How to prove set containments on power sets?

Question

$P(A-B) \not =P(A) - P(B)$

How can we apply here the definition of a power set?

$P(A)=\{S|S \text{ is a subset of }A\}$

$P(B)=\{S|S \text{ is a subset of }B\}$

$P(A-B)=\{S|S \text{ is a subset of }A \text{ and not of } B \}$

Answer 1

This is obviously True if $A \cap B \not= \emptyset$. In case that $A \cap B = \emptyset $, $P(A-B)=P(A)-P(B).$

Also, $ \emptyset \in P(A-B)$ unless $A-B = \emptyset$, but $\emptyset \not\in P(A)−P(B)$ and that is because - $\emptyset \in P(A),P(B)$ (unless $P(A)=P(B) = \emptyset)$.

A more accurate definition of $P(A-B)$ might be $\{S\,|\,\forall s\in S:s\in A \,\land s\not\in B\}$.

Answer 2

Let $A=\{1,2\}$ and $B=\{2\}$. Then $A-B=\{1\}$ and $$P(A)=\{\emptyset, \{1\},\{2\}, \{1,2\}\}, \qquad P(B)=\{\emptyset,\{2\}\}$$ Now compare the sets.

Answer 3

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

$B = \{1,2\}$

$A-B = \{3\}$

$\{1,3\}$ is the example you need.

Answer 4

yYur definition of $P(A \setminus B)$ is wrong. Let's take $\mathbb{N}$ and $\mathbb{N}^+$ as an example.

$\mathbb{N} \setminus \mathbb{N}^+ = \{0\}$

As such, it's power set will be $\{\emptyset, \{0\}\}$. Following your definition of $P(A \setminus B)$, it will contain any set which has a zero in it.

Answer 5

The set constructors you require are:

$\begin{align} \mathcal P(A)\smallsetminus\mathcal P(B) & =\{S\mid S\in\mathcal P(A)\wedge S\notin \mathcal P(B)\} & \textsf{definition of set difference} \\ &=\{S\mid S\subseteq A\wedge S\nsubseteq B\}\tag 1 & \textsf{definition of power set} \\ &=\{S\mid (\forall x\in S.x\in A)\wedge(\exists x\in S.x\notin B)\} & \textsf{definition of subset(or equal)} \\[2ex] \mathcal P(A\smallsetminus B) & =\{S\mid S \subseteq (A \smallsetminus B) \}\tag 2 & \textsf{definition of power set} \\ & =\{S\mid \forall x\in S. x\in (A\smallsetminus B)\} & \textsf{definition of subset(or equal)} \\ &= \{S\mid \forall x\in S. (x\in A\wedge x\notin B)\} & \textsf{definition of set difference} \\ &= \{S\mid (\forall x\in S.x\in A)\wedge(\forall x\in S.x\notin B)\} & \textsf{quantifier distribution} \end{align}$

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Of note, $\emptyset \subseteq A$, $\emptyset \subseteq B$, and $\emptyset \subseteq (A\smallsetminus B)$, therefore ...