How many sets in the power set containing a given integer?

Question

Let $\mathcal{J}\equiv \{1,...,J\}$ and let $\mathcal{C}$ be the power set of $\mathcal{J}$ (with cardinality $2^{J}$).

Question: take any $j\in \mathcal{J}$. How many elements of $\mathcal{C}$ (sets) contain $j$?

For example: if $J=3$ and $j=1$, then $\mathcal{C}\equiv \Big\{\emptyset, \{1,2,3\}, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}\Big\}$ and there are $4$ sets in $\mathcal{C}$ containing $1$.

Answer 1

Hint: each set containing $j$ can be paired nicely with a certain set not containing $j$ and vice versa, so the answer is "half of them".

Answer 2

Can you find a bijective function that maps each element of the power set of $\mathcal{J}-\{j\}$ to the set that you want to count?

Answer 3

Let $\mathcal{J}\equiv \{1,...,J\}$ and let $\mathcal{C}$ be the power set of $\mathcal{J}$ (with cardinality $2^{J}$).

The reason the cardinality of $\mathcal{C}$ is $2^{J}$ is that for each element you will have a set of subsets in $\mathcal{C}$ without some given element $a$ and another, otherwise identical set of subsets that do contain $a$. This applies to each member of $\mathcal{J}$ giving the multiplication of the $2$ states (present/absent) for each base set element in the powerset elements, hence $2^{J}$ cardinality.