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I found the following statement in a book after the usual definition of preorder and preordered set and I am not sure about how I should interpret it.

"While a preordered set $(X, \succsim)$ is not a set, it is convenient to talk as if it is a set when referring to properties that apply only to $X$. For instance, by a 'finite preordered set', we understand a preordered set $(X, \succsim)$ with $|X|< \infty$."

So the question (probably trivial) is, in the end is the preordered set a set or not?

My interpretation is that here the only set is $X$, and even if we put a preorder on it, and we call the result a preordered set, still the only set around is $X$, even if we had some characteristics (i.e. the predorder) that describe the behavior of the elements of $X$ with more details.

Is this the right interpretation of this paragraph?

PS I was not sure about the title of this question. I hope it is ok.

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up vote 2 down vote accepted

This is a small, but sometimes relevant, key issue. Are ordered pairs sets?

Recall that preordered sets are ordered pairs, $(X,R)$ such that $X$ is a [non-empty] set and $R$ is a preorder of $X$. If you treat ordered pairs as non-set objects, then preorders are not sets for that very reason.

But when we talk about the preordered set $(X,R)$, we are saying that $X$ is a set, and it is preordered by $R$. This is how you should interpret this paragraph. Often, however, it is very good to confuse between $(X,R)$ and $X$. It helps us read things better, as the example about finiteness proposes.

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Thanks a lot for the clarification. –  Kolmin May 30 '13 at 19:31
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