MHB Which property does it satisfy?

[evinda]

Hey again! (Nerd)

According to my notes, the Kuratowski definition for the ordered pair is the following:

Let $a,b$ sets. We define the ordered pair of $a,b$ like that:

$$<a,b>=\{ \{a \}, \{ a, b \} \}$$

So, when $x \in <a,b>$, which property does it satisfy? (Thinking)

 

[I like Serena]

Hey again! (Nerd)

According to my notes, the Kuratowski definition for the ordered pair is the following:

Let $a,b$ sets. We define the ordered pair of $a,b$ like that:

$$<a,b>=\{ \{a \}, \{ a, b \} \}$$

So, when $x \in <a,b>$, which property does it satisfy? (Thinking)

Hi! :)

What are the possibilities for $x$? (Wondering)

 

[Evgeny.Makarov]

Let $a,b$ sets. We define the ordered pair of $a,b$ like that:

$$<a,b>=\{ \{a \}, \{ a, b \} \}$$

So, when $x \in <a,b>$, which property does it satisfy?

Strictly speaking, you need to use the axiom of pair to determine what $x\in\langle a,b\rangle$ is equivalent to.

I am not sure if this is what the exercise has in mind, but I would say $x\in\langle a,b\rangle$ is usually understood in a different, less formal way. The things is that, using programming speech, $\{ \{a \}, \{ a, b \} \}$ is just one data structure (a hack, really) that implements $\langle a,b\rangle$, i.e., behaves like we want $\langle a,b\rangle$ to behave. But as every concrete data structure, it has some incidental properties, which should rather be hidden from the "user". For a similar example, the disjoint union $A\sqcup B$ of two sets $A$ and $B$ can be implemented as $A\times\{1\}\cup B\times\{2\}$. This implementation does provide the injection functions $i_A: A\to A\sqcup B$ and $i_B: B\to A\sqcup B$ defined as $i_A(x)=\langle x,1\rangle$ and $i_B(x)=\langle x,2\rangle$, which are essential for the concept of disjoint union. But it has has some incidental properties, for example, the fact that it uses 1 and 2 and not, say, 9 and 10. This fact should be hidden because it is not intrinsic to disjoint union.

What I am trying to say is that to most people, an ordered pair is first of all a pair, so $x\mathrel{\boldsymbol{\in}}\langle a,b\rangle$ should be equivalent to $x=a\lor x=b$. But this predicate $\boldsymbol{\in}$ is different from the regular set-theoretic $\in$. If necessary, one can define it so that it works with this particular implementation of $\langle a,b\rangle$ as expected.

 

[evinda]

So, what is the difference between an ordered pair and a non ordered pair? (Thinking)

What do we know for $x$, when $x \in \{ a, b \}$ and what when $x \in <a,b>$ ?

 

[evinda]

When $x \in \{ a, b \} \Rightarrow x =a \text{ or } x=b$.
So, in this case $x$ is a set, right? (Thinking)

When $x \in <a,b> \Rightarrow x \in \{ \{ a \}, \{ a, b \} \} \Rightarrow x= \{ a \} \text{ or } x=\{ a, b \} $.
So, is $x$ in this case a set of a set? (Thinking)

 

[Evgeny.Makarov]

So, what is the difference between an ordered pair and a non ordered pair?

Informally, we want an ordered pair to contain two (possibly the same) elements as well as the information saying which element is the first and which is the second. That is, if we denote the desired set of ordered pairs where the first element comes from $A$ and the second from $B$ by $A\times B$, we want an operation $\langle\cdot,\cdot\rangle$ that takes an element of $A$ and an element of $B$ and returns an element of $A\times B$. We also want the inverse operations: projections $\pi_1: A\times B\to A$ and $\pi_2:A\times B\to B$ such that
\begin{align}
\pi_1\langle a,b\rangle&=a&&\text{for any }b\in B\qquad(1)\\
\pi_2\langle a,b\rangle&=b&&\text{for any }a\in A\qquad(2)
\end{align}
This implies that $\langle\cdot,\cdot\rangle$ is injective:
\[
\text{If }\langle a_1,b_1\rangle=\langle a_2,b_2\rangle,\text{ then }a_1=a_2\text{ and }b_1=b_2.\qquad{(3)}
\]
Indeed,
\[
\langle a_1,b_1\rangle=\langle a_2,b_2\rangle\implies a_1\overset{(1)}{=}\pi_1\langle a_1,b_1\rangle=\pi_1\langle a_2,b_2\rangle\overset{(1)}{=}a_2
\]
and similarly $b_1=b_2$. Note that if $C$ is the set of unordered pairs of elements from $A$ and $B$, then there is again a pairing function $\{\cdot,\cdot\}$ mapping $A$ and $B$ to $C$, but it has the property
\[
\{a,b\}=\{b,a\}\qquad(4)
\]
(there is only one pair with given elements, as you have shown in another thread). Therefore, there cannot be inverse functions because once two elements are packed into a unordered pair, the information which elements is the first and which is the second is lost. If there were such functions $\rho_1$ and $\rho_2$, then
\[
a=\rho_1\{a,b\}\overset{(4)}{=}\rho_1\{b,a\}=b
\]
which is a contradiction if $a\ne b$.

When $x \in \{ a, b \} \Rightarrow x =a \text{ or } x=b$.

Yes.

So, in this case $x$ is a set, right?

In axiomatic set theory, everything is a set.

When $x \in <a,b> \Rightarrow x \in \{ \{ a \}, \{ a, b \} \} \Rightarrow x= \{ a \} \text{ or } x=\{ a, b \} $.

Yes.

So, is $x$ in this case a set of a set?

Yes, but the important point is that $a$ and $b$, along with the information which of them is the first can be recovered from $\langle a,b\rangle$.

Property (3) is usually considered essential and proved for ordered pairs.