[Math] Power set of a set with three elements

elementary-set-theory

I need help with the set operation.

for example I got
$$A=\{0,1,2\}$$
and *

the question is asking how many elements in the power set are proper
subsets of A?

*

my answer is zero because there's no proper subset of itself. Am I right?

and what do they mean by how many elements in the power set are nonempty subsets of A?

****please explain… thank you 🙂

Best Answer

Power set of $A$: $P(A) = \{\emptyset,\{0\},\{1\},\{2\},\{0,1\},\{0,2\},\{1,2\},\{0,1,2\} \}$

Next up: with a proper subset of $A$, they mean a subset of $A$ that is not equal to $A$. This makes $\{0,1,2\}$ ineligible. All other elements of $P(A)$ are proper subsets.

If you count, you'll find that the solution is $7$. In general, for a set with $n$ elements, the power set contains $2^n$ elements. Only one of those elements is not a proper subset of our original set (because it is equal to that set), so for any $n$, the answer is $2^n - 1$.

And regarding your edit: with a "nonempty subset", they mean any subset that is not the empty set. If you take $P(A)$ and remove the empty set, that leaves $7$ other elements.