List all subsets of the 3 -element set $A={1,2,3}$. How many subsets does a set with $n$ elements have? How many of these subsets have at most two elements?

The subsets of the 3-element set $A={1,2,3}$ are:

$\emptyset$ (empty set)
${1}$
${2}$
${3}$
${1,2}$
${1,3}$
${2,3}$
${1,2,3}$

So, there are 8 subsets in total.

The number of subsets that a set with $n$ elements has is $2^n$. This can be understood by considering that each element in the set can either be included or excluded in a subset, giving 2 choices for each element. By multiplying these choices together for all $n$ elements, we get $2^n$ possible subsets.

To find the number of subsets with at most two elements, we can count the number of subsets with 0, 1, and 2 elements separately and add them together.

Subsets with 0 elements: There is only one such subset, which is the empty set $\emptyset$.

Subsets with 1 element: There are $n$ choices for the single element to include in the subset. In this case, $n=3$, so there are 3 subsets with 1 element.

Subsets with 2 elements: There are $\binom{n}{2}$ ways to choose 2 elements from a set with $n$ elements. In this case, $\binom{3}{2} = 3$, so there are 3 subsets with 2 elements.

Adding these counts together, we have $1+3+3=7$ subsets with at most two elements.

Simplify descriptions of the following sets. These sets depend on subsets $A, B$ of a universal set $X$, so that $A^{\prime}=X \backslash A$, and so on.
(a) $A^{\prime} \cup A, A^{\prime} \cap A, \quad\left(A^{\prime} \cup A^{\prime}\right)^{\prime} \cup\left(A^{\prime} \cap A\right)$.
(b) $(A \cap B) \cup(A \cup B), \quad\left(A \cup B^{\prime}\right) \cap\left(A^{\prime} \cap B\right)$,
$(A \cap B) \backslash B, \quad(A \cup B) \backslash B, \quad(A \cap B) \cup(A \backslash B)$

(a) Simplified descriptions of the sets are as follows:

$A^{\prime} \cup A$:
This set is the universal set $X$ because $A^{\prime} \cup A$ includes all elements that are either in $A^{\prime}$ or in $A$. Since $A^{\prime}$ contains all elements not in $A$ and $A$ contains all elements in $A$, the union of these two sets covers all elements in the universal set $X$.

$A^{\prime} \cap A$:
This set is the empty set $\emptyset$ because the intersection of $A^{\prime}$ and $A$ includes only the elements that are both in $A^{\prime}$ and in $A$. However, these two sets are disjoint, meaning they have no common elements, so their intersection is empty.

$(A^{\prime} \cup A^{\prime})^{\prime} \cup (A^{\prime} \cap A)$:
This set is the universal set $X$ because $(A^{\prime} \cup A^{\prime})^{\prime} \cup (A^{\prime} \cap A)$ combines the complement of the union of two disjoint sets ($A^{\prime} \cup A^{\prime}$) with the intersection of two disjoint sets ($A^{\prime} \cap A$). Since the complement of the empty set is the universal set $X$, and the union of the universal set $X$ with any set also gives the universal set $X$, the resulting set is the universal set $X$.

(b) Simplified descriptions of the sets are as follows:

$(A \cap B) \cup (A \cup B)$:
This set is equal to $A \cup B$ because $(A \cap B) \cup (A \cup B)$ combines the intersection of $A$ and $B$ with the union of $A$ and $B$. Since the union of two sets includes all elements that are in either set, the resulting set is $A \cup B$.

$(A \cup B^{\prime}) \cap (A^{\prime} \cap B)$:
This set is equal to $\emptyset$ because $(A \cup B^{\prime}) \cap (A^{\prime} \cap B)$ combines the union of $A$ and the complement of $B$ with the intersection of the complement of $A$ and $B$. However, these two sets have no common elements, so their intersection is empty.

$(A \cap B) \backslash B$:
This set is equal to $A \cap B^{\prime}$ because $(A \cap B) \backslash B$ removes all elements that are in both $A$ and $B$, resulting in only the elements that are in $A$ but not in $B$.

$(A \cup B) \backslash B$:
This set is equal to $A$ because $(A \cup B) \backslash B$ removes all elements that are in $B$ from the union of $A$ and $B$, leaving only the elements that are in $A$.

$(A \cap B) \cup (A \backslash B)$:
This set is equal to $A$ because $(A \cap B) \cup (A \backslash B)$ combines the intersection of $A$ and $B$ with the elements that are in $A$ but not in $B$, resulting in all elements that are in $A$.

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