Let β(π) be the collection of all subsets of π, and β(π) the collection of all proper subsets of π.
Which of the following hold for every set π?
β(π) β β(π)
β(π) β β(π)
β(π) β β(π)
β(π) = β(π)
What's exactly wrong with this thought process?
If π = {1,2,3}, then
β(π) = {β ,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
β(π) = {β ,{1},{2},{3},{1,2},{1,3},{2,3}}
β(π) β β(π) false. β(π) is not a subset of β(π)
β(π) β β(π) false. β(π) is not a superset of β(π)
β(π) β β(π) true. β(π) is a proper superset of β(π), and β(π) is not equal to β(π)
β(π) = β(π) false. β(π) and β(π) sets are not subsets of each other.
2 Answers
$\begingroup$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 $\Bbb Q(S)$.
According to my point of view, if we take a same example If π = {1,2,3}, then
β(π) = {β ,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
β(π) = {β ,{1},{2},{3},{1,2},{1,3},{2,3}}
False -> β(π) β β(π) , β(π) is not a subset of β(π)
True -> β(π) β β(π) , Because Q(S) is a subset of P(S)e.g Q(S) β P(S), So that`s why P(S) is a superset of Q(S) e.g P(S) β Q(S)
True -> β(π) β β(π) , β(π) is a proper superset of β(π), and β(π) is not equal to β(π)
False -> β(π) = β(π) , β(π) and β(π) sets are not subsets of each other.
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