Topology

The study of topological spaces.

Answers to Burt Mendelson's text - "Introduction to topology - third edition"

I was reading this book and thought it would be a good idea to provide answers to it. I added a few questions myself. These were either based on proofs shown in the text, proofs left to the reader, or proofs I used myself to clarify things.

My proof style will probably change as you read through the text. This is inevitable because I should have became more advanced in skill as I practiced more and progressed through the text. And in anycase, my proof style tends to change with my mood.

The nature of the questions are that it is generally clear to the proponent if he is correct or not, since the arguments which are the substance of a proof speak for themselves, in which case the need for a set of answers seems lacking. Nethertheless, perhaps the answers can serve a struggling student who needs extra clarification (and instead will end up with their mind boggled by my convoluted proofs), or perhaps the student desires a different perspective. In any case, reading and checking proofs is always a good exercise.

If you spot that I have made a mistake, please let me know and I will endeavor to correct it. In the case of you spotting a typo, your name will be listed in an appended section entitled "Typo Spotters". In the case of you spotting an error, I expect you to provide an alternative, correct proof. I will then attempt to excise my stupidity with my own alternative correct proof. Both your proof and my alternative proof will occur in the text, my proof will follow yours, and your name will occur in the text at the point of your proof (unless you wish to remain anonymous) accompanied by an acknowledgement that I made a mistake (this should incite me to double check my proofs the first time round).

Enjoy yourself doing the questions.

Questions

Theory of Sets

Indexed families of sets

Let be an indexed family of subsets of a set . Prove that
1. $C(\cup_{\alpha \in I} A_\alpha) = \cap_{\alpha \in I} C(A_\alpha)$
2. $C(\cap_{\alpha \in I} A_\alpha) = \cup_{\alpha \in I} C(A_\alpha)$
Let , be two indexed families of subsets of a set . Prove the following:
1. For each $\beta \in I$ , $A_\beta \subset \cup_{\alpha \in I} A_\alpha$
2. For each $\beta \in I$ , $\cap_{\alpha \in I} A_\alpha \subset A_\beta$
3. $\cup_{\alpha \in I}(A_\alpha \cup B_\alpha) = (\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in I} B_\alpha)$
4. $\cap_{\alpha \in I}(A_\alpha \cap B_\alpha) = (\cap_{\alpha \in I} A_\alpha) \cap (\cap_{\alpha \in I} B_\alpha)$
If for each , then
1. $\cup_{\alpha \in I} A_\alpha \subset \cup_{\alpha \in I} B_\alpha$ ,
2. $\cap_{\alpha \in I} A_\alpha \subset \cap_{\alpha \in I} B_\alpha$
Left . Then
1. $\cup_{\alpha \in I}(A_\alpha \cap D) = (\cup_{\alpha \in I} A_\alpha) \cap D$ ,
2. $\cap_{\alpha \in I}(A_\alpha \cup D) = (\cap_{\alpha \in I} A_\alpha) \cup D$
Let , , . Then
1. $A \cap (B \cap D) = (A \cap B) \cup (A \cap D)$ ,
2. $A \cup (B \cup D) = (A \cup B) \cap (A \cup D)$ ,
Let be an indexed family of subsets of a set . Let . Prove that
1. $\cap_{\alpha \in J} A_\alpha \supset \cap_{\alpha \in I} A_\alpha$ .
2. $\cup_{\alpha \in J} A_\alpha \subset \cup_{\alpha \in I} A_\alpha$ .
Let be an indexed family of subsets of a set . Let . Prove that
1. $B \subset \cap_{\alpha \in I} A_\alpha$ if and only if for each $\beta \in I$ , $B \subset A_\beta$ .
2. $\cup_{\alpha \in I} A_\alpha \subset B$ if and only if for each $\beta \in I$ , $A_\beta \subset B$ .
Let be the set of real numbers that are greater than 0. For each , let be the open interval . Prove that
1. $\cap_{x \in I} A_x = \O$ .
2. $\cup_{x \in I} A_x = I$ .
For each , let be the closed interval . Prove that
1. $\cap_{x \in I} B_x = \{0\}$ .
2. $\cup_{x \in I} B_x = I \cup \{0\}$ .

Proof reasoning

Explain why it is necessary to prove both ``if $B$ then $A$ " and ``if $A$ then $B$ " in order to prove `` $A$ if and only if $B$ ".

Sets and subsets

Prove that for any set $A$ , $\O \subset A$
Prove that $\O \neq \{\O\}$
Prove that for any set $A$ , $A \subset A$ .
Determine whether each of the following statements is true or false:
1. For each set $A$ , $A \in 2^A$ .
2. For each set non-empty set $A$ , $A \subset 2^A$ .
3. For each set $A$ , $\{A\} \subset 2^A$ .
4. For each set $A$ , $\O \in 2^A$ .
5. For each set $A$ , $\O \subset 2^A$ .
6. There are no members of the set $\{\O\}$ .
7. Let $A$ and $B$ be sets. If $A \subset B$ , then $2^A \subset 2^B$ .
8. There are two distinct objects that belong to the set $\{\O,\{\O\}\}.$
Let $A$ , $B$ , $C$ be sets. Prove that if $A \subset B$ and $B \subset C$ then $A \subset C$ .
Let $A_1,...,A_n$ be sets. Prove that if $A_1 \subset A_2, A_2 \subset A_3, ..., A_{n-1} \subset A_n$ and $A_n \subset A_1$ , then $A_1 = A_2 = ... = A_n$ .

Set operations: Union, Intersection, and Compliment

Prove that $C(A \cup B) = C(A) \cap C(B)$
Prove that $C(A \cap B) = C(A) \cup C(B)$
Let , . Prove the following:
1. $A \subset B$ if and only if $A \cup B = B$ .
2. $A \subset B$ if and only if $A \cap B = A$ .
3. $A \subset C(B)$ if and only if $A \cap B = \O$ .
4. $C(A) \subset B$ if and only if $A \cup B = S$ .
5. $A \subset B$ if and only if $C(B) \subset C(A)$ .
6. $A \subset C(B)$ if and only if $B \subset C(A)$ .
Let . Prove the following:
1. $C_Y(X) \subset C_Z(X)$ .
2. $Z - (Y - X) = X \cup (Z - Y)$

Questions with answers

Theory of Sets

Proof reasoning

Explain why it is necessary to prove both ``if $B$ then $A$ " and ``if $A$ then $B$ " in order to prove `` $A$ if and only if $B$ ".

In order to prove `` $A$ if and only if $B$ " (aka `` $A$ iff $B$ ") one must prove both ``if $B$ then $A$ " and ``if $A$ then $B$ ". But why both? The necessity of ``if $B$ then $A$ " is obvious from the statement of the question and corresponds to the ``if" part. The ``only if" part refers to the notion that $A$ is true only when $B$ is true, i.e that $A$ is not true when $B$ is false, this is proven by showing that whenever $A$ is true, $B$ is also true and hence that $B$ is never false when $A$ is true, hence the expression ``only if".

Sets and subsets

Prove that for any set $A$ , $\O \subset A$

Assume $\O \not \subset A$ then $\exists_{x \in \O} : x \not \in A$ , but this is impossible because $\O$ is empty, hence our assumption is false and it is true that $\O \subset A$ .
Prove that $\O \neq \{\O\}$
Assume $\O = \{\O\}$ then $\O \subset \{\O\}$ and $\{\O\} \subset \O$ . Now $\O \in \{\O\}$ but $\O \not \in \O$ because $\O$ contains nothing, therefore $\{\O\} \not \subset \O$ so the assumption is false and it is true that $\O \neq \{\O\}$ .
Prove that for any set $A$ , $A \subset A$ .
Clearly $x \in A \implies x \in A$ so that $A \subset A$ . $A$ is called an improper subset of $A$ .
Determine whether each of the following statements is true or false:
1. For each set $A$ , $A \in 2^A$ .
  True. By definition $2^A$ is the power set of $A$ and contains all subsets of $A$ . Since $A \subset A$ it follows that $A \in 2^A$ .
2. For each non-empty set $A$ , $A \subset 2^A$ .
  %False. For $A \subset 2^A$ to be true it must hold that if $x \in A$ then $x \in 2^A$ . But if $A$ is any non-empty set it contains elements and an element of $A$ is not a subset of $A$ , it is therefore not in $2^A$ and hence $A \not \subset 2^A$ .
  
  False. Assume $A \subset 2^A$ , then $\forall_{x \in A} (x \in A \implies x \in 2^A)$ . Consider any such $x$ , $x \in A$ but $x \not \subset A$ . Therefore the assumption, namely that $A \subset 2^A$ , is false.
3. For each set $A$ , $\{A\} \subset 2^A$ .
  True. $\{A\}$ contains only the element $A$ and $A \subset A$ so $A \in 2^A$ . Everything in $\{A\}$ is thus in $2^A$ so it follows from the definition of subset that $\{A\} \subset 2^A$ .
4. For each set $A$ , $\O \in 2^A$ .
  True. It was proven above that for any set $A$ , $\O \subset A$ . $2^A$ contains all subsets of $A$ by definition, $\O \subset A$ , therefore $\O \in 2^A$ .
  
  %Since $2^A$ contains all subsets of $A$ it clearly contains $\O$ .
5. For each set $A$ , $\O \subset 2^A$ .
  True. $\O$ is a subset of any set $A$ therefore $\O \subset 2^A$ irrespective of $A$ .
6. There are no members of the set $\{\O\}$ .
  False. The set $\{\O\}$ contains the element $\O$ .
7. Let $A$ and $B$ be sets. If $A \subset B$ , then $2^A \subset 2^B$ .
  %True. Assume that it is not true that if $A \subset B$ , then $2^A \subset 2^B$ . It follows that there exists a subset $X$ of $A$ which is not a subset of $B$ , but since any element in $A$ is also in $B$ , this is impossible. By the same construction steps that we apply to the elements of $A$ to construct $X$ we can construct $X$ from the elements of $A$ as they occur in $B$ . Therefore the assumption is false and it is true that if $A \subset B$ , then $2^A \subset 2^B$ .
  
  True. Let $A \subset B$ and assume $2^A \not \subset 2^B$ . By the assumption, $\exists_{X \in 2^A} : X \not \in 2^B$ . Consider this $X$ , by definition of $2^A$ , $X \subset A$ and therefore $\forall_{x \in X} (x \in X \implies x \in A)$ but $A \subset B$ which means $\forall_{x \in A} (x \in A \implies x \in B)$ , therefore $\forall_{x \in X} (x \in X \implies x \in A \implies x \in B)$ which by definition means that $X \subset B$ .
  
  Now if $X \subset B$ then by definition of $2^B$ , $X \in 2^B$ but this is a contradiction since it was claimed that $X \not \in 2^B$ . Therefore the assumption, namely that $2^A \not \subset 2^B$ given $A \subset B$ , is false, and it is therefore true that if $A \subset B$ , then $2^A \subset 2^B$ .
8. There are two distinct objects that belong to the set $\{\O,\{\O\}\}$ .
  True since $\O$ and $\{\O\}$ are distinct objects and they belong to the set.
Let $A$ , $B$ , $C$ be sets. Prove that if $A \subset B$ and $B \subset C$ then $A \subset C$ .
If $B \subset C$ , then $\forall_{x \in B} (x \in C)$ , and if $A \subset B$ , then $\forall_{x \in A} (x \in B)$ . It follows from the former, given the latter, that $\forall_{x \in A} (x \in C)$ , and hence $A \subset C$ .

Alternative proof: If $A \subset B$ then $x \in A \implies x \in B$ , and if $B \subset C$ then $x \in B \implies x \in C$ . Thus $x \in A \implies x \in B \implies x \in C$ and hence $x \in A \implies x \in C$ thus $A \subset C$ .

Alternative proof (by contradiction): Assume that $A \subset B$ and $B \subset C$ but $A \not \subset C$ , this means that $\exists_{x \in A} : x \not \in C$ . Taking this $x$ since $A \subset B$ , $x \in B$ , and since $B \subset C$ , $x \in C$ which is a contradiction. Therefore the assumption is false, namely that $A \not \subset C$ , and hence it is true that if $A \subset B$ and $B \subset C$ , then $A \subset C$ .
Let $A_1,...,A_n$ be sets. Prove that if $A_1 \subset A_2, A_2 \subset A_3, ..., A_{n-1} \subset A_n$ and $A_n \subset A_1$ , then $A_1 = A_2 = ... = A_n$ .
Proof by induction on $n$ .

The base case, $n=1$ is trivially true since $A_1 = A_1$ . %For the base case $n=1$ , $A_1 \subset A_2$ and $A_2 \subset A_1$ , therefore by definition of set equality $A_1 = A_2$ .

Now it is required to show that for any $n \geq 1$ , if the claim holds for $n$ then it holds for $n+1$ . For $n+1$ we have

$A_1 \subset A_2 \subset ... \subset A_n \subset A_{n+1}$

and $A_{n+1} \subset A_1$ . Now $A_1 \subset A_2 \subset .. \subset A_n$ and $A_n \subset A_{n+1} \subset A_1$ which means that $A_n \subset A_1$ , therefore $A_1 \subset A_2 \subset ... \subset A_n$ and $A_n \subset A_1$ both hold. Given that the claim holds for $n$ it follows that

$A_1 = A_2 = ... = A_n$

and therefore because $A_1 = A_n$ that $A_1 \subset A_{n+1}$ because $A_n \subset A_{n+1}$ . Given that it is also true that $A_{n+1} \subset A_1$ , this means that $A_{n+1} = A_1$ and since $A_1 = A_2 = ... = A_n$ it follows that

$A_1 = A_2 = ... = A_{n+1}$

and the induction step holds.

Set operations: Union, Intersection, and Compliment

Prove that $C(A \cup B) = C(A) \cap C(B)$

If $x \in C(A \cup B)$ then $x \not \in A \cup B$ so $x \not \in A$ and $x \not \in B$ , which means that $x \in C(A)$ and $x \in C(B)$ so that $x \in C(A) \cap C(B)$ , therefore

$C(A \cup B) \subset C(A) \cap C(B)$

If $x \in C(A) \cap C(B)$ then $x \not \in A$ and $x \not \in B$ so $x \not \in A \cup B$ , hence $x \in C(A \cup B)$ , therefore

$C(A) \cap C(B) \subset C(A \cup B)$

it follows that

$C(A \cup B) = C(A) \cap C(B)$
Prove that $C(A \cap B) = C(A) \cup C(B)$
It is easy to prove this from the previous proof (as noted by Bert himself), namely that $C(A \cup B) = C(A) \cap C(B)$ , taking the complements of $A$ and $B$

$C(C(A) \cup C(B)) = C(C(A)) \cap C(C(B)) = A \cap B$

taking complements of the first and last terms

$C(C(C(A) \cup C(B))) = C(A \cap B)$

so that

$C(A) \cup C(B) = C(A \cap B)$

Alternatively, it can be proved from scratch:

If $x \in C(A \cap B)$ then $x \not \in A \cap B$ , so at least one of $x \in C(A)$ or $x \in C(B)$ must be true, thus $x \in C(A) \cup C(B)$ and

$C(A \cap B) \subset C(A) \cup C(B)$

If $x \in C(A) \cup C(B)$ then at least one of $x \in C(A)$ or $x \in C(B)$ must be true, therefore it is never true that $x \in A \cap B$ and hence it is always true that $x \in C(A \cap B)$ and

$C(A) \cup C(B) \subset C(A \cap B)$

so that

$C(A) \cup C(B) = C(A \cap B)$

Using the same technique as before, this relation can be used to prove that $C(A \cup B) = C(A) \cap C(B)$ . Taking the complements of $A$ and $B$ in $C(A \cap B) = C(A) \cup C(B)$ gives

$C(C(A) \cap C(B)) = C(C(A)) \cup C(C(B)) = A \cup B$

and taking complements of the first and last terms gives

$C(C(C(A) \cap C(B))) = C(A \cup B)$

and it follows that $C(A) \cap C(B) = C(A \cup B)$ .
Let , . Prove the following:
1. $A \subset B$ if and only if $A \cup B = B$ .
  First, the if part. Let $A \cup B = B$ and assume $A \not \subset B$ . From the assumption it follows that $\exists_{x \in A} : x \not \in B$ and consequently $\exists_{x \in A \cup B} : x \not \in B$ . This means $A \cup B \not \subset B$ and it follows that $A \cup B \neq B$ which is a contradiction. Thus the assumption is false and in fact $A \subset B$ .
  
  Second, the only if part. Let $A \subset B$ and assume $A \cup B \neq B$ . From the assumption, since $B \subset A \cup B$ , to prevent equality it follows that $A \cup B \not \subset B$ which means $\exists_{x \in A \cup B} : x \not \in B$ . Consider such an $x$ , $x \not \in B$ but $x \in A \cup B$ , therefore $x \in A$ and $\exists_{x \in A} : x \not \in B$ which means $A \not \subst B$ which is a contradiction. Thus the assumption is false and in fact $A \cup B = B$ .
2. $A \subset B$ if and only if $A \cap B = A$ .
  Let $A \cap B = A$ and assume $A \not \subset B$ . From the assumption it follows that $\exists_{x \in A} : x \not \in B$ , but this means that $\exists_{x \in A} : x \not \in A \cap B$ which means $A \cap B \neq A$ which is a contradiction. Hence the assumption is false and in fact $A \subset B$ .
  
  If $A \subset B$ , then $A \cap B = A$ . Let $A \subset B$ and assume that $A \cap B \neq A$ . From the assumption it follows, since $A \cap B \subset A$ , that $A \not \subset A \cap B$ . Thus $\exists_{x \in A} : x \not \in A \cap B$ , and since $x \in A$ , $\exists_{x \in A} : x \not \in B$ . Hence $A \not \subset B$ which is a contradiction. Therefore the assumption is false and in fact $A \cap B = A$ .
3. $A \subset C(B)$ if and only if $A \cap B = \O$ .
  If $A \cap B = \O$ then $\not\!\exists_{x \in A} : x \in B$ therefore $\forall_{x \in A} (x \not \in B)$ and $A \subset C(B)$ .
  
  If $A \subset C(B)$ then $\forall_{x \in A}(x \not \in B)$ and therefore $A \cap B = \O$ .
4. $C(A) \subset B$ if and only if $A \cup B = S$ ( $S$ is the universe).
  If $A \cup B = S$ then $x \not \in A \implies x \in B$ and therefore $C(A) \subset B$ .
  
  If $C(A) \subset B$ then $x \not \in A \implies x \in B$ so for any $x \in S$ either $x \in A$ or $x \in B$ , or in other words $A \cup B = S$ .
  
  Alternative proof:
  
  Let $A \cup B = S$ , and assume $x \in C(A)$ , then $x \in (S - A = B)$ , and so $C(A) \subset B$ .
  
  Let $C(A) \subset B$ , and assume $A \cup B \neq S$ . $S$ is the universe so $A \cup B \subset S$ and thus it follows from the assumption that $S \not \subset A \cup B$ , therefore $\exists_{x \in S} : x \not \in A \cup B$ , taking this $x$ , since $x \not \in A$ , by $C(A) \subset B$ , it follows that $x \in B$ , but this is a contradiction since $x \not \in A \cup B$ . Therefore the assumption is false and in fact $A \cup B = S$ .
5. $A \subset B$ if and only if $C(B) \subset C(A)$ .
  Let $C(B) \subset C(A)$ and assume that $A \not \subset B$ then $\exists_{x \in A} : x \not \in B$ , but this means that $\exists_{x \not \in B} : x \in A$ , or in other words $\exists_{x \in C(B)} : x \not \in C(A)$ and thus $C(B) \not \subset C(A)$ which is a contradiction, therefore the assumption is false and in fact $A \subset B$ .
  
  Let $A \subset B$ and assume $C(B) \not \subset C(A)$ . From the assumption, $\exists_{x \in C(B)} : x \not \in C(A)$ and thus $\exists_{x \in C(B)} : x \in A$ , or in other words $\exists_{x \in A} : x \not \in B$ and hence $A \not \subset B$ which is a contradiction. Hence the assumption is false and in fact $C(B) \subset C(A)$ .
  
  Alternative proof:
  
  Let $C(B) \subset C(A)$ . Assume $A \not \subset B$ , then $\exists_{x \in A} : x \not \in B$ , consider this $x$ , since $x \not \in B$ , by $C(B) \subset C(A)$ , $x \not \in A$ , which is a contradiction. Therefore, the assumption is false and in fact $A \subset B$ .
  
  Let $A \subset B$ . Assume $C(B) \not \subset C(A)$ then $\exists_{x \not \in B} : x \in A$ , consider this $x$ , since $x \in A$ , by $A \subset B$ , $x \in B$ , which is a contradiction. Therefore, the assumption is false and in fact $C(B) \subset C(A)$ .
6. $A \subset C(B)$ if and only if $B \subset C(A)$ .
  Let $B \subset C(A)$ . Assume $A \not \subset C(B)$ . From the assumption $\exists_{x \in A} : x \not \in C(B)$ or in other words $\exists_{x \not \in C(B)} : x \in A$ which in other words says $\exists_{x \in B} : x \not \in C(A)$ and hence $B \not \subset C(A)$ which is a contradiction. Therefore the assumption is false and in fact $A \subset C(B)$ .
  
  Let $A \subset C(B)$ . Assume $B \not \subset C(A)$ . From the assumption $\exists_{x \in B} : x \not \in C(A)$ or in other words $\exists_{x \not \in C(B)} : x \in A$ which in other words says $\exists_{x \in A} : x \not \in C(B)$ and hence $A \not \subset C(B)$ which is a contradiction. Therefore the assumption is false and in fact $B \subset C(A)$ .
  
  Alternative proof:
  
  Let $B \subset C(A)$ . Assume $A \not \subset C(B)$ , then $\exists_{x \in A} : x \in B$ . Consider this $x$ , since $B \subset C(A)$ , $x \not \in A$ which is a contradiction. Therefore the assumption is false and in fact $A \subset C(B)$ .
  
  Let $A \subset C(B)$ . Assume $B \not \subset C(A)$ , then $\exists_{x \in B} : x \in A$ . Consider this $x$ , since $A \subset C(B)$ , $x \not \in B$ , which is a contradiction. Therefore the assumption is false and in fact $B \subset C(A)$ .
Let . Prove the following:
1. $C_Y(X) \subset C_Z(X)$ .
  If $x \in C_Y(X)$ then $x \not \in X$ and $x \in Y$ , but since $Y \subset Z$ then $x \in Z$ , and given $x \not \in X$ it follows that $x \in C_Z(X)$ and hence $C_Y(X) \subset C_Z(X)$ .
2. $Z - (Y - X) = X \cup (Z - Y)$
  I was humbled when a former colleague Michael Pfeiffer showed me a trivial proof for this, after I had written a one page proof. So here it is:
  
  Let $A = Z - (Y - X)$ and let $B = X \cup (Z - Y)$
  
  $z \in A \implies z \in Z, z \not \in Y - X$ , so $Z - (Y - X) = Z \cap C(Y-X)$ , and
  
  $Z \cap C(Y-X) = Z \cap C(Y \cap C(X))$
  
  and (the first step is by Demorgan's laws):
  
  $Z \cap C(Y \cap C(X)) = Z \cap (C(Y) \cup X) = (Z \cap C(Y)) \cup (Z \cap X)$
  
  since $Z \cap C(Y) = Z - Y$ and $Z \cap X = X$ , it follows that
  
  $Z \cap C(Y-X) = (Z - Y) \cup X$
  
  Now here is the drawn out proof I began with:
  
  Preliminaries: generally $A - B = A \cap C(B)$ and thus $(A \cup B) - C = (A \cup B) \cap C(C)$ . Generally $(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$ and thus $(A \cup B) - C = (A \cup B) \cap C(C) = (A \cap C(C)) \cup (B \cap C(C)) = (A - C) \cup (B - C)$ which means that generally
  
  \begin{equation} \label{distributivesetsubtraction} (A \cup B) - C = (A - C) \cup (B - C) \end{equation}
  
  It is required to prove that $Z - (Y - X) = X \cup (Z - Y)$ .
  
  Let $x \in Z - (Y - X)$ , then $x \in Z$ and $x \not \in (Y - X)$ . Another way to write $x \in Z$ is to write $x \in (Z - Y) \cup Y$ since $(Z - Y) \cup Y = Z$ (when $Y \subset Z$ as is the case here). Therefore $x \in (Z - Y) \cup Y$ and $x \not \in (Y - X)$ , which is alternatively written $x \in ((Z - Y) \cup Y) - (Y - X)$ . Now, since generally $(A \cup B) - C) = (A - C) \cup (B - C)$ , it follows that $x \in ((Z - Y) - (Y - X)) \cup (Y - (Y - X))$ . Now since $(Y - X) \subset Y$ it follows that $(Z - Y) - (Y - X) = Z - Y$ , and since $X \subset Y$ , $Y - (Y - X) = X$ , hence $((Z - Y) - (Y - X)) \cup (Y - (Y - X)) = Z - Y \cup X$ and hence
  
  $Z - (Y - X) \subset X \cup (Z - Y)$
  
  Now let $x \in X \cup (Z - Y)$ , then either $x \in X$ or $x \in (Z - Y)$ .
  
  First case, let $x \in X$ , then since $X \subset Y \subset Z$ , $X \subset Z$ and therefore $x \in Z$ . Now assume $x \not \in Z - (Y - X)$ , since $x \in Z$ for $x \not \in Z - (Y - X)$ to be true it must follow that $x \in (Y - X)$ , but this leads to a contradiction because if $x \in (Y - X)$ , then $x \in Y \wedge x \not \in X$ but it was stated that $x \in X$ . So the assumption is false and in fact $x \in Z - (Y - X)$ and
  
  $X \subset Z - (Y - X)$
  
  Second case, let $x \in Z - Y$ . It is true that $x \in Z$ since $Z - Y \subset Z$ . Assume that $x \not \in Z - (Y - X)$ , then since $x \in Z$ it must be true that $x \in (Y - X)$ , but $(Y - X) \subset Y$ and so $x \in Y$ , but this is a contradiction since $x \in Z - Y$ means that $x \not \in Y$ . Hence the assumption is false and in fact $x \in Z - (Y - X)$ . Thus
  
  $(Z - Y) \subset Z - (Y - X)$
  
  so putting the cases together, if $x \in X$ , then $x \in Z - (Y - X)$ , and if $x \in (Z - Y)$ then $x \in Z - (Y - X)$ , it follows that if $x \in X \cup (Z - Y)$ then $x \in Z - (Y - X)$ and hence
  
  $X \cup (Z - Y) \subset Z - (Y - X)$
  
  and so
  
  $Z - (Y - X) = X \cup (Z - Y)$

Indexed families of sets

Let be an indexed family of subsets of a set . Prove that
1. $C(\cup_{\alpha \in I} A_\alpha) = \cap_{\alpha \in I} C(A_\alpha)$
  If $x \in C(\cup_{\alpha \in I} A_\alpha)$ then $x \not \in \cup_{\alpha \in I} A_\alpha$ , which means $\not\!\exists_{\alpha \in I} : x \in A_\alpha$ , therefore $\forall_{\alpha \in I}(x \in C(A))$ and thus $x \in \cap_{\alpha \in I} C(A_\alpha)$ meaning
  
  $C(\cup_{\alpha \in I} A_\alpha) \subset \cap_{\alpha \in I} C(A_\alpha)$
  
  If $x \in \cup_{\alpha \in I} C(A_\alpha)$ then $\forall_{\alpha \in I} (x \in C(A_\alpha))$ and hence $\not\!\exists_{\alpha \in I} : x \in A_\alpha$ therefore $x \not \in \cup_{\alpha \in I} A_\alpha$ and in fact $x \in C(\cup_{\alpha \in I} A_\alpha)$ meaning
  
  $\cap_{\alpha \in I} C(A_\alpha) \subset C(\cup_{\alpha \in I} A_\alpha)$
  
  so
  
  $C(\cup_{\alpha \in I} A_\alpha) = \cap_{\alpha \in I} C(A_\alpha)$
2. $C(\cap_{\alpha \in I} A_\alpha) = \cup_{\alpha \in I} C(A_\alpha)$
  If $x \in C(\cap_{\alpha \in I} A_\alpha)$ then $x \not \in \cap_{\alpha \in I} A_\alpha$ which means that $\exists_{\alpha \in I} : x \not \in A_\alpha$ which means that $\exists_{\alpha \in I} : x \in C(A_\alpha)$ and hence $x \in \cup_{\alpha \in I} C(A_\alpha)$ , therefore
  
  $C(\cap_{\alpha \in I} A_\alpha) \subset \cup_{\alpha \in I} C(A_\alpha)$
  
  If $x \in \cup_{\alpha \in I} C(A_\alpha)$ then $\exists_{\alpha \in I} : x \in C(A_\alpha)$ therefore $\exists_{\alpha \in I} : x \not \in A_\alpha$ which means $x \not \in \cap_{\alpha \in I} A_\alpha$ , and thus $x \in C(\cap_{\alpha \in I} A_\alpha)$ and
  
  $\cup_{\alpha \in I} C(A_\alpha) \subset C(\cap_{\alpha \in I} A_\alpha)$
  
  so it follows that
  
  $C(\cap_{\alpha \in I} A_\alpha) = \cup_{\alpha \in I} C(A_\alpha)$
Let , be two indexed families of subsets of a set . Prove the following:
1. For each $\beta \in I$ , $A_\beta \subset \cup_{\alpha \in I} A_\alpha$
  Clearly if $x \in A_\beta$ then $x \in \cup_{\alpha \in I} A_\alpha$ because $\beta \in I$ .
2. For each $\beta \in I$ , $\cap_{\alpha \in I} A_\alpha \subset A_\beta$
  If $x \in \cap_{\alpha \in I} A_\alpha$ then $\forall_{\alpha \in I} (x \in A_\alpha)$ therefore $x \in A_\beta$ since $\beta \in I$ and so $\cap_{\alpha \in I} A_\alpha \subset A_\beta$ .
3. $\cup_{\alpha \in I}(A_\alpha \cup B_\alpha) = (\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in I} B_\alpha)$
  If $x \in \cup_{\alpha \in I}(A_\alpha \cup B_\alpha)$ then $\exists_{\beta \in I} : x \in A_\beta \cup B_\beta$ . If $x \in A_\beta$ then $x \in (\cup_{\alpha \in I} A_\alpha)$ , and if $x \in B_\beta$ then $x \in (\cup_{\alpha \in I} B_\alpha)$ therefore $x \in (\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in I} B_\alpha)$ . Therefore
  
  $\cup_{\alpha \in I}(A_\alpha \cup B_\alpha) \subset (\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in I} B_\alpha)$
  
  If $x \in (\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in I} B_\alpha)$ then $x \in (\cup_{\alpha \in I} A_\alpha)$ or $x \in (\cup_{\alpha \in I} B_\alpha)$ therefore $\exists_{\beta \in I} : x \in A_\beta$ or $\exists_{\beta \in I} : x \in B_\beta$ , in either case the implication is that $\exists_{\beta \in I} : x \in A_\beta \cup B_\beta$ and it follows that $x \in \cup_{\alpha \in I} (A_\alpha \cup B_\alpha)$ therefore
  
  $(\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in I} B_\alpha) \subset \cup_{\alpha \in I}(A_\alpha \cup B_\alpha)$
  
  so that
  
  $\cup_{\alpha \in I}(A_\alpha \cup B_\alpha) = (\cup_{\alpha \in I} A_\alpha) \cup (\cup_{\alpha \in I} B_\alpha)$
4. $\cap_{\alpha \in I}(A_\alpha \cap B_\alpha) = (\cap_{\alpha \in I} A_\alpha) \cap (\cap_{\alpha \in I} B_\alpha)$
  If $x \in \cap_{\alpha \in I}(A_\alpha \cap B_\alpha)$ then $\forall_{\alpha \in I} (x \in A_\alpha \cap B_\alpha)$ but this means that $x \in A_\alpha$ and $x \in B_\alpha$ for any $\alpha \in I$ and hence $(x \in \cap_{\alpha \in I} A_\alpha)$ and $(x \in \cap_{\alpha \in I} B_\alpha)$ , therefore $x \in (\cap_{\alpha \in I} A_\alpha) \cap (\cap_{\alpha \in I} B_\alpha)$ and
  
  $\cap_{\alpha \in I}(A_\alpha \cap B_\alpha) \subset (\cap_{\alpha \in I} A_\alpha) \cap (\cap_{\alpha \in I} B_\alpha)$
  
  If $x \in (\cap_{\alpha \in I} A_\alpha) \cap (\cap_{\alpha \in I} B_\alpha)$ then $x \in (\cap_{\alpha \in I} A_\alpha)$ and $x \in (\cap_{\alpha \in I} B_\alpha)$ but this means that $\forall_{\alpha \in I} (x \in A_\alpha)$ and $\forall_{\alpha \in I} (x \in B_\alpha)$ , therefore for any $\alpha \in I$ , $x \in A_\alpha$ and $x \in B_\alpha$ so that $x \in A_\alpha \cap B_\alpha$ and thus $\forall_{\alpha \in I} (x \in A_\alpha \cap B_\alpha)$ , hence $x \in \cap_{\alpha \in I}(A_\alpha \cap B_\alpha)$ so that
  
  $(\cap_{\alpha \in I} A_\alpha) \cap (\cap_{\alpha \in I} B_\alpha) \subset \cap_{\alpha \in I}(A_\alpha \cap B_\alpha)$
  
  it follows then that
  
  $\cap_{\alpha \in I}(A_\alpha \cap B_\alpha) = (\cap_{\alpha \in I} A_\alpha) \cap (\cap_{\alpha \in I} B_\alpha)$
If for each , then
1. $\cup_{\alpha \in I} A_\alpha \subset \cup_{\alpha \in I} B_\alpha$ ,
  If $x \in \cup_{\alpha \in I} A_\alpha$ then $\exists_{\alpha \in I} : x \in A_\alpha$ and since for each $\alpha \in I$ , $A_\alpha \subset B_\alpha$ it follows that $\exists_{\alpha \in I} : x \in B_\alpha$ and in fact $x \in \cup_{\alpha \in I} B_\alpha$ .
2. $\cap_{\alpha \in I} A_\alpha \subset \cap_{\alpha \in I} B_\alpha$
  If $x \in \cap_{\alpha \in I} A_\alpha$ then $\forall_{\alpha \in I} (x \in A_\alpha)$ and since $\forall_{\alpha \in I} (A_\alpha \subset B_\alpha)$ this means $\forall_{\alpha \in I}(x \in B_\alpha)$ and thus $x \in \cap_{\alpha \in I} B_\alpha$ .
Let . Then
1. $\cup_{\alpha \in I}(A_\alpha \cap D) = (\cup_{\alpha \in I} A_\alpha) \cap D$
  Direct proof:
  
  Let $x \in \cup_{\alpha \in I} (A_\alpha \cup D)$ , then $\exists_{\alpha \in I} : x \in A_\alpha \cup D$ , now since $A_\alpha \subset \cup_{\alpha \in I} A_\alpha$ $x \in (\cup_{\alpha \in I} A_\alpha) \cap D$ and
  
  $\cup_{\alpha \in I}(A_\alpha \cap D) \subset (\cup_{\alpha \in I} A_\alpha) \cap D$
  
  Let $x \in (\cup_{\alpha \in I} A_\alpha) \cap D$ , then $x \in D$ and $x \in \cup_{\alpha \in I} A_\alpha$ . From the latter it follows that $\exists_{\alpha \in I} : x \in A_\alpha$ and thus, given $x \in D$ , $\exists_{\alpha \in I} : x \in A_\alpha \cap D$ from which it follows that $x \in \cup_{\alpha \in I} (A_\alpha \cap D)$ and
  
  $(\cup_{\alpha \in I} A_\alpha) \cap D \subset \cup_{\alpha \in I}(A_\alpha \cap D)$
  
  leading to the conclusion
  
  $\cup_{\alpha \in I}(A_\alpha \cap D) = (\cup_{\alpha \in I} A_\alpha) \cap D$
  
  Alternative proof, proof by contradiction:
  
  Let $x \in \cup_{\alpha \in I}(A_\alpha \cap D)$ , it follows that $x \in D$ and $\exists_{\alpha \in I} : x \in A_\alpha$ . Assume that $x \not \in (\cup_{\alpha \in I} A_\alpha) \cap D$ , then either $x \not \in D$ or $x \not \in (\cup_{\alpha \in I} A_\alpha)$ . $x \in D$ so the former is impossible, thus the latter must be true which means that $\not\!\exists_{\alpha \in I} : x \in A_\alpha$ but this is a contradiction. Therefore the assumption is false and in fact $x \in (\cup_{\alpha \in I} A_\alpha) \cap D$ .
  
  Let $x \in (\cup_{\alpha \in I} A_\alpha) \cap D$ . Then $x \in D$ and $\exists_{\alpha \in I} : x \in A_\alpha$ . Assume that $x \not \in \cup_{\alpha \in I}(A_\alpha \cap D)$ then $\not\!\exists_{\alpha \in I} : x \in A_\alpha \cap D$ . Since $x \in D$ , it follows that $\not\!\exists_{\alpha \in I} : x \in A_\alpha$ , which is a constradiction. Therefore the assumption is false and in fact $x \in \cup_{\alpha \in I}(A_\alpha \cap D)$
2. $\cap_{\alpha \in I}(A_\alpha \cup D) = (\cap_{\alpha \in I} A_\alpha) \cup D$
  Let $x \in \cap_{\alpha \in I}(A_\alpha \cup D)$ . Assume $x \not \in (\cap_{\alpha \in I} A_\alpha) \cup D$ then $x \not \in D$ and $x \not \in (\cap_{\alpha \in I} A_\alpha)$ , the latter means $\exists_{\alpha \in I} : x \not \in A_\alpha$ , and it follows from $x \not \in D$ that $\exists_{\alpha \in I} : x \not \in A_\alpha \cup D$ , therefore $x \not \in \cap_{\alpha \in I}(A_\alpha \cup D)$ but this is a contradiction. Therefore it is not true that $x \not \in (\cap_{\alpha \in I} A_\alpha) \cup D$ and it follows that
  
  $\cap_{\alpha \in I}(A_\alpha \cup D) \subset (\cap_{\alpha \in I} A_\alpha) \cup D$
  
  Let $x \in (\cap_{\alpha \in I} A_\alpha) \cup D$ . Assume $x \not \in \cap_{\alpha \in I}(A_\alpha \cup D)$ , then $\exists_{y \in I} : x \not \in A_y \cup D$ , considering this $y$ , $x \not \in A_y \cup D$ which means that $x \not \in A_y$ and $x \not \in D$ . Since $x \not \in D$ , it follows from $x \in (\cap_{\alpha \in I} A_\alpha) \cup D$ that $x \in (\cap_{\alpha \in I} A_\alpha)$ but $x \not \in A_y$ so this is a contradiction. Therefore the assumption is false and in fact $x \in \cap_{\alpha \in I}(A_\alpha \cup D)$ so that
  
  $(\cap_{\alpha \in I} A_\alpha) \cup D \subset \cap_{\alpha \in I}(A_\alpha \cup D)$
  
  leading to the conclusion that
  
  $\cap_{\alpha \in I}(A_\alpha \cup D) = (\cap_{\alpha \in I} A_\alpha) \cup D$
Let , , . Then
1. $A \cap (B \cup D) = (A \cap B) \cup (A \cap D)$ ,
  If $x \in A \cap (B \cup D)$ then $x \in A$ and $x \in (B \cup D)$ , therefore $x \in A$ and $x \in B$ or $x \in D$ , if $x \in B$ then $x \in (A \cap B)$ hence $x \in (A \cap B) \cup (A \cap D)$ , and if $x \in D$ then $x \in (A \cap D)$ and $x \in (A \cap B) \cup (A \cap D)$ so in either case $x \in (A \cap B) \cup (A \cap D)$ so
  
  $A \cap (B \cup D) \subset (A \cap B) \cup (A \cap D)$
  
  If $x \in (A \cap B) \cup (A cap D)$ then either $x \in A \cap B$ or $x \in A \cap D$ . If $x \in A \cap B$ then $x \in A$ and $x \in B$ , from the latter it follows that $x \in B \cup D$ and hence $x \in A \cap (B \cup D)$ . And if $x \in A \cap D$ then $x \in A$ and $x \in D$ , and from the latter $x \in (B \cup D)$ and thus $x \in A \cap (B \cup D)$ . So
  
  $(A \cap B) \cup (A \cap D) \subset A \cap (B \cup D)$
  
  and
  
  $A \cap (B \cup D) = (A \cap B) \cup (A \cap D)$
2. $A \cup (B \cap D) = (A \cup B) \cap (A \cup D)$
  If $x \in A \cup (B \cap D)$ then $x \in A$ or $x \in (B \cap D)$ . If $x \in A$ then $x \in (A \cup B)$ and $x \in (A \cup D)$ so $x \in (A \cup B) \cap (A \cup D)$ , if $x \in (B \cap D)$ then $x \in B$ so $x \in (A \cup B)$ and $x \in D$ so $x \in (A \cup D)$ , therefore $x \in (A \cup B) \cap (A \cup D)$ and
  
  $A \cup (B \cap D) \subset (A \cup B) \cap (A \cup D)$
  
  If $x \in (A \cup B) \cap (A \cup D)$ then $x \in (A \cup B)$ and $x \in (A \cup D)$ , it is clear that if $x \in A$ then $x \in (A \cup B)$ and $(A \cup D)$ so that $x \in A$ satisfies the relation. If $x \not \in A$ then $x$ must be in $B$ in the first clause and $x$ must be in $D$ in the second clause so that $x$ is in both clauses. Therefore either $x \in A$ or $x \in (B \cap D)$ hence $x \in A \cup (B \cap D)$ and
  
  Let $x \in (A \cup B) \cap (A \cup D)$ and assume $x \not \in A \cup (B \cap D)$ . It follows from the assumption that $x \not \in A$ and $x \not \in (B \cap D)$ . $x \not \in A$ , yet from the premise $x \in (A \cup B)$ , therefore $x \in B$ , similarly $x \in (A \cup D)$ and hence $x \in D$ . Therefore $x \in B$ and $x \in D$ and thus $x \in B \cap D$ which is a contradiction. Therefore the assumption is false and in fact $x \in A \cup (B \cap D)$ so
  
  $(A \cup B) \cap (A \cup D) \subset A \cup (B \cap D)$
  
  and
  
  $A \cup (B \cap D) = (A \cup B) \cap (A \cup D)$
Let be an indexed family of subsets of a set . Let . Prove that
1. $\cap_{\alpha \in J} A_\alpha \supset \cap_{\alpha \in I} A_\alpha$ .
  If $x \in \cap_{\alpha \in I} A_\alpha$ then $\forall_{\alpha \in I}(x \in A_\alpha)$ and since $J \subset I$ , $\forall_{\alpha \in J}(x \in A_\alpha)$ , hence $x \in \cap_{\alpha \in J} A_\alpha$ . Therefore $\cap_{\alpha \in I} A_\alpha \subset \cap_{\alpha \in J} A_\alpha$ , or in other words $\cap_{\alpha \in J} A_\alpha \supset \cap_{\alpha \in I} A_\alpha$ .
2. $\cup_{\alpha \in J} A_\alpha \subset \cup_{\alpha \in I} A_\alpha$ .
  Let $x \in \cup_{\alpha \in I} A_\alpha$ , then $\exists_{\alpha \in J} : x \in A_\alpha$ but since $J \subset I$ , $\exists_{\alpha \in I} : x \in A_\alpha$ , hence $x \in \cup_{\alpha \in I} A_\alpha$ and $\cup_{\alpha \in J} A_\alpha \subset \cup_{\alpha \in I} A_\alpha$ .
Let be an indexed family of subsets of a set . Let . Prove that
1. $B \subset \cap_{\alpha \in I} A_\alpha$ if and only if for each $\beta \in I$ , $B \subset A_\beta$ .
  Let $B \subset A_\beta, \forall_{\beta \in I}$ and assume $B \not \subset \cap_{\alpha \in I} A_\alpha$ . From the assumption it follows that $\exists_{x \in B} : x \not \in \cap_{\alpha \in I} A_\alpha$ , consider this $x$ , since $x \not \in \cap_{\alpha \in I} A_\alpha$ , $\exists_{\alpha \in I} : x \not \in A_\alpha$ , thus considering this $\alpha$ it follows that $x \in B$ and $x \not \in A_\alpha$ which means $B \not \subset A_\alpha$ which contradicts the premise. Therefore the assumption is false and
  
  $B \subset A_\beta, \forall_{\beta \in I} \implies B \subset \cap_{\alpha \in I} A_\alpha$
  
  Now let $B \subset \cap_{\alpha \in I} A_\alpha$ and assume $\forall_{\beta \in I},B \subset A_\beta$ is not true. From the assumption it follows that $\exists_{\beta \in I} : B \not \subset A_\beta$ , considering this $\beta$ , $\exists_{x \in B} : x \not \in A_\beta$ , and hence $\exists_{x \in B} : x \not \in \cap_{\alpha \in I} A_\alpha$ which means $B \not \subset \cap_{\alpha \in I} A_\alpha$ but this contradicts the premise, hence the assumption is false and
  
  $B \subset \cap_{\alpha \in I} A_\alpha \implies B \subset A_\beta, \forall_{\beta \in I}$
  
  and therefore
  
  $B \subset \cap_{\alpha \in I} A_\alpha \Longleftrightarrow B \subset A_\beta, \forall_{\beta \in I}$
2. $\cup_{\alpha \in I} A_\alpha \subset B$ if and only if for each $\beta \in I$ , $A_\beta \subset B$ .
  Let $A_\beta \subset B$ , $\forall_{\beta \in I}$ and assume $\cup_{\alpha \in I} A_\alpha \not \subset B$ . From the assumption it follows that $\exists_{x \in U_{\alpha \in I} A_\alpha} : x \not \in B$ . Consider this $x$ , since $x \in \cup_{\alpha \in I} A_\alpha$ , it follows that $\exists_{\alpha \in I} : x \in A_\alpha$ , yet $A_\beta \subset B, \forall_{\beta \in A}$ and hence $x \in B$ , but this is a contradiction. Therefore the assumption is false and in fact
  
  $A_\beta \subset B, \forall_{\beta \in I} \implies \cup_{\alpha \in I} A_\alpha \subset B$
  
  Now let $\cup_{\alpha \in I} A_\alpha \subset B$ and assume that $\forall_{\beta \in I}, A_\beta \subset B$ is false. From the assumption it follows that $\exists_{\beta \in I} : A_\beta \not \subset B$ . Considering this $\beta$ it follows that $\exists_{x \in A_\beta} : x \not \in B$ , but this means $\exists_{x \in \cup_{\alpha \in I} A_\alpha} : x \not \in B$ and hence $\cup_{\alpha \in I} A_\alpha \not \subset B$ which contradicts the premise. Therefore the assumption is false and in fact
  
  $\cup_{\alpha \in I} A_\alpha \subset B \implies A_\beta \subset B, \forall_{\beta \in I}$
  
  which means
  
  $A_\beta \subset B, \forall_{\beta \in I} \Longleftrightarrow \cup_{\alpha \in I} A_\alpha \subset B$
Let be the set of real numbers that are greater than 0. For each , let be the open interval . Prove that
1. $\cap_{x \in I} A_x = \O$ .
  $A_x = (0,x)$ hence if $y \in A_x$ , since $x \in \dR$ and $x>0$ , it follows that $y \in I$ . Assume that $\cap_{x \in I} A_x \neq \O$ , then $\exists_y : y \in \cap_{x \in I} A_x$ . Consider this $y$ . Clearly $y \not \in ((0,y) = A_y)$ , yet $y \in I$ which means $y \not \in \cap_{x \in I} A_x$ , but this is a contradiction. Therefore, the assumption is false and infact $\cap_{x \in I} A_x = \O$ .
2. $\cup_{x \in I} A_x = I$ .
  Let $y \in \cup_{x \in I} A_x$ , then $\exists_{x \in I} : y \in A_x$ and hence $y \in I$ since $A_x \subset I$ , thus $\cup_{x \in I} A_x \subset I$ .
  
  Let $y \in I$ , clearly $y \in (A_{y+1} = (0,y+1))$ and since $y+1 \in I$ it follows that $y \in \cup_{x \in I} A_x$ and so $I \subset \cup_{x \in I} A_x$ .
  
  Therefore $\cup_{x \in I} A_x = I$ .
Let be the set of real numbers that are greater than . For each , let be the closed interval . Prove that
1. $\cap_{x \in I} B_x = \{0\}$ .
  $0 \in \cap_{x \in I} B_x$ because $0 \in (B_x = [0,x]), \forall_{x \in I}$ . Thus, since $0$ is the only element of $\{0\}$ , $y \in \{0\} \implies y \in \cap_{x \in I} B_x$ and thus
  
  $\{0\} \subset \cap_{x \in I} B_x$
  
  I say that if $y \in \cap_{x \in I} B_x$ , then $y=0$ . Proof, assume that $\exists_{y\not =0} : y \in \cap_{x \in I} B_x$ and consider this $y$ . $y \not \in (B_{\frac{y}{2}} = [0,\frac{y}{2}])$ yet $\frac{y}{2} \in I$ and hence $y \not \in \cap_{x \in I} B_x$ , but this is a contradiction. Therefore, the assumption is false and in fact $\not\!\exists_{y \not = 0} : y \in \cap_{x \in I} B_x$ and hence if $y \in \cap_{x \in I} B_x$ then $y = 0$ , consequently $y \in \cap_{x \in I} B_x \implies y \in \{0\}$ and so
  
  $\cap_{x \in I} B_x \subset \{0\}$
  
  which leads to the conclusion
  
  $\{0\} = \cap_{x \in I} B_x$
2. $\cup_{x \in I} B_x = I \cup \{0\}$ .
  Let $y \in \cup_{x \in I} B_x$ then $\exists_{x \in I} : y \in B_x$ and hence $y \in [0,x]$ for some $x \in I$ . $y$ is therefore either $0$ , in which case $y \in I \cup \{0\}$ or $y$ is a real number which is greater than $0$ and less than or equal to $x$ , in which case $y \in I$ and thus $y \in I \cup \{0\}$ , therefore
  
  $\cup_{x \in I} B_x \subset I \cup \{0\}$
  
  Let $y \in I \cup \{0\}$ then either $y \in \{0\}$ , in which case $y \in \cup_{x \in I} B_x$ ( $0$ is an element of any $B_x$ , where $x \in I$ ), or $y \in I$ in which case $y \in B_y$ and thus $y \in \cup_{x \in I} B_x$ , therefore
  
  $I \cup \{0\} \subset \cup_{x \in I} B_x$
  
  so that
  
  $\cup_{x \in I} B_x = I \cup \{0\}$