Each element count from original set in power set

Question

Let say we have a set A = {x, y, z}. Than power set will be:

P(A) = {{}, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}}

When I look at the P(A) I see that each of the n elements from A will be contained in half of the subsets:

• Element $x: 4$ times
• Element $y: 4$ times
• Element $z: 4$ times

Question: how to prove it?

Answer 1

Let $S$ be the subset of $A$. Then we are seeking the number of such subsets so that $x \in S$. So put $x$ in $S$. Then for $y$ and $z$, there are two options each (either in $S$ or not). So there are $2\cdot2 = 4$ such subset $S$.

By renaming the objects, same argument is valid for $y$ and $z$.

More generally, if we have a set $A$ with $|A| = n$, number of subsets such that certain element is in that subset is $2^{n-1}$ because it is as same as finding the number of subsets of a set with size $n-1$ (Because that certain element is already chosen to be icluded in the subset).

Answer 2

Proposition: If $A$ is a set with $n$ elements ($n\in\mathbb N$) and $a\in A$, then $a$ belongs to half of the elements of $\mathcal{P}(A)$.

Proof: The map from $\{S\in\mathcal{P}(A)\,|\,a\in S\}$ into $\{S\in\mathcal{P}(A)\,|\,a\notin S\}$ defined by $\psi(S)=S\setminus\{a\}$ is a bijaction; its inverse is $S\mapsto S\cup\{a\}$. Since there is a bijection between $\{S\in\mathcal{P}(A)\,|\,a\in S\}$ and its complement in $\mathcal{P}(A)$, both sets have the same number of elements: $\frac12\#\mathcal{P}(A)=2^{n-1}$. So, $a$ belongs to half of the elements of $\mathcal{P}(A)$.

Answer 3

Pick an arbitrary element, $x$ from your set.

Classify the elements of the power set as" with $x$" and "without $x$."

Note that there is a one-to-one correspondence between the two classes.

If you adjoin $x$ to a set without $x$ you get a set with $x.$

If you remove $x$ from a set with $x$, you get a set without $x$.

Therefore the number of elements in each class is the same.

Thus $x$ appears in half of the sets.

Answer 4

Imagine a set $A$ with one or more elements. Put one element b aside. Consider all subsets of $A$'s other elements. Now take each of these subsets and add the excluded element. The number of subsets is the same. The collection of subsets with $b$ and without $b$ are all of the subsets of $A$.