Assignment 4

This assignment has been graded.

Questions

1. Choose one of the following statements and provide a counterexample to show that it is false.

  • For all sets $A$, $B$, and $C$, $(A \cap B) \cup C = A \cap (B \cup C)$.
  • For all sets $A$, $B$, and $C$, $A \setminus (B \cup C) = (A \setminus B) \cup (A \setminus C)$.
  • For all sets $A$ and $B$, $\mathcal P(A \cup B) = \mathcal P(A) \cup \mathcal P(B)$.
  • For all sets $A$ and $B$, $\overline{ A \setminus B } = \overline A \setminus \overline B$.
  • For all sets $A$ and $B$, $A \cup B \not= A \cap B$.
  • For all sets $A$ and $B$, $A \cup B$ contains more elements than does $A$.

2. Choose one of the following sets, and give a brief English language description of what the set contains. (For example, $\overline A$ contains all elements that are not in $A$.) Make sure your description is unambiguous!

  • $A \cap \overline B$
  • $A \setminus (B \cap C)$
  • $\overline A \cap \overline B \cap \overline C$
  • $\overline{ A \cap B \cap C }$
  • $\overline A \cup \overline B \cup \overline C$
  • $\overline{ A \cup B \cup C }$

3. Choose one of the following sentences, and replace each $\square$ with either $\in$ or $\subseteq$ so that it becomes a statement that is true whenever $A$ is a set and $X$ is any object (set or otherwise).

  • If $X \square A$, then $\{X\} \square A$.
  • If $X \square A$, then $X \square \mathcal P(A)$.
  • If $X \square A$, then $\{X\} \square \mathcal P(A)$.
  • If $X \square \mathcal P(A)$, then $X \square A$.
  • If $\{X\} \square A$, then $X \square A$.
  • If $\{X\} \square A$, then $\{X\} \square \mathcal P(A)$.

4. Below are some totally fun families of indexed sets!

  • For each $n \in \mathbb Z^+$, $A_n = \left[ \frac 12, \frac 12 + \frac 1n \right)$.
  • For each $n \in \mathbb Z^+$, $B_n = \left( \frac 12 - \frac 1n, \frac 12 + \frac 1n \right)$.
  • For each $n \in \mathbb Z^+$, $C_n = \left[ 1 + \frac 1n, 3 - \frac 1n \right]$.

Choose one of the families above, and then choose one of the following things to do.

  • Write a big scary expression (like the $\bigcap_{t \in \mathbb R} C_t$ of Exercise 2.3.15 in Chapter Zero) for the intersection of all the sets in the family. Then explain what that intersection actually consists of.
  • Write a big scary expression (like the $\bigcup_{t \in \mathbb R} C_t$ of Exercise 2.3.15 in Chapter Zero) for the union of all the sets in the family. Then explain what that union actually consists of.

Team B1

1. * For all sets $A$ and $B$, $A \cup B$ contains more elements than does $A$.

Counterexample:

Let Set A= {1,2,3,4,5,6,7,8,9} which contains 9 elements.
Let Set B= {2,3,4,5} which contains 4 elements.

The Union of Set A and B will contain {1,2,3,4,5,6,7,8,9} which is also 9 elements.

And since 9=9, we can not say that the Union of Set A and B contains more elements than does Set A.

Very nice! (And the union of A and B will be the set {1,2,3,4,5,6,7,8,9}, which contains 9 elements.)

2. * $A \cap \overline B$ contains all of the elements that Set A and the Complement of Set B have in common.
A contains all of the elements in Set A. B Complement contains all of the elements that are not in Set B. And the intersection of the two contains the elements that appear under both conditions (in set A and in the complement of set B).

This looks fine. Does everything you said distill down to saying "$A \cap \overline B$ contains all elements that are in A but not in B"?

3. If X◻A, then X◻P(A) should be written as: If $X \in A$, then $X \in P(A)$

You have to be careful with this. Suppose A={1,2}. Then $1 \in A$, but is it the case that $1\in\mathcal P(A)$? Actually, no, since the only elements of the power set of A are {}, {1}, {2}, and {1,2}. And 1 is not one of those 4 things. (Be careful not to confuse 1 with {1}.)

4. $\cap _{n \in \mathbb{Z^+}} \text{,} A_n = [\frac{1}{2} \text{,} \frac{1}{2} + \frac{1}{n})$

This intersection contains $\emptyset$. As n increases, the value of $\frac{1}{n}$ moves towards zero, and therefore the value of $\frac{1}{2} + \frac{1}{n}$ moves towards $\frac{1}{2}$. This leaves the intersection to include only $[\frac{1}{2} \text{,} \frac{1}{2})$.

I'm trying to decide what you mean by "contains $\emptyset$". The intersection of all those sets is a set, so are you saying that that set is the empty set? Or was that a typo, and you meant to say that it contains the number zero? I'm not sure.

Meanwhile, your reasoning about the behavior of the sets An as n varies does seem to make sense, and I think it's right! You say that the intersection will "include only $[\frac{1}{2} \text{,} \frac{1}{2})$", which is a weird looking interval. I guess, though, that by the definitions on page 41 of Chapter Zero, we'd have
$[\frac{1}{2} \text{,} \frac{1}{2}) = \{ x \in \mathbb R : \frac 12 \le x < \frac 12 \}$, and since the defining property of that set, namely $1/2 \le x < 1/2$, cannot be satisfied for any $x \in \mathbb R$, I guess the set is… the empty set. So maybe you are saying the intersection is the empty set. But I think the set does contain one number, since $\forall n\in \mathbb Z^+,\, \frac 12 \in [\frac 12, \frac 12 + \frac 1n)$, right? So actually I think the intersection is just the set {1/2}.

So… correct analysis, slightly mistaken conclusion?

Comments:


Team B2

1. Statement: For all sets A and B, A∪B≠A∩B.
Counterexample: If set A= {2,5,8,} and set B={2,5,8}, then the union of sets A and B is {2,5,8} and the intersection of sets A and B is also {2,5,8}. This shows that that the statement "For all sets A and B, A∪B≠A∩B" is false because there can be two sets that contain only the same elements, and these 2 sets will have the same union and intersection.

Nice!!

2. $\overline{ A \cap B \cap C }$ is the set we chose. This set represents the complement of the intersection of sets A, B, and C. This set contains all the elements that are not in sets A and B and C. Therefore, in terms of a venn diagram, this set would have everything shaded in besides the small piece in the middle where sets A, B, and C "intersect."

Bah, fine. Cheat by making a reference to Venn diagrams. Of course, saying "not in sets A and B and C" is kind of ambiguous… right? Well, maybe not… It's ambiguous as to whether or not it's ambiguous? I give up. By the way, I like how "intersect" is in scare quotes.

3. If {X}◻A, then X◻A. These blanks can be filled in so that the sentence turns into the following statement:
If {x}⊆A, then x∈A. This statement is true because x must be in A in order for x to be a subset of A.

Wow — correct and crystal clear!

And you know what else? This reminds me of the question from Team Assignment 2 about the many ways an implication can be phrased; as you point out here, saying "If X, then Y" is the same as saying "Y must be true in order for X to be true".

4. $\bigcap_{t \in Z^+}C_n$
$C_n$= $\left [1+ \frac 1n , 3-\frac 1n \right]$

This intersection consists of the closed interval $[2,2]$. When $n=1$ all sets in the family of sets consists of only the integer 2 which means the sets are intersecting.

I think your answer is right. And you have part of the explanation, namely that C1=[2,2]. But you also need to note that for any n > 1, Cn contains [2,2], so that intersecting C1 with all of the other sets doesn't exclude anything in [2,2].

Moreover, the statement "all sets in the family of sets consists of only the integer 2 which means the sets are intersecting" isn't precise enough to be meaningful. (I really didn't know what to make of it.)

Comments:


Team B3

1.) * For all sets $A$, $B$, and $C$, $(A \cap B) \cup C = A \cap (B \cup C)$.

Counterexample:
Let set $A$ : {1,2,3}
Let set $B$ : {3,4,5}
Let set $C$ : {1,3,5}

$(A \cap B) \cup C = {1,3,5}$
$A \cap (B \cup C) = {3}$

Clearly $(A \cap B) \cup C$ does not equal $A \cap (B \cup C)$. Therefore, the sets provided are a counterexample.

Hey, this is excellent! Very clearly presented, and it feels like nothing more than just what's needed for a counterexample. If you wanted to spell things out even more, you could say what $A \cap B$ and $B \cup C$ are separately, too, to make it easier to see how it is that $(A \cap B) \cup C$ and $A \cap (B \cup C)$ come out to be what you claim.

And one final note: To get the curly braces in LaTeX, you can use \{ and \}.

2.) $\overline A \cap \overline B \cap \overline C$
The complement of A contains all of the elements not in A which intersects all the elements that are not in B and intersects all the elements not in C.

Important advice: Always talk about what a set is in terms of what it means for one element to be in the set, as opposed to trying to describe "the elements of the set" in general terms. So here, try saying that this set contains "every element which is not in A and not in B and not in C", and things may get easier. As it is, when you wrote "which intersects all" I think you were trying to make that connection to "and" but it was more difficult to think about without the advice above.

3.) If $X \square A$, then $\{X\} \square A$.
If $X \in A$, then $\{X\} \subseteq A$.

Yes, for sure.

4.) $\bigcup_{n \in \mathbb Z^+} A_n$ = $\left[ \frac 12, \frac 12 + \frac 1n \right)$

The union of set $A_n$ consists of all numbers between $\frac 12$ and $\frac 32$. This is because when $n=1$ , $\left[ \frac 12 + \frac 1n \right)$ = $\frac 32$. Then as n gets bigger and bigger, $\frac 1n$ gets smaller and smaller, forcing $\left[ \frac 12 + \frac 1n \right)$ to get closer and closer to $\frac 12$. Therefore the Union is as follows:

$[\frac 12 , \frac 12 < x \leq\frac 32)$

This is perfect, except for two minor notation issues. One is that you seemingly wrote $\left[ \frac 12 + \frac 1n \right)$ (which is sort of meaningless) when you wanted to write just the number $\frac 12 + \frac 1n$. The second is that the very last thing you wrote was probably supposed to be the half-open interval $[\frac 12 , \frac 32)$ — see page 41 of Chapter Zero to see if that set is indeed what you intended.

Comments:


Team B4

1. For all sets $A$ and $B$, $\mathcal P(A \cup B) = \mathcal P(A) \cup \mathcal P(B)$.
Let set A={1,2} and Let set B={3,4}
The power set for A∪B would then be { {1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4},{empty} }
P(A)∪P(B)= { { empty } }
Since these are not the same, this proves the first statement false.

Ah, the power set one! Looks like you nailed the power set of $A\cup B$. (The wiki seems to ignore two {'s typed in a row, so I fixed it for you.)

One minor quibble would be that the empty set is written $\emptyset$, or if you're lazy, even just as "{}". The bigger problem is that I think your expression for $\mathcal P(A) \cup \mathcal P(B)$ is not correct — what you wrote down (namely, the set containing only the empty set) is the intersection of $\mathcal P(A)$ and $\mathcal P(B)$, not the union.

2. $\overline{ A \cup B \cup C }$ Would be the set of all elements that are not in $A$ or $B$ or $C$. In other words, if $A, B$ and $C$ were represented by a venn diagram, there would be no sections "Shaded in."

Oh fine, resort to describing the Venn diagram. It certainly is easier than trying to come up with English phrasing that is less ambiguous than "not in $A$ or $B$ or $C$". But let's try anyway! How about this: "…the set of every element that is in none of A or B or C." Ha!

3. "If $X \in\mathcal P(A)$, then $X \subset A$." If $X$ is a set, then by being in $\mathcal P(A)$, $X$ is a subset of $A$. By being in $\mathcal P(A)$, $X$ is an element, of $\mathcal P(A)$, but every element in $\mathcal P(A)$ is a set, and specifically, a subset of $A$. So, "if $X \in\mathcal P(A)$, then $X \subset A$" is true when $A$ is a set and when $X$ is a set or an element (Because $X$ can be considered an element and a set!)

Awesome! Very right, and well-explained. X is a set… that is an element of another set. It's sick, really.

Oh, before I forget, though: The symbol you want in LaTeX is \subseteq instead of \subset, since \subseteq (as the "eq" suggests) allows the possibility that the sets are equal. (Although it used to be that $\subset$ just meant "subset" and that was the only symbol, so be prepared for it to mean that if you ever look at really old books.)

4. $\bigcap_{n \in \mathbb Z^+} C_n$ = {$\emptyset$}

The intersection of these sets has only the empty set because combining all of the sets would only allow possible elements in their intersection to be between the interval $\left(1, \frac 32\right)$, as it must be at least 1 + 1 over some fraction (as stated by set C) and less than 1.5 (as stated by set A). However, the first set A has no elements in this interval, as it has the element 1 but none greater than 1 and also less than 1.5. So therefore, there could be no element in all of the sets except the empty set.

I think you mean that the intersection of the sets "is" (not "has") the empty set. In any case, it's not too hard to see that $2\in C_n$ for all $n\in\mathbb Z^+$, so I don't think the intersection can be empty, right?

I tried to figure out where your reasoning went wrong, but it was hard to understand the language you used… for example, does "combining" mean intersecting? Or combining in some other way? And when you say "it must be at least 1+1", what do you mean by "it"? A number that is in the intersection? That I agree with, and also that any such number must be less than 1.5. But this all seems okay so far. But now you refer to "the first set A"! I thought you had chosen to work with the set $C_n$. So I'm not sure if perhaps the question was somehow misinterpreted, or what.

The moral here may be to make sure you get all the brains on your team to think through the solution before you finalize it.

Comments:


Team B5

1. For all sets $A$ and $B$, $\overline{ A \setminus B } = \overline A \setminus \overline B$

A counter example to make the previous statement false would be:

Let set $\overline {A}$ = {1,2,3,4,5}
Let set $\overline {B}$ = {6,7}

This means that $\overline {A \setminus B}$ is false since the sets are not relative complements, making $\overline A \setminus \overline B$ false through the Distribution Property of Sets.

What does it mean to say "$\overline {A \setminus B}$ is false"? $\overline {A \setminus B}$ is a set, and how do you think of a set as being false? What does it mean to say that sets are "relative complements"? What is the "Distribution Property of Sets"? (I don't think that phrase appears in either of our books.) You also didn't identify what your sets A and B were, but gave what their complements are supposed to be. This does indirectly specify A and B, assuming our universe is the integers, say. In that case, I don't think this choice of A and B works, since we'd have $\overline A \setminus \overline B = \{1,2,3,4,5\}\setminus\{6,7\} = \{1,2,3,4,5\}$ and I don't think this can be equal to $\overline{ A \setminus B }$ here, since the latter set will contain, e.g., the integer 100. (Without having the sets A and B written out, that's not easy to see, however.)

2. $\overline{ A \cup B \cup C }$
This statement is asking for the complements to the elements of A and B and C

A description of what is going on would be:
$\overline{ A \cup B \cup C }$ contains elements that are not in the union A,B,C. In terms of a venn diagram, the shaded area would be the area outside of the circles since there are no elements in ${ A \cup B \cup C }$ .

Unfortunately, it appears Team B4 had addressed this choice earlier.

3. If $\{X\} \square A$, then $\{X\} \square \mathcal P(A)$.

If $\{X\} \subseteq A$, then $\{X\} \subseteq \mathcal P(A)$
If the X is a subset of A, then X is a subset of P(A). The power set, P(A) of set A is the set of all subsets of A by the definition of a power set. Therefore, if X is a subset of A, then it is therefore a subset of P(A) because all subsets of A are in the power set, P(A) by the definition.

Yes, you are correct that "all subsets of A are in the power set P(A) by the definition" — that is perfect. However, here we have to be precise about what it means to be "in" a set. Something is "in" a set if it is an element of that set. This is distinct from what it means to be a "subset" of a set. The distinction is somewhat subtle, but is made clear when you consider the precise definition of "subset". (See Definition 2.2.1 in Chapter Zero.) Hence, if {X} is a subset of A, then it is in P(A), not a subset of P(A). (Actually, it could be both, if you have something totally weird like A={ 1, { 1 } } and X={ {1} }, but it certainly doesn't have to be.)

Here's an example. A={X,Y}. Then X is in A. But P(A)={ {}, {X}, {Y}, {X,Y} } and so X is not in P(A) — sure, {X} is, but {X} and X are not the same thing…

4. $\bigcup_{n \in \mathbb Z^+} B_n$ =$\left( \frac 12 - \frac 1n, \frac 12 + \frac 1n \right)$.
for this family when n=1 the set will be $\left( \frac 12 - \frac 11, \frac 12 + \frac 11 \right)$ which is $\left( \frac 12, \frac 32\right)$ where the $( \frac12 )$ is negative. This is the maximum set for the family whereas the others are approaching
$\left( \frac 12 , \frac 12\right)$ because as n gets exponentially larger the fraction will get infinitely smaller thus having less and less of an effect on the $( \frac12)$
The union of these sets therefore are all the sets between $\left( \frac 12, \frac 32\right)$ where the $( \frac12 )$ is negative and $\left( \frac 12 , \frac 12\right)$

There's some good stuff here… And your final answer is correct! Was there some fear about the notation, though? Why the hesitation to write just $(-\frac 12, \frac 32)$ instead of this weird stuff about " $\left( \frac 12, \frac 32\right)$ where the $( \frac12 )$ is negative". There's no need to fear interval notation — it is defined precisely on page 41 of Chapter Zero any time you need to refer to it.

Also, the final statement about "The union of these sets therefore are all the sets between…" is hard to interpret. The union of all the sets gives just one set (the union) so the "are all" confused me. Then, the "all the sets between" is even stranger; how is one set between two other sets? Anyway, I think mostly the reasoning here was all right, but the presentation has problems.

Comments:


Team B6

1. For all sets $A$, $B$, and $C$, $A \setminus (B \cup C) = (A \setminus B) \cup (A \setminus C)$.

A counterexample would be the following:

Let set $A$ : {1,2,3,4}
Let set $B$ : {1,2}
Let set $C$ : {3,4}

My guess is that you want the symbol "=" in the places where you have colons. In any case, you did not show that your choice of sets is a counterexample. Is it? Well, we'll have $B \cup C = \{1,2,3,4\}$ and so $A \setminus (B \cup C) = \emptyset$. Meanwhile, $A \setminus B = \{3,4\}$ and $A \setminus C = \{3,4\}$. So it appears that in fact $(A \setminus B) \cup (A \setminus C) = \{3,4\}\cup\{3,4\}=\{3,4\} \not= \emptyset = A \setminus (B \cup C)$. So it does appear you have a counterexample, but the expectation when you present a counterexample is that you will explain how it does in fact show the theorem false!

2. $\overline A \cup \overline B \cup \overline C$ This set contains everything not in A, combined with everything not in B, combined with everything not in C. So basically, if an element is not in A, B, or C then it is in this set.

This perfectly illustrates my advice from Team B3's solution! Your "combined with" is pretty ambiguous, but when you put it in terms of "an element", you are able to state it very precisely.

3. If $X \square A$, then $\{X\} \square \mathcal P(A)$. If x is a subset of A then {x} is in P(A)

The problem said to use the symbols $\subseteq$ and $\in$. You think you're too good for those symbols, is that what's going on here?

4. For each $\bigcap_{n \in \mathbb Z^+}$$B_n = \left( \frac 12 - \frac 1n, \frac 12 + \frac 1n \right)$.

This can be read as the intersection of Bn. The set is unique because it's a collection of numbers which appear in every interval between -1/2 and 3/2 (not inclusive). How did I come to these end values? Since the n is only involved in fractions in the set, as n gets bigger the numbers get smaller, ultimately shrinking the range size. With that being said when n = 1 then our values are at their largest and therefore we can establish the largest and smallest end value. Since this set calculates values by adding or subtracting fractions starting from 1/2, the ONLY number in this intersection would have to be 1/2.

Yes: $\bigcap_{n \in \mathbb Z^+} B_n = \{\frac 12 \}$ I think some of your wording (e.g., "involved in fractions" and "range size") could be made more precise, but ultimately this is quite good.

Comments:


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