# Set Operations with Empty sets

I have a question regarding performing set operations on empty sets.

For example let A = ∅, Let B = {A, ∅}, Let C = {A, B}.

Would B = {∅, ∅} and C = {∅, {∅,∅}}? or would B = {∅} and C = {∅, {∅}} since {∅, ∅} reduces to {∅}?

Now if I wanted to do $A \cup C$ would the answer be $∅ \cup C$ -> C = {∅, {∅}}?

Now what if I wanted to do D = C - B, would the result be {{∅}}?

One last question. If P = {∅}, Q = {P}, R = {∅,P}, S = {∅, P, Q, R};

What would X = {$x : (x ∈ R) \wedge ( x ⊆ S)$}?

My thoughts are that Q = {{∅}}, R = {∅, {{∅}}} and S = { ∅, {∅}, {{∅}}, {∅, {{∅}}} }

• x ∈ R would mean x = ∅ and x = {{∅}}
• x ⊆ S would mean x is a set containing any combination of ∅, {∅}, {{∅}}, {∅, {{∅}}} as members

Therefore X = {{{∅}}} as {{∅}} is the only member that fits both those conditions?

I've tried looking at some lectures online on set theory but I want to make sure my understanding is correct so far.

On a more general note, I am also wondering if lets say D = {1, 2} and E = {D, 5}. Would E = {1,2,5} or E = {{1,2},5}?

Thanks

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$\{ \emptyset, \emptyset \} = \{ \emptyset \} \neq \emptyset$ – Hurkyl Jul 2 '12 at 11:49
@Hurkyl, yep sorry that was a typo. edited. – BYS2 Jul 2 '12 at 11:51
I wonder if it might be easier to deal with these questions if we write 0=∅, 1={∅}, 2={0,1}, and in general $n=\{0\ldots n-1\}$ for each number $n$. Then we have $A=0, B=\{0,0\} = \{0\} = 1, C=\{A,B\} = \{0, 1\} = 2$. We also have $x\cup y = \max(x,y)$ and $x\cap y = \min(x,y)$. This saves many confusing nested curly braces, although it doesn't save all of them, since $D=\{1\}$. But it does simplify the notation quite a bit. – MJD Jul 2 '12 at 16:27
@MarkDominus.. yes this is similar to the proof for the construction of natural numbers which is what brought me to operations on empty sets in the first place :p – BYS2 Jul 3 '12 at 6:18

$B$ contains a single element which is an empty set (or, equivalently, is equal to $A$): $B = \{\emptyset\} = \{A\}$

$C$ contains two elements, one of which is an empty set (or, equivalently, is equal to $A$), and another is $B$: $C = \{\emptyset, \{\emptyset\}\}$.

$A \cup C$ contains all elements $A$ contains (there are no such elements, as $A$ is empty) and all elements $C$ contains: $A \cup C = C = \{\emptyset, \{\emptyset\}\}$.

$C = \{A\} \sqcup \{B\}$, and $B = \{A\}$, hence $C - B = (\{A\} \sqcup \{B\}) - \{A\} = \{B\} = \{\{A\}\} = \{\{\emptyset\}\}$.

As for the last question, $x \in R$ is equal to $x = \emptyset \vee x = P$ per $R$ definition. Empty set is obviously a subset of any set (including $S$); and $P \subseteq S$ because all elements $P$ contains (there is only one of them - an empty set) are also elements of $S$. So $X = R$.

On a more general note, I am also wondering if lets say D = {1, 2} and E = {D, 5}. Would E = {1,2,5} or E = {{1,2},5}?

$E$ is a two-element set (as long as $D$ is not equal to $5$): one element is $5$ and another is $D$. Thus, $E = \{\{1, 2\}, 5\}$.

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Ok thanks for that :). what about the X = {x:(x∈R)∧(x⊆S)} question? Are you able to provide your thoughts of whether I did it correctly? – BYS2 Jul 3 '12 at 5:17
@BYS2 Read the third lowest paragraph: "As for the last question ..." – penartur Jul 3 '12 at 5:59
ooh yes sorry, missed that for some reason.. Thanks! – BYS2 Jul 3 '12 at 6:22