Calculate Relations with Hey! (Nerd)

[evinda]

Hey! (Nerd)

Given the relation: $R=\{ \langle\{ \{ \varnothing \} \}, \varnothing \rangle, \langle \varnothing, \{ \varnothing \}\rangle, \langle \{ \varnothing \},\{ \{ \varnothing \} \}\rangle \}$, I want to calculate $R^{-1}[\{ \varnothing \}], R \circ R, \mathcal{P}R$.

That's what I have tried:

$$R^{-1}[\{ \varnothing \}]=\{ x: \exists y \in \{ \varnothing\}:xRy \}=\{x: xR \varnothing\}=\{ \{ \{ \varnothing \} \}\}$$

$$R \circ R=\{ \langle \{ \varnothing \},\varnothing\rangle, \langle \{ \{ \varnothing \} \},\{ \varnothing\} \rangle, \langle \varnothing, \{ \{ \varnothing\} \}\rangle \}$$

$\mathcal{P}R$ will have $2^3=8$ elements, right? Will $\mathcal{P}R$ contain the elements $\{\{ \{ \varnothing \} \}\}, \{ \varnothing \},\{ \{ \varnothing \}\}$?

 

[Evgeny.Makarov]

$$R^{-1}[\{ \varnothing \}]=\{ x: \exists y \in \{ \varnothing\}:xRy \}=\{x: xR \varnothing\}=\{ \{ \{ \varnothing \} \}\}$$

Correct.

$$R \circ R=\{ \langle \{ \varnothing \},\varnothing\rangle, \langle \{ \{ \varnothing \} \},\{ \varnothing\} \rangle, \langle \varnothing, \{ \{ \varnothing\} \}\rangle \}$$

Correct.

$\mathcal{P}R$ will have $2^3=8$ elements, right?

Yes.

Will $\mathcal{P}R$ contain the elements $\{\{ \{ \varnothing \} \}\}, \{ \varnothing \},\{ \{ \varnothing \}\}$?

No, it will contain sets of ordered pairs.

To make things easier, I recommend renaming the sets as follows.
\begin{align}
\varnothing&\mapsto0\\
\{\varnothing\}&\mapsto1\\
\{\{\varnothing\}\}&\mapsto2
\end{align}

 

[evinda]

Correct.

Correct.

Yes.

(Happy)

No, it will contain sets of ordered pairs.

To make things easier, I recommend renaming the sets as follows.
\begin{align}
\varnothing&\mapsto0\\
\{\varnothing\}&\mapsto1\\
\{\{\varnothing\}\}&\mapsto2
\end{align}

So, suppose that we have the relation:

$$R=\{ \langle 2,0\rangle, \langle 0,1 \rangle, \langle 1,2 \rangle\}= \{ \{ \{2\}, \{2,0\} \}, \{ \{0\},\{0,1\} \},\{ \{ 1\}, \{1,2\} \}\}$$

$\mathcal{P}R$ contains these elements:

• $\varnothing$
  $$$$
• $\{ \{ \{2\}, \{2,0\} \}, \{ \{0\},\{0,1\} \},\{ \{ 1\}, \{1,2\} \}\}$
  $$$$
• $ \{ \{ \{2\}, \{2,0\} \} \}$
  $$$$
• $ \{ \{ \{0\},\{0,1\} \} \}$
  $$$$
• $\{ \{ \{ 1\}, \{1,2\} \} \}$
  

Right? How can I find the remaining three elements, that $\mathcal{P}R$ contains?

 

[Evgeny.Makarov]

It helps to write all triples consisting of 0 and 1 in a systematic order.

000
001
010
011
100
101
110
111

Note that they represent binary numbers from 0 to 7. Also, instead of expanding ordered pairs (as in $\langle0,1\rangle=\{\{0\},\{0,1\}\}$), let's give them names, e.g., $\langle0,1\rangle=a$, $\langle0,2\rangle=b$ and $\langle1,2\rangle=c$. This is because when we compute $\mathcal{P}R$, the pairs are not taken apart, but occur in the resulting set as they are. Now, for each of the 8 triples above, if the first element is 0, don't include $a$ in the set, and if the first element is 1, do include $a$ in the set. The second element of the triple similarly controls the inclusion of $b$ and the third one controls the inclusion of $c$. Thus, each triple of 0s and 1s corresponds to a set containing from 0 to 3 elements. All these sets taken together form $\mathcal{P}R$.

 

[evinda]

It helps to write all triples consisting of 0 and 1 in a systematic order.

000
001
010
011
100
101
110
111

Note that they represent binary numbers from 0 to 7. Also, instead of expanding ordered pairs (as in $\langle0,1\rangle=\{\{0\},\{0,1\}\}$), let's give them names, e.g., $\langle0,1\rangle=a$, $\langle0,2\rangle=b$ and $\langle1,2\rangle=c$. This is because when we compute $\mathcal{P}R$, the pairs are not taken apart, but occur in the resulting set as they are. Now, for each of the 8 triples above, if the first element is 0, don't include $a$ in the set, and if the first element is 1, do include $a$ in the set. The second element of the triple similarly controls the inclusion of $b$ and the third one controls the inclusion of $c$. Thus, each triple of 0s and 1s corresponds to a set containing from 0 to 3 elements. All these sets taken together form $\mathcal{P}R$.

So, we have: $R=\{a,b,c\}$, where $a=\langle0,1\rangle$, $b=\langle0,2\rangle$ and $c=\langle1,2\rangle$

So, is it like that? (Thinking)

$$\mathcal{P}R=\{ \varnothing, \{a,b,c\}, \{a\},\{b\},\{c\},\{a,b\}, \{a,c\},\{b,c\}\}$$

Or have I understood it wrong?

 

[Evgeny.Makarov]

This is correct. I think you also asked in another thread why $\varnothing\in\mathcal{P}B$ for any $B$. You should be able to answer this now.

 

[evinda]

This is correct. I think you also asked in another thread why $\varnothing\in\mathcal{P}B$ for any $B$. You should be able to answer this now.

Yes, I see why it is like that.. (Nod) Thanks a lot! (Smile)