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So, first of all I'm gonna start with stating that I'm a complete noob in mathematics that started doing this as a hobby. I encounter problems that I can't figure out on a regular basis, like a few times per page but I usually find someone to help me. This time those same people were of no help so I came here.

We have an ordered pair $(a,b)$, okay?

And if we wanted to show the world that ordered pair using set theory it would look like this: $\{\{a\},\{a,b\}\}$.

Yeah, I don't get that at all. As I've came to understand in the ordered pair order does matter. So how come it doesn't look like this when using set theory: $\{\{a\},\{b\}\}$? I know my question is probably retarded but this seems awfully confusing to me. How come the second coordinate can be either a or b? Logic makes me believe that my previous way of thinking can't be true and that I'm understanding it completely wrong. Can someone help me with this using simple language?

Improvements: Now I understand why $\{\{a\},\{b\}\}$ doesn't apply. Thanks for the first step helpers.

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  • $\begingroup$ $\{\{a\};\{b\}\}$ is by definition the set whose elements are those equal to $\{a\}$ or to $\{b\}$: the set whose elements are those equal to $\{b\}$ or to $\{a\}$: $\{\{b\};\{a\}\}$. So $\{\{a\};\{b\}\}$ doesn't specify the order. $\endgroup$ – nombre Dec 14 '15 at 13:15
  • $\begingroup$ Maybe this might help, although nothing new I think (looking at other answers). $\endgroup$ – quapka Dec 14 '15 at 13:31
  • $\begingroup$ This might be useful as well. $\endgroup$ – Asaf Karagila Dec 14 '15 at 13:44
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Two sets are equal if and only if they share the same elements. Thus there is no distinction between the sets $\{\{a\},\{b\}\}$ and $\{\{b\},\{a\}\}.$ That's why we need a different trick to create a mathematical object involving $a$ and $b$ in some particular order so that $(a,b)\neq(b,a)$ unless $a=b.$ The model $\{\{a\},\{a,b\}\}$ that you cite is the most common one but not the only possibility.

In fact the full requirement for our model is: $(a,b)=(c,d)$ if and only if $a=c$ AND $b=d.$ It is not difficult to verify that the model in your book satisfies that requirement, using the criterion that sets are equal if and only if they contain the same elements: let us show that equality of the ordered pairs implies equality of the corresponding elements.

If the sets $\{\{a\},\{a,b\}\}$ and $\{c\},\{c,d\}\}$ are identical then we must have

$$(\{a\}=\{c\}\hbox{ OR }\{a\}=\{c,d\})\hbox{ AND }(\{a,b\}=\{c\}\hbox{ OR }\{a,b\}=\{c,d\})$$

and

$$(\{c\}=\{a\}\hbox{ OR }\{c\}=\{a,b\})\hbox{ AND }(\{c,d\}=\{a\}\hbox{ OR }\{c,d\}=\{a,b\}).$$

The first two equalities imply that $c$ is an element of the singleton $\{a\},$ so their OR implies $a=c.$

The last two equalities on the first line imply that $b=c$ or $b=d.$

The last two equalities on the second line imply that $d=a$ or $b=d.$

So we either have $b=d$ or $d=a=c=b.$ In either case, $a=c$ and $b=d.$

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  • $\begingroup$ Woah there! That was some swift response, thanks man. Now I understand why my proposed model up there is completely wrong but I still do not understand why this one would apply. Also, I was not aware that other models do exist but I'm particulary interested in this one because it is the one that appears in my book. Could you or anyone else elaborate why this one is coo' to go? $\endgroup$ – Damjan Babić Dec 14 '15 at 13:16
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    $\begingroup$ Simply, because it has the desired property: if $a$ and $b$ are distinct elements then $(a,b)$ and $(b,a),$ defined thus, are also distinct. $\endgroup$ – Justpassingby Dec 14 '15 at 13:18
  • $\begingroup$ Sorry, but I simply fail to understand this. I just don't get it how do you create order by using that model. Maybe I have the wrong perception of what the part "{a,b}" in that model means. Could you focus on explaining that exact bit? $\endgroup$ – Damjan Babić Dec 14 '15 at 13:23
  • $\begingroup$ It's not that ordered pairs must be implemented in this way; it's that this particular construction is a model of ordered pairs. For example, sets of the form $\{ \{ 0, a \}, \{ 1, b \} \}$ also act like ordered pairs. The key property of ordered pairs $(a, b)$ is the "full requirement" that Justpassingby describes. Here's how to see it satisfies the requirement. The if part is obvious, let's think about the only if. Consider two ordered pairs $\{ \{ a \}, \{ a, b \} \}$ and $\{ \{ c \}, \{ c, d \} \}$. If these are equal, they must have the same elements. In particular, (continued) $\endgroup$ – Henry Swanson Dec 14 '15 at 13:38
  • $\begingroup$ $\{ a \}$ is equal to either $\{ c \}$ or $\{c,d\}$. We know that $\{c, d\}$ has two elements*, and so $\{ a \}$ cannot possibly equal $\{ c,d \}$. So $\{ a \} = \{ c \}$, implying $a = c$. We also know $\{ a, b \}$ is equal to either $\{ c \}$ or $\{ c, d \}$. Again, because one of these sets has the wrong number of elements**, $\{ a, b \} = \{ c, d \}$, and since we know $a = c$, this lets us conclude that $b = d$, as desired. $$ $$ * If $c = d$, then $\{c,d\} = \{ c \}$, and so $a = c = d$. Then $\{ a, b \}$ must equal $\{ c \}$ or $\{c,d\}$, In either case, $b = c$. $$ $$ ** Similar to *. $\endgroup$ – Henry Swanson Dec 14 '15 at 13:42
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A short answer... but hopefully useful!

Indeed, the aim of an ordered pair, is that the order matters. Then the target is to define the ordered pair using classical "set constructions": union, intersection...

The issue with $\{\{a\},\{b\}\}$, is that you cannot differentiate the two elements of the set. You don't create "an order".

With $\{\{a\},\{a,b\}\}$ it is much better. One of the elements of the set $\{a\}$ has only ONE element, while the second $\{a,b\}$ has TWO elements. A good way to differentiate both of them... and create the "order" which was desired.

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  • $\begingroup$ What does the second set having two elements have with creating order? I see that you understand why and how but I'm completely blind here. $\endgroup$ – Damjan Babić Dec 14 '15 at 13:24
  • $\begingroup$ An "order" is a kind of inappropriate word. This is why I put the words in quotes. The fundamental topic is that you want to differentiate both elements. The fact that one has only ONE element and the second one TWO is a way to differentiate. $\endgroup$ – mathcounterexamples.net Dec 14 '15 at 13:32
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suppose $a$ and $b$ belong to a set $S$. then $$ (a,b) =\{ \{a\}, \{a,b\}\} \in \mathfrak{P}^2(S) $$ this implies the identification $$ (a,a) = \{\{a\}\} $$

let $T=\{1,2\}$ and define the collection $P_S$ of p-maps on S as the maps $t:T \to \mathfrak{P}(S)$ satisfying: $$ 1 \le |t(k)| \le k $$ and $$ j \lt k \Rightarrow t(j) \subseteq t(k) $$ then every ordered pair is the image of a p-map and vice versa

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It's because a set isn't concerned by order so the set $\{a, b\}$ is the same as the set $\{b, a\}$, but that's not all that's needs to be addressed. A set is not concerned with multiplicity either so the sets $\{a, a\}$ and $\{a\}$ are also the same. This is the reason one need to use something more clever than just putting the elements into a set to construct an ordered pair.

What we need is both a way to construct a set from an ordered pair, but also a way to from the set be able to extract the first and the second element.

Now there's multiple ways that would do to define an ordered set with the required properties. The way it's chosen here has a subtle "problem", if the elements are the same you would end up with $(a,a) = \{\{a\}, \{a, a\}\} = \{\{a\}\}$, but that's not really a problem.

The recipe for extracting the first element from the pair is to in the set find (precicely) a set containing only one element. This element it contains is the first element.

The recipe for extracting the second element is that if there's only one set in the set, it contains the second element, otherwise there's a set containing two elements one of them which is the first - the other element in that set is the second element.

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