Exercises
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Exercises
Determine whether $A\subseteq B$, $B\subseteq A$ or neither. Is either set a proper subset of the other?
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If $A=\{2, 4, 6, 8\}$, $B=\{1, 3, 5, 7, 9\}$, $C=\{1, 2, 3, 7, 8, 9\}$ and $U=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, determine what the following sets are.
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If $A=\{x\in\mathbb{R} | 0 \lt x \lt 4\}$, $B=\{x\in \mathbb{R} | 3 \le x \le 5\}$, $C=\{x\in \mathbb{R} | 4 \lt x \lt 5\}$ and $U=\{x\in\mathbb{R} | 0 \lt x \lt 6\}$, determine what the following sets are.
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Suppose $A_i=\{x\in\mathbb{R} | (1/n, 1]\}$ where $U=\{x\in\mathbb{R} | 0 \le x \le 1\}$. determine the following sets.
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- Are the sets mutually disjoint?
Suppose $A_i=\{x\in\mathbb{R} | (1/(n+1), 1/n]\}$ where $U=\{x\in\mathbb{R} | 0 \le x \le 1\}$. Determine the following sets.
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- Are the sets mutually disjoint?
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Determine if one of $A=\{1, 2, 3\}$ and $B=\{\{1\}, \{2\}, \{3\}\}$ is a subset (or proper subset) of the other
Neither set is a subset (or a proper subset) of the other. The elements of A are numbers where the elements of B are sets (each with one element).
Determine if one of $A=\emptyset$ and $B=\emptyset$ is a subset (or proper subset) of the other
Both $A\subseteq B$ and $B\subseteq A$ because the empty set is a subset of every set. Because each is a subset of the other they must be equal which means one can't be a proper subset of the other.
Alternatively, you could argue that, if one was a proper subset of the other then the second one would have to have at least one element that the first doesn't. But this can't be the case because that set is empty.
Determine if one of $A = \{(x, y) | x, y\in\mathbb{R} \text { and } x^2 + y^2 = 1\}$ and $B=\{(x, y) | x=r\cos(\theta), y=r\sin(\theta), 0\lt\theta\lt 2\pi\}$ is a subset (or proper subset) of the other
Both sets describe a unit circle but A is complete where B is missing the point $(0, 0)$. This means that $B \subseteq A$ and, since A has a point that B doesn't, $B \subset A$.
Determine if one of $A = \{\emptyset\}$ and $B=\emptyset$ is a subset (or proper subset) of the other
The key point here is that A isn't empty. It has a single element - the empty set. Therefore, $B\subseteq A$ because the empty set is a subset of every set and $B\subset A$ because A has an element that B doesn't. However, $A\nsubseteq B$ because A has elements where the empty set doesn't.
Determine if one of $A = \{1, 4, 9, 16\}$ and $B=\{1^2, 2^2, 3^2, 4^2\}$ is a subset (or proper subset) of the other
It's the value of the elements that we care about, not how they're formatted so $A\subseteq B$ and $\B\subseteq A$. In fact $A = B$ so neither is a proper subset of the other.
If $A=\{2, 4, 6, 8\}$, $B=\{1, 3, 5, 7, 9\}$, $C=\{1, 2, 3, 7, 8, 9\}$ and $U=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, what is $A-B$.
$$A - B = \{2, 4, 6, 8\} - \{1, 3, 5, 7, 9\} = \{2, 4, 6, 8\} = A $$
If $A=\{2, 4, 6, 8\}$, $B=\{1, 3, 5, 7, 9\}$, $C=\{1, 2, 3, 7, 8, 9\}$ and $U=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, what is $A\cap B$.
$$A \cap B = \emptyset $$
If $A=\{2, 4, 6, 8\}$, $B=\{1, 3, 5, 7, 9\}$, $C=\{1, 2, 3, 7, 8, 9\}$ and $U=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, what is $A\cup B \cup C$.
$$A\cup B \cup C\ = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} $$
If $A=\{2, 4, 6, 8\}$, $B=\{1, 3, 5, 7, 9\}$, $C=\{1, 2, 3, 7, 8, 9\}$ and $U=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, what is $\overline{A\cup C}$.
$$\overline{A\cup C} = \{5, 10\} $$
If $A=\{2, 4, 6, 8\}$, $B=\{1, 3, 5, 7, 9\}$, $C=\{1, 2, 3, 7, 8, 9\}$ and $U=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, what is $\overline{A} \cup \overline{C}$.
$$\overline{A} \cup \overline{C} = \{1, 3, 4, 5, 6, 7, 9, 10\} $$
If $A=\{2, 4, 6, 8\}$, $B=\{1, 3, 5, 7, 9\}$, $C=\{1, 2, 3, 7, 8, 9\}$ and $U=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, what is $\overline{A} \cap \overline{B} \cap \overline{C}$.
$$\overline{A} \cap \overline{B} \cap \overline{C} = \{10\} $$
If $A=\{x\in\mathbb{R} | 0 \lt x \lt 4\}$, $B=\{x\in \mathbb{R} | 3 \le x \le 5\}$, $C=\{x\in \mathbb{R} | 4 \lt x \lt 5\}$ and $U=\{x\in\mathbb{R} | 0 \lt x \lt 6\}$, what is $A-B$.
$$A-B = \{x\in\mathbb{R} | 0 \lt x \lt 3\}$$
If $A=\{x\in\mathbb{R} | 0 \lt x \lt 4\}$, $B=\{x\in \mathbb{R} | 3 \le x \le 5\}$, $C=\{x\in \mathbb{R} | 4 \lt x \lt 5\}$ and $U=\{x\in\mathbb{R} | 0 \lt x \lt 6\}$, what is $A \cup B$.
$$A\cup B = \{x\in\mathbb{R} | 0 \lt x \le 5\}$$
If $A=\{x\in\mathbb{R} | 0 \lt x \lt 4\}$, $B=\{x\in \mathbb{R} | 3 \le x \le 5\}$, $C=\{x\in \mathbb{R} | 4 \lt x \lt 5\}$ and $U=\{x\in\mathbb{R} | 0 \lt x \lt 6\}$, what is $\overline{C}$.
$$\overline{C} = \{x\in\mathbb{R} | 0 \lt x \le 4 \text{ or } 5 \le x \lt 6\}$$
If $A=\{x\in\mathbb{R} | 0 \lt x \lt 4\}$, $B=\{x\in \mathbb{R} | 3 \le x \le 5\}$, $C=\{x\in \mathbb{R} | 4 \lt x \lt 5\}$ and $U=\{x\in\mathbb{R} | 0 \lt x \lt 6\}$, what is $\overline{A} \cup \overline{B}$.
$$\overline{A} \cup \overline{B} = \{x\in\mathbb{R} | 0 \lt x \lt 3 \text{ or } 4 \le x \lt 6\}$$
If $A=\{x\in\mathbb{R} | 0 \lt x \lt 4\}$, $B=\{x\in \mathbb{R} | 3 \le x \le 5\}$, $C=\{x\in \mathbb{R} | 4 \lt x \lt 5\}$ and $U=\{x\in\mathbb{R} | 0 \lt x \lt 6\}$, what is $overline{A-C}$.
$$\overline{A-C} = \overline{A} = \{x\in\mathbb{R} | 5 \le x \lt 6\}$$
If $A=\{x\in\mathbb{R} | 0 \lt x \lt 4\}$, $B=\{x\in \mathbb{R} | 3 \le x \le 5\}$, $C=\{x\in \mathbb{R} | 4 \lt x \lt 5\}$ and $U=\{x\in\mathbb{R} | 0 \lt x \lt 6\}$, what is $A \cup B \cup C$.
$$A \cup B \cup C = \overline{A} = \{x\in\mathbb{R} | 0 \lt x \le 5\}$$
Suppose $A_i=\{x\in\mathbb{R} | (1/n, 1]\}$ where $U=\{x\in\mathbb{R} | 0 \le x \le 1\}$. What is $\bigcup\limits_{i=1}^{\infty} A_i$?
Try writingo out the first several terms in the union then look for a pattern in terms of which ones are contained in the others.
Suppose $A_i=\{x\in\mathbb{R} | (1/n, 1]\}$ where $U=\{x\in\mathbb{R} | 0 \le x \le 1\}$. What is $\bigcup\limits_{i=1}^{\infty} A_i$?
It can help to start by writing out some of the first few elements of the union:
$$\bigcup\limits_{i=1}^{\infty} A_i = (1, 1] \cup (1/2, 1] \cup (1/3, 1] . . . (1/n, 1] . . .$$First, notice that the first set, $(1, 1]$, is the empty set since there are no numbers between 1 and 1 that are both equal to 1 and not equal to it. Since adding the union of the empty set and any other set is just the other set, we can effectively discard that from the union.
$$\bigcup\limits_{i=1}^{\infty} A_i = (1/2, 1] \cup (1/3, 1] . . . (1/n, 1] . . .$$From here, you can see that the sets get professively bigger, each one containing all the ones that came before it. As n gets large, 1/n gets arbitrarily close to 0 but never actually equals it. That makes the unionn $(0, 1]$ or $\{x\in\mathbb{R} | 0 \lt x \le 1\}$.
Suppose $A_i=\{x\in\mathbb{R} | (1/n, 1]\}$ where $U=\{x\in\mathbb{R} | 0 \le x \le 1\}$. What is $\bigcap\limits_{i=2}^{\infty} A_i$?
Try writingo out the first several terms in the union then look for a pattern in terms of which ones are contained in the others.
Also, notice that the starting index here is different from the first question's.
Suppose $A_i=\{x\in\mathbb{R} | (1/n, 1]\}$ where $U=\{x\in\mathbb{R} | 0 \le x \le 1\}$. What is $\bigcap\limits_{i=2}^{\infty} A_i$?
Notice how the starting index is different here from the previous question, starting at 2 rather than 1. That avoids having $(1,1]$ in the intersection. Since that set is the empty set, that would have made the intersection the empty set. What we do have here is
$$\bigcap\limits_{i=2}^{\infty} A_i = (1/2, 1] \cap (1/3, 1] . . . (1/n, 1] . . .$$As in the previous example, the sets get progressively bigger (wider) with each one containing all of the sets that came before it. That makes the first set in the list the "smallest" one that is also contained in all the others so:
$$\bigcap\limits_{i=2}^{\infty} A_i = (1/2, 1] = \{x\in\mathbb{R} | 1/2 \lt x \le 1\}$$Suppose $A_i=\{x\in\mathbb{R} | (1/n, 1]\}$ where $U=\{x\in\mathbb{R} | 0 \le x \le 1\}$. What is $\bigcup\limits_{i=2}^{\infty} \overline{A_i}$?
$$\bigcup\limits_{i=1}^{\infty} \overline{A_i}\ = [0, 1/2] = \{x\in\mathbb{R} | 0 \le x \le 1/2\}$$
Suppose $A_i=\{x\in\mathbb{R} | (1/n, 1]\}$ where $U=\{x\in\mathbb{R} | 0 \le x \le 1\}$. What is $\bigcap\limits_{i=1}^{\infty} \overline{A_i}$?
$$\bigcap\limits_{i=1}^{\infty} \overline{A_i} = \{0\}$$
Are the sets mutually disjoint?
Is there a number that's contained in all the sets?
Are the sets mutually disjoint?
The sets aren't mutually disjoint since 1/2 is an element of all of them.
Suppose $A_i=\{x\in\mathbb{R} | (1/(n+1), 1/n]\}$ where $U=\{x\in\mathbb{R} | [0,1]\}$. What is $\bigcup\limits_{i=1}^{\infty} A_i$?
It can help to start by writing out some of the first few elements of the union:
$$\begin{aligned}\bigcup\limits_{i=1}^{\infty} A_i $= (1/2, 1] \cup (1/3, 1/2] \cup (1/4, 1/3] . . . (1/(n+1), 1/n] . . . \\ . . . (1/(n+1), 1/n] . . . (1/4, 1/3] \cup (1/3, 1/2] \cup (1/2, 1] \end{aligned}$$In the second line, I changed the order so that the starting/ending endpoints were in ascending order. Seen that way, you can see that each interval ends where the next one begins so ulimtately, every number in the form $1/n$ is contained in the union along with all the values in between so we must have:
$$\bigcup\limits_{i=1}^{\infty} A_i = (0, 1] = \{x\in\mathbb{R} | 0 \lt x \le 1\}$$Suppose $A_i=\{x\in\mathbb{R} | (1/(n+1), 1/n]\}$ where $U=\{x\in\mathbb{R} | [0,1]\}$. What is $\bigcap\limits_{i=1}^{\infty} A_i$?
I'll start by writing this out the same way that we did in the previous question:
$$\begin{aligned}\bigcup\limits_{i=1}^{\infty} A_i $= (1/2, 1] \cup (1/3, 1/2] \cup (1/4, 1/3] . . . (1/(n+1), 1/n] . . . \\ . . . (1/(n+1), 1/n] . . . (1/4, 1/3] \cup (1/3, 1/2] \cup (1/2, 1] \end{aligned}$$Again, in this format, it's easy to see that none of the individual elements overlap, including at their end points, so
$$\bigcap\limits_{i=1}^{\infty} A_i = (0, 1] = \emptyset$$Suppose $A_i=\{x\in\mathbb{R} | (1/(n+1), 1/n]\}$ where $U=\{x\in\mathbb{R} | [0,1]\}$. What is $\bigcup\limits_{i=1}^{\infty} \overline{A_i}$?
$$\bigcup\limits_{i=1}^{\infty} \overline{A_i} = [0, 1] = U$$
Suppose $A_i=\{x\in\mathbb{R} | (1/(n+1), 1/n]\}$ where $U=\{x\in\mathbb{R} | [0,1]\}$. What is $\bigcap\limits_{i=1}^{\infty} \overline{A_i}$?
$$\bigcap\limits_{i=1}^{\infty} \overline{A_i} = \{0\}$$
Are the sets mutually disjoint?
We say when we were calculating the intersection that there was no overlap between the sets so they are mutually disjoing.
