Why is it the case that $\rm\langle Socrates\rangle = Socrates$, while $\rm\{Socrates\}\neq Socrates$? In Nolt's book called Logics (part III, chapter 7, § 7.1 Sets and $n$-tuples) one can read:


  Ordered one-tuples, unlike unit sets, are the same things as the object they contain. Thus , for example, the one-tuple $\rm (Socrates) = Socrates$; and more generally, for any object $x$, $(x)  = x$.


Could you please explain Nolt's (too) short explanation (at least for me)? 

Is this explanation a way of saying that a "$1$-tuple" is not a well defined object? 

Is there a way to derive the formula $(x) = x$ from the set theoretic definition of an $n$-tuple?
The explanation identifies a $1$-tuple with its object. So while $x$ is generally not the same as $\{x\}$, the $1$-tuple $(x)$ is identified with $x$ itself.

This is perfectly reasonable to do, and so people do it. But does that clash with the set theoretic definition? Well, that depends on your definition of a $1$-tuple to begin with. Since the language of set theory only has $\in$, any definition of $1$-tuples is just a way to interpret this concept of a sequence of length $1$.

Generally, we think about $n$-tuples as functions from $\{0,\ldots,n-1\}$, so a $1$-tuple would be a function from $\{0\}$ into something. But this is not something which is hardcoded into set theory, it's just one way to interpret $n$-tuples. You can do it just as well in a different way, or even explicitly decide this is fine for $n&gt;1$, and for $n=1$, $(x)$ is just $x$ itself (which then raises the question what is a $1$-tuple whose object is a $3$-tuple?).

This is also why we can't require $\{x\}=x$ in general. Since we understand $\{x\}$ as a set whose unique member is $x$. And then $\varnothing\neq\{\varnothing\}$, since exactly one of these has an element. But tuples are looser, since as we see, they are not hardcoded into set theory like singletons.
Your book's author was being careless and accidentally contradicted set theory. There are two ways tuples are defined, and neither of them allow for $\{x\} = x$.


A function $T : \mathbb{N} \to S$ together with a collection of projection functions $\pi_i : \{i\} \to \mathbb{N}$ such that $\pi_i \circ T$ is the ith element of $T$. Here obviously $(x) \neq x$.
Some set theoretic definition involving the Kuratowski pair. And in these definitions, $(a) = \{a\}$ which according to the axiom of regularity cannot be equal to $a$.


So no, your book has made a mistake.