# How can an ordered pair be a set assuming there is no order in a set?

The following propositions, I think, are generally considered as true ( though they may not all have the same level of rigor).

(1) In a set there is no order ( due to the extensionality axion) : $$\{a,b\} = \{b,a\}$$

(2) In an ordered pair there is an order : $$(a, b)$$ is not equal to $$(b,a)$$.

(3) An ordered pair is a set: $$(a,b) = \{ \{a\} , \{a,b\} \}$$.

Does the problem lie in proposition (2) : should one say, instead of " in an ordered pair there is an order" that " an ordered pair is an order"?

How to formulate rigorously these propositions in order to make them compatible?

You formally define $$(a,b)$$ to be equal to $$\{\{a\}, \{a,b\}\}$$.

That's the only formal definition.

The rest is our interpretation. We notice that using the definition above, $$(a,b)$$ is not equal to $$(b,a)$$ if $$a\neq b$$ (*), and therefore, in the newly introduced symbol $$(.,.)$$, order matters. That's why we call this newly introduced symbol an "ordered pair".

In other words, (2) is not a proposition, it is a consequence of (3).

(*) Note that the only thing you need to prove $$(a,b)\neq (b,a)$$ is that $$a\neq b$$. This is because, if $$a\neq b$$, then the set $$\{a\}$$ is an element of $$(a,b)$$ (because $$(a,b)=\{\{a\}, \{a,b\}\}$$, but it is not an element of $$(b,a)$$ (because $$\{a\}\neq \{b\}$$ and $$\{a\}\neq\{b,a\}$$ and there are no other elements in $$(b,a)$$.

An order is a relation between elements, not the elements themselves. Propositions 1 and 2 are fine as they stand.

(2) is a definition on "reader-level", and defines a piece of notation we would like to make use of and how to think about its use intuitively. (3) is a formal, "lower-level" definition of that same notaiton that makes sure we haven't really made any new set theory in the process, but rather that we are still (under the hood) using only the regular (unordered) sets we already had.

There's no incompatibility to be fixed. In fact $$(a,b)=\{ \{a\} , \{a,b\} \} =\{ \{a,b\} ,\{a\} \}=\{ \{a\} , \{b,a\} \} =\{ \{b,a\} ,\{a\} \}.$$