$\{x|x$ is a positive integer definable in one line of type $\}$ $\{x|x$ is a positive integer definable in one line of type $\}$
I found this example in Enderton's book of set theory !
What does it mean ? I have no idea about "definable" and "in one line of type"
 A: For simplicity, assume that one line of type means "at most 80 characters of English, including mathematical symbols". You may also assume that a "line of type" defines an integer if it's a term denoting that integer (e.g. "$17$", "$1729*452$"), or if it uniquely identifies that integer (e.g. "the least prime number greater than $10^{10^{10}}$", which has at least 10 billion digits — far more than 80).
Although the set is not that small, and contains some perhaps surprisingly large integers, there are after all only finitely many "lines of type" consisting of mathematical English, because there are only finitely many letters, including symbols, in the alphabet of mathematical English. Thus the set is finite.

However, you're right to be suspicious of this "definition" of a set. As Asaf points out in his comment, there's a paradox hovering over this would-be "definition" — Berry's Paradox. Consider 
$$
\text{"the least integer not definable in one line of type"}.\tag{*}
$$
An awkward question arises: is this a valid "definition" of some integer $n$? If $n$ isn't definable in one line of type, as per the definition (*); then it is definable in one line of type after all, because (*) is such a definition. So $n$ is definable in one line of type, and (*) does not define it. 

A version of the paradox is even nearer to hand, arising from the (alleged) definition of the set. Consider:
$$
\text{"the least integer not in $\{x\in\Bbb N_+\mid\text{x is definable in one line of type}\}$"}.\tag{P}
$$
Let $S$ be this set we've been talking about:
$$
S = \{x\in\Bbb N_+\mid\text{x is definable in one line of type}\}.
$$ 
Because $S$ is finite, there are integers not in $S$, so (P) seems to define an integer, $n_0$, in one line of type. 
Question: is $n_0\in S$? Well, now we have a problem: $n_0\in S \iff n_0\notin S$.
