@Khalli First think of a different situation : Say you're given the n-object multiset $\{\{0, 1\}, \{0, 1\}, \cdots, \{0, 1\}\}$. Consider that your object is to pick up one element from each of the $\{0, 1\}$ set and form another set of $n$ elements.
Since there are $2$ elements in each set and you have $n$ such 2-sets, you can form $2 \times 2 \times 2 \times \cdots \times 2 = 2^n$ such $n$-elt sets.
Just pick one element from the first set, say $0$. Another from the second, say $1$ and another from the third, say $1$. Form a set $\{0, 1, 1\}$
That is what I mean by pick and you are asked to count such sets you can form.
I can see loads of such multisets, certainly more than 6. $\{0, 1, 1\}, \{1, 1, 1\}, \{0, 0, 0\}, \{1, 0, 1\}, \{1, 1, 0\}, \{0, 1, 0\}, \{1, 0, 0\}$ are 7 examples.
@Khallil So for each element, there are two choices for different subsets, namely {inside, outside}. Since there are $n$ elements, the number of subset is then, as per out previous experiment with the binary sets, $2^n$. QED.
"Consider a set $S$ with $k+1$ members. Mentally divide it into a set with $k$ members plus one new member $a$. Now consider all the subsets of $S$. We know that there are $2^k$ subsets of the $S – {a}$, so there are $2^k$ subsets of $S$ that do not include $a$, namely $T_i \cup \{a\}$ ..."
@BalarkaSen Here's the full link. I didn't quite get the introduction of $T_i$. (people.umass.edu/partee/409/h15.pdf) Oh, I get it. It's the same thing you were talking about, kinda.
@BalarkaSen It's referred to the set difference I believe, but if it's like normal subtraction, then we'll simply have a set $S-A$ that does contain $a$.
In other words, there'd be $2(2^k)$ i.e. $2^{k+1}$ subsets of $\mathcal{P}(A) = 2^{k+1}$ so long as $A$ contains $k+1$ elements where the extra element is what we've been referring to as $a$ the whole time.
If I want to find all $X$ s.t. $\left| X \right| \leqslant 1$ that's an element of the power set, $\mathcal{P}(\{1,2,3\})$, is it as simple as listing the subsets of the original set with cardinality less than or equal to $1$?
If that's the case, would it be true that $\{ X \in \mathcal{P}(\{ 1, 2, 3 \}) : \left| X \right| \leqslant 1 \} = \{ \{1\}, \{2\}, \{3\}, \varnothing \}$?
@BalarkaSen Cool. Would it also be the case that $\{ X \in \mathcal{P}(\{ 1, 2, 3 \}) : \left| X \right| \leqslant 1 \} = \{ X \subseteq \mathcal{P}(\{ 1, 2, 3 \}) : \left| X \right| \leqslant 1 \}$?