还在面临现代代数的学习挑战吗？别担心！我们的algebra-guide团队专业为您解决群论、环论、域论、线性代数等现代代数的问题。我们拥有深厚的专业背景和丰富的经验，能帮您完成高水平的作业和论文，让您的学习之路一路顺风！

以下是一些我们可以帮助您解决的问题：

现代代数基础：群、环、域的定义、性质和基本定理。

群论：正规子群、同态、同构、群作用等相关内容。

环论：环的理想、同态、同构、多项式环等知识点。

域论：域的扩张、域的结构、域的特征等内容。

线性代数：向量空间、线性变换、矩阵代数、特征值和特征向量等问题。

无论您面临的问题是什么，我们都会尽力为您提供专业的帮助，确保您的学习之旅顺利无阻！

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 $A$.

**E-mail:** help-assignment@gmail.com **微信:**shuxuejun

**help-assignment™是一个服务全球中国留学生的专业代写公司专注提供稳定可靠的北美、澳洲、英国代写服务专注于数学，统计，金融，经济，计算机科学，物理的作业代写服务**