Topology: An Inquiry-Based Approach

Let $A\text{,}$ $B\text{,}$ and $C$ be subsets of a set $X\text{.}$ Express each of the following sets in mathematical notation using the symbols $\cup \text{,}$ $\cap \text{,}$ and $\setminus \text{.}$

Let $X\subset Y\subset Z\text{.}$ Prove or disprove.

Let $A$ and $B$ be subsets of a universal set $U\text{.}$ Prove the associative and distributive laws. That is, prove each of the following.

Prove DeMorgan’s Laws. That is, let $\{A_{\alpha }\}$ be a collection of sets indexed by a set $I$ in some universal set $U\text{.}$ Prove that

If $A$ is a set, the power set of $A\text{,}$ denoted $2^A$ is the collection of all subsets of $A\text{.}$

If $A$ is a set, the power set of $A\text{,}$ denoted $2^A$ is the collection of all subsets of $A\text{.}$ (See Exercise 6.) Critique each of the following statements. Doe the statement make sense or not? If not, explain why and then correct the statement to something that is true (and non-trivial).

For each of the following, answer true if the statement is always true. If the statement is only sometimes true or never true, answer false and provide a concrete example to illustrate that the statement is false. If a statement is true, explain why. As an example of a true statement, consider the statement

Then $B\cap C\ne \mathrm{\varnothing }\text{.}$ We can justify the truth of this statement with a short argument. Since $A\cap B\ne \mathrm{\varnothing }\text{,}$ there is an element $x\in A\cap B\text{.}$ Then $x\in B\text{.}$ Since $A\cap B=A\cap C\text{,}$ we also must have $x\in A\cap C\text{,}$ which implies that $x\in C\text{.}$ Thus, $x\in B\cap C$ and $B\cap C\ne \mathrm{\varnothing }\text{.}$ As an example of a false statement, consider the statement

Then $B=C\text{.}$ We can show that this statement is false by providing a counterexample. For example, let $A=\{0,1\}\text{,}$ $B=\{1\}\text{,}$ and $C=\{1,2\}\text{.}$ Then $A\cap B=\{1\}=A\cap C\text{,}$ but $B\ne C\text{.}$