Meaning of the term faithful enumeration

Can someone please explain what the term "faithful enumeration" of an infinite set means?

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In what context did you encounter the term? –  Alex Becker Mar 23 '12 at 6:39

In my experience it most often means simply a bijective enumeration, one without repetitions. A faithful enumeration of $\Bbb Q$, for instance, would be one of the form $\Bbb Q=\{q_n:n\in\omega\}$, where $q_n\ne q_m$ whenever $n\ne m$, while a faithful enumeration of $\Bbb R$ would be one of the form $\Bbb R=\{x_\xi:\xi<2^\omega\}$, where $x_\xi\ne x_\zeta$ whenever $\xi\ne\zeta$. The domain of the enumeration is typically the cardinality of the set; here that’s $\omega$ in the case of the countably infinite set $\Bbb Q$ and $2^\omega$ in the case of $\Bbb R$.
If the set $S$ already bears a well-ordering $\prec$, the term might possibly be used for a bijective enumeration $S=\{s_\xi:\xi<\alpha\}$, where $\alpha$ is the order type of $\langle S,\prec\rangle$, and $s_\xi\prec s_\zeta$ whenever $\xi<\zeta<\alpha$.
I've encountered this term here: "Let $X$ be an (infinite) topological space and let $\{a_{n}\}$ where $n \in \mathbb{N}$ is an injective (faithful enumeration) sequence of elements of $X$. My question is why can we always choose such sequence? can we simply let $J \subseteq X$ be countable and pick a countable injective sequence of elements of $J$ or why? I mean $X$ is not assumed to be countable. –  user6495 Mar 25 '12 at 7:07
@user6495: If $X$ is an infinite set, then there is an injection from $f:\Bbb N\to X$. Technically this requires (some part of) the axiom of choice, but it’s a routine assumption in most of mathematics. The intuitive idea is simple. Let $f(0)$ be any element of $X$, $f(1)$ be any element of $X\setminus\{f(0)\}$, and so on; since $X$ is infinite, you’re never unable to find a new element to pick. To return to the original question, in this context faithful enumeration simply means injective. –  Brian M. Scott Mar 26 '12 at 20:33