!the first question is the one relevant: you need to explain/prove if the definitions given are correct] (https://i.stack.imgur.com/xZOtI.jpg)
I need to determine whether the first 3 definitions are equivalent to the definition of ordered couples, and to explain why.
i'm taking the class of "introduction to set theory" for math b.a students, so the question is not under the assumption of ZFC
On
Consider the two cases $$ a_1 = \{\{x\}\}, b_1 = x\\ a_2 = \{x\}, b_2 = \{x\} $$ I have used one of the rows above to make one of your "ordered pairs" $(a_i, b_i)$, and the result I get is $$ (a_i, b_i) = \{\{x\}, \{\{x\}\}\} $$ Can you tell me whether I've used $i = 1$ or $i = 2$?
On
If $a=\{y\}$ and $x=\{b\}$ then $\{a,\{b\}\}=\{x,\{y\}\}$ but it is evidently not necessary that $a=x$ and $b=y$.
On
I need to determine whether the first 3 definitions are equivalent to the definition of ordered couples, and to explain why.
For that you have to know what an "ordered couple" actually means and how those different definitions can be verified? I assume this is the property you are looking at:
Ordered pair equality: $(a,b) = (c,d)\text{ if and only if }a=c\text{ and }b=d$
Note the subtle difference between this property and equality between sets:
Set equality: $\{a, b\}=\{c, d\}\text{ if and only if }(a = c\text{ and }b=d)\textbf{ or }(a = d\text{ and }b=c)$
This is not a standard definition, more like a consequence of "$A\subseteq B$ and $B\subseteq A$" which is easier to understand when both sets have $1-2$ elements. Anyway it will be useful later.
Another important thing is that the underlying axiomatic system matters. We often assume ZF (or sometimes ZFC) implicitly but I have a tendency to be more explicit. Because you see, ZF is one of the most popular models. And in ZF the equality $x=\{x\}$ cannot hold, no matter what $x$ is. But there are systems in which $x=\{x\}$ holds, most notably ZF without the axiom of foundation. This will become important as we will see soon. And it is far from a trivial observation.
Everything that follows is assumed to be under ZF.
So let's have a look at those 3 cases:
(1) $(a,b) := \{\{0,a\}, \{1, b\}\}$
I assume here that $0\neq 1$, e.g. $0=\emptyset$ and $1=\{\emptyset\}$. It really doesn't matter what $0, 1$ are as long as they are distinct.
Assume that $(a,b)=(c,d)$. Then $\{\{0,a\}, \{1, b\}\}=\{\{0,c\}, \{1, d\}\}$. So under that assumption the equivalent condition for the equality is:
$$\{0,a\}=\{0,c\}\text{ and }\{1,b\}=\{1,d\}$$ $$\textbf{or}$$ $$\{0,a\}=\{1,d\}\text{ and }\{1,b\}=\{0,c\}$$
The first "or" case implies that $a=c$ and $b=d$. What about the other one? It implies that $a=1, d=0$ and $b=0, c=1$ which means $a=c$ and $b=d$.
So this one is good, all paths lead to $a=c$ and $b=d$.
This definition is also known as the Hausdorff's ordered pair.
(2) $(a,b) := \{a, \{b\}\}$
As others noted this fails since $(\{x\}, \{x\})=(\{\{x\}\}, x)$ and we can never have $x=\{x\}$ for $x=\emptyset$. Also in ZF this cannot happen for any other set (due to the axiom of foundation).
(3) $(a,b) := \{ \{a\}, \{a, \{b\}\}\}$
Assume that $(a,b)=(c,d)$. In other words we have
$$\{a\} = \{c\}\text{ and }\{a, \{b\}\} = \{c, \{d\}\}$$ $$\textbf{or}$$ $$\{a\} = \{c, \{d\}\}\text{ and }\{a, \{b\}\} = \{c\}$$
The first "or" case obviously implies that $a=c$ and $b=d$. The other "or" case implies that $c=\{d\}$ (first "and" equality) and $a=\{b\}$ (second "and" equality). And then $a=c$ and $b=d$ follows. So this one is good as well.
Also note that the last one is very similar to the Kuratowski's definition of the ordered pair $(a,b)=\{\{a\},\{a,b\}\}$.
