Let ℙ(𝑆) be the collection of all subsets of 𝑆, and ℚ(𝑆) the collection of all proper subsets of 𝑆.
Which of the following hold for every set 𝑆?
ℙ(𝑆) ⊆ ℚ(𝑆)
ℙ(𝑆) ⊇ ℚ(𝑆)
ℙ(𝑆) ⊃ ℚ(𝑆)
ℙ(𝑆) = ℚ(𝑆)
What's exactly wrong with this thought process?
If 𝑆 = {1,2,3}, then
ℙ(𝑆) = {∅,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
ℚ(𝑆) = {∅,{1},{2},{3},{1,2},{1,3},{2,3}}
ℙ(𝑆) ⊆ ℚ(𝑆)
false. ℙ(𝑆) is not a subset of ℚ(𝑆)
ℙ(𝑆) ⊇ ℚ(𝑆)
false. ℙ(𝑆) is not a superset of ℚ(𝑆)
ℙ(𝑆) ⊃ ℚ(𝑆)
true. ℙ(𝑆) is a proper superset of ℚ(𝑆), and ℚ(𝑆) is not equal to ℙ(𝑆)
ℙ(𝑆) = ℚ(𝑆)
false. ℙ(𝑆) and ℚ(𝑆) sets are not subsets of each other.
Best Answer
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).