Relationship between a power set of subsets and a power set of proper subsets Let ℙ() be the collection of all subsets of , and ℚ() the collection of all proper subsets of .
Which of the following hold for every set ?
ℙ() ⊆ ℚ()
ℙ() ⊇ ℚ()
ℙ() ⊃ ℚ()
ℙ() = ℚ()

What's exactly wrong with this thought process?
If  = {1,2,3}, then
ℙ() = {∅,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
ℚ() = {∅,{1},{2},{3},{1,2},{1,3},{2,3}}
ℙ() ⊆ ℚ() false. ℙ() is not a subset of ℚ()
ℙ() ⊇ ℚ() false. ℙ() is not a superset of ℚ()
ℙ() ⊃ ℚ() true. ℙ() is a proper superset of ℚ(), and ℚ() is not equal to ℙ()
ℙ() = ℚ() false. ℙ() and ℚ() sets are not subsets of each other.
 A: The notation $A\subseteq B$ is usually defined as $A\subset B$ or $A=B$. If one of the two things $A\subset B$ or $A=B$ is true, then $A\subseteq B$ must be true as well. Therefore it is impossible that $\Bbb P(S)\supseteq \Bbb Q(S)$ is false when $\Bbb P(S)\supset \Bbb Q(S)$ is true.

Apart from that, your reasoning has another error: you don't need an example set (in your case you use $S=\{1,2,3\}$) to show whether the assertions are true or false. In fact, you shouldn't use an example in a proof, since you then just prove a single special case, instead of a general principle. Of course examples are useful to gain intuition, and I highly encourage to first work out a few examples before you start proving something. But for the proof itself, you have to keep things general.
To show the assertions in a general case, we note that for any set $S$, the set $S$ itself is a subset of $S$ that is not proper. This follows from the definition of what a proper subset is. Furthermore, we know that any proper subset is also a subset (once again by definition of proper subset).


*

*$\Bbb P(S)\not\subseteq \Bbb Q(S)$, since the set $S$ is an element of $\Bbb P(S)$ that is not a subset of $\Bbb Q(S)$.

*$\Bbb P(S)\supseteq\Bbb Q(S)$, since any proper subset is also a subset.

*$\Bbb P(S)\supset\Bbb Q(S)$, since the set $S$ is an element of $\Bbb P(S)$ that is not an element of $\Bbb Q(S)$, therefore $\Bbb P(S)\neq \Bbb Q(S)$.  By the previous point $\Bbb P(S)\supseteq \Bbb Q(S)$, and remember that $\supseteq$ is defined as "$\Bbb P(S)\supset \Bbb Q(S)$ or $\Bbb P(S)= \Bbb Q(S)$". Since the latter is false, the former must be true.

*$\Bbb P(S)\neq \Bbb Q(S)$, as we saw because $S$ is an element in $\Bbb P(S)$ and not in $\Bbb Q(S)$.

A: According to my point of view, if we take a same example
If  = {1,2,3}, then
ℙ() = {∅,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
ℚ() = {∅,{1},{2},{3},{1,2},{1,3},{2,3}}
False -> ℙ() ⊆ ℚ() ,  ℙ() is not a subset of ℚ()
True ->  ℙ() ⊇ ℚ() , Because Q(S) is a subset of P(S)e.g Q(S) ⊆ P(S), So that`s why  P(S) is a superset of Q(S) e.g P(S) ⊇ Q(S)
True -> ℙ() ⊃ ℚ()  , ℙ() is a proper superset of ℚ(), and ℚ() is not equal to ℙ()
False -> ℙ() = ℚ() , ℙ() and ℚ() sets are not subsets of each other.
