Question

Suppose that \(|A|=m\) and \(|B|=n .\) Find the following cardinalities. $$ |\\{X \subseteq \mathscr{P}(A):|X| \leq 1\\}| $$

Short answer

The cardinality of the set of subsets taken from the power set of A such that the cardinality of each subset is less than or equal to 1, is \(2^m + 1\).

Step-by-step solution

Identify Cardinality of Power Set

Recall that the power set \(\mathscr{P}(A)\) of a set A with m elements is the set of all possible subsets of A. The number of elements in the power set, also known as its cardinality, is \(2^m\). This means that \(\mathscr{P}(A)\) will have \(2^m\) subsets.

Count Suitable Subsets

We are trying to find the number of subsets X of the power set such that the cardinality of each X is less than or equal to 1. This means it can have 0 or 1 element. So the subsets X of \(\mathscr{P}(A)\) that we are counting are the empty set and all one-element sets. The empty set counts as 1, and the number of one-element sets is equal to the number of elements in \(\mathscr{P}(A)\), which is \(2^m\). So adding these up, we get \(2^m + 1\) subsets that satisfy the condition.

Conclude

Therefore, the cardinality of the set of subsets X of \(\mathscr{P}(A)\) such that \(|X| \leq 1\) is \(2^m + 1\).