \cap B$ are precisely the sets that are included by both $A$ and $B$. Draw a diagram, and you’ll see its quite obvious.

On the other hand, the statement $$P(A \cup B)=P(A) \cup P(B)$$

implies

$$P(A \cup B) \subseteq P(A) \cup P(B)$$

which is just saying that if a set is included by $A \cup B$, then its included by at least one of $A$ or $B$. However, $A \cup B$ is itself a counterexample, unless $A$ and $B$ are comparable. Again, draw a Venn diagram, and you’ll see its fairly obvious, as long as you don’t draw one ball completely encircled by the other.

In general, its useful to try to understand what something is actually saying. If you can work out what its saying, often it becomes obvious.

For proving equality of sets (e.g. $S = T$), it’s often helpful to prove $S \subset T$ and $T \subset S$. That is, start by choosing an arbitrary element $s \in S$, and show it must be in $T$, then pick $t \in T$, and show it is in $S$.

In the first case, you want to show that an arbitrarily chosen subset of $A\cap B$ is a subset of $A$ and a subset of $B$, and then vice versa (an arbitrarily chosen set that is a subset of both $A$ and $B$ must be a subset of $A \cap B$.)

For the second problem, it may be useful to decide, before attempting to prove, which (if not both) of the inclusions fail, i.e. whether

$$\mathcal{P}(A \cup B) \not\subset \mathcal{P}(A)\cup\mathcal{P}(B),$$

or

$$\mathcal{P}(A \cup B) \not\supset \mathcal{P}(A)\cup\mathcal{P}(B)$$

or possibly both.

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