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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.

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  • $\begingroup$ P(S) is a powerset of Q(S). Proper superset is, in particular, a superset $\endgroup$ – Wojowu Sep 14 at 20:21
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    $\begingroup$ How can $\mathbb{P}(S)$ be a proper superset of $\mathbb{Q}(S)$ but not be a superset? $\endgroup$ – Mark Kamsma Sep 14 at 20:21
  • $\begingroup$ @Wojowu You mean "superset", not "powerset". $\endgroup$ – Mark Kamsma Sep 14 at 20:22
  • $\begingroup$ @MarkKamsma Are you saying that both statements are correct? ℙ(𝑆) ⊇ ℚ(𝑆) and ℙ(𝑆) ⊃ ℚ(𝑆)? $\endgroup$ – Gabriel H. Sep 14 at 20:31
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    $\begingroup$ @MarkKamsma Right, thanks $\endgroup$ – Wojowu Sep 14 at 20:36
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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 $\Bbb Q(S)$.
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