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In the rules for set complements theorem, we prove that the complement of a union is the intersection of the complements, and we proved that the complement of an intersection was the union of the complements. These are often called De Morgan's laws. The next problem has you prove De Morgan's laws are true regardless of the number of sets involved in the union or intersection.
Problem.De Morgan's Laws for Sets
Suppose that for each $j$ in some nonempty set $J$ that $A_j$ is a set, and also suppose that $B$ is a set. Pick one of the statements below and prove that it is true. The other is very similar.
- $\ds B\setminus \left(\bigcup_{j\in J}A_j\right) = \bigcap_{j\in J}\left(B\setminus A_j\right) $
- $\ds B\setminus \left(\bigcap_{j\in J}A_j\right) = \bigcup_{j\in J}\left(B\setminus A_j\right) $
Problem 81: (Additional Properties Of Cartesian Products)
Let $A$, $B$, $C$, and $D$ be sets. Prove or disprove each statement.
- $(A\times B)\cap(C\times D) = (A\cap C)\times(B\cap D)$
- $(A\times B)\cup(C\times D) = (A\cup C)\times(B\cup D)$
Defintion.Interior Closure
Let $A$ be a subset of the real numbers.
- We defined $A'$ to be the set of limit points of $A$.
- The closure of $A$, written $\text{cl}(A)$, is the union of $A$ and it's limit points, so $\cl(A) = A\cup A'$.
- The interior of $A$, written $\text{int}(A)$, is the collection of interior points of $A$.
Problem.The closure of $A$ is a closed set.
Prove that the closure of $A$ is a closed set.
Problem.The interior is the union of every open set inside a set
Let $S^\circ$ be the union of every open set contained in $S$. Prove that $\text{int} (S)=S^\circ$.
Problem. The closure is the intersection of every closed set containing a set
Let $\bar A$ be the intersection of every closed set that contains $A$. Prove that $\bar A=\text{cl}(A)$.
- Boundary points???? Maybe not.
- Define Isolated point???? Maybe not. Sounds like a great final exam question. Maybe give them a specific set and these defintions, and then have them prove that something is what it is.
Limits of functions
- Define limit of $f$ at $c$.
Exercise.Limits Of Sequences Are Unique
Suppose that the sequence $(a_n)$ converges to $A$ and converges to $B$. Prove that $A=B$.
Let $\varepsilon>0$ be given. We will show that $|A-B|<\varepsilon$, which shows that $A=B$. The key in this problem is to realize that by adding zero and using the triangle inequality, we have $$|A-B|=|A+0-B|=|A+(-a_n+a_n)-B|=|A-a_n+a_n-B|\leq |A-a_n|+|a_n-B| =|a_n-A|+|a_n-B|.$$ The reason we want to rewrite the difference in this way is because both $|a_n-A|$ and $|a_n-B|$ show up in the definition of converges. To apply the definition, we must make these terms appear in our work.
We have assumed that $(a_n)$ converges to $A$ and converges to $B$. Since $(a_n)$ converges to $A$, we know that given $\varepsilon_1>0$, there exists an $N_1$ such that $|a_n-A|<\varepsilon_1$ for every $n\geq N_1$. Since $(a_n)$ converges to $B$, we know that given $\varepsilon_2>0$, there exists an $N_2$ such that $|a_n-B|<\varepsilon_2$ for every $n\geq N_2$. We can pick $\varepsilon_1$ and $\varepsilon_2$ to be any positive real numbers we want, and once we've done that we can find $N_1$ and $N_2$ so that whenever $n$ is greater than both, we would have $$|A-B|=|A-a_n+a_n-B|\leq |A-a_n|+|a_n-B| = |a_n-A|+|a_n-B| < \varepsilon_1+\varepsilon_2.$$ We want to show that $|A-B|<\varepsilon$, so if we pick $\varepsilon_1 =\varepsilon/2$ and $\varepsilon_2=\varepsilon/2$, then we would be able to replace $\varepsilon_1+\varepsilon_2$ with $\epsilon$ and we'd succeed in showing what's needed. All of the above work is scratch work that leads the proof below.
Let $\varepsilon>0$ be given. We will show that $|A-B|<\varepsilon$, and since this holds for any $\varepsilon>0$ we have then shown that $A=B$. Let $\varepsilon_1=\varepsilon_2=\varepsilon/2$. Since $(a_n)$ converges to both $A$ and $B$, pick $N_1$ and $N_2$ so that if $n\geq N$ then $|a_n-A|<\varepsilon_1$, and if $n\geq N_2$ then $|a_n-B|<\varepsilon_2$. Let $N=\max\{N_1,N_2\}$. Then for any $n\geq N$ we have $$|A-B|=|A-a_n+a_n-B|\leq |A-a_n|+|a_n-B| = |a_n-A|+|a_n-B| < \varepsilon_1+\varepsilon_2 = \varepsilon.$$ This show that $|A-B|<\varepsilon$ as desired.
Here is an alternate proof. Suppose that $A\neq B$. Let $\varepsilon=|A-B|/2$. Since $(a_n)$ converges to both $A$ and $B$, pick $N_1$ and $N_2$ so that if $n\geq N$ then $|a_n-A|<\varepsilon$, and if $n\geq N_2$ then $|a_n-B|<\varepsilon$. Let $N=\max\{N_1,N_2\}$. Then for any $n\geq N$ we have $$|A-B|=|A-a_n+a_n-B|\leq |A-a_n|+|a_n-B| = |a_n-A|+|a_n-B| < \varepsilon+\varepsilon = |A-B|.$$ This shows that $|A-B|<|A-B|$, a contradiction. Hence we must have $A=B$.
Problem.
Give an example of sequences $(a_n)$ and $(b_n)$ so that $(a_n+b_n)$ converges, but neither $(a_n)$ nor $(b_n)$ converges.
For more problems, see AllProblems