# Ordered pairs in sets

So I am asked to explain why $(x,y,z)=(x,(y,z))$ is a reasonable definition of an ordered triple. However, I am in a limbo at the moment let alone with trying to explain an ordered pair,(x,y), intuitively and why it is so except it is excepted by Mathematicians. So an order pair states that $(a,b)=\{\{a\},\{a,b\}\}$ with an unordered pair $\{a,b\}$. I understand the definition but not intuitively. I have looked at some postings on M.SE such as How can an ordered pair be expressed as a set? but the answer provided seemed a little vague, for me at least, since I am now beginning to learn Set Theory. So if anyone could explain it in English and demonstrate your point with notations, that would be great.

Also, the way I tried viewing or explaining it to myself "intuitively" is that since $(x,y)$ is an ordered pair on a coordinate plane, $\{\{a\}\}$ can be viewed as a set of points up to the x-coordinate and $\{\{a,b\}\}$ as counting the elements of the set up to the x-coordinate and upwards up to the y-coordinate. Not sure if this makes sense to anyone else but somehow this explanation is not satisfying even if it is correct.

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math.stackexchange.com/questions/308422/… would be more useful to you, I believe. –  Asaf Karagila May 21 '13 at 4:45

$$(x,y)=(u,v)\iff x=u\textbf{ and } y=v.$$ (For triplets, of course, we want equality in all three coordinates.)
Assuming that we have a reasonable definition of an ordered pair, i.e. one that satisfies the above, then defining $(x,y,z)=(x,(y,z))$ is a good definition for a triplet. To see why, you need to unwind the definition of an ordered pair twice and conclude that two triplets are equal if and only if their elements are [pointwise-]equal -- which is a task I will leave to you to verify.