MHB Calculate Relations with Hey! (Nerd)

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The discussion focuses on calculating properties of the relation R, defined as a set of ordered pairs. The inverse relation R^{-1} for the empty set is confirmed to be { { {∅} } }, while the composition R ∘ R yields a specific set of ordered pairs. The participants clarify that the power set of R, denoted as P(R), will contain 8 elements, but it specifically consists of sets of ordered pairs rather than individual elements. A systematic approach is suggested for determining the elements of P(R) by using binary representations to include or exclude pairs. The final understanding confirms that P(R) includes the empty set and combinations of the ordered pairs.
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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 $\{\{ \{ \varnothing \} \}\}, \{ \varnothing \},\{ \{ \varnothing \}\}$?
 
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evinda said:
$$R^{-1}[\{ \varnothing \}]=\{ x: \exists y \in \{ \varnothing\}:xRy \}=\{x: xR \varnothing\}=\{ \{ \{ \varnothing \} \}\}$$
Correct.

evinda said:
$$R \circ R=\{ \langle \{ \varnothing \},\varnothing\rangle, \langle \{ \{ \varnothing \} \},\{ \varnothing\} \rangle, \langle \varnothing, \{ \{ \varnothing\} \}\rangle \}$$
Correct.

evinda said:
$\mathcal{P}R$ will have $2^3=8$ elements, right?
Yes.

evinda said:
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}
 
Evgeny.Makarov said:
Correct.

Correct.

Yes.

(Happy)

Evgeny.Makarov said:
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? :confused:
 
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 $\mathcal{P}R$.
 
Evgeny.Makarov said:
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? :confused:
 
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.
 
Evgeny.Makarov said:
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)
 

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