Trying to show that set of all one-element subsets of a denumerable set is denumerable Let $A$ be a denumerable and put $X = \\{ B : B \subset A, \; \; |B| =1 \\} $. Then $X$ is denumerable: I know there is a bijection $f: A \to \mathbb{N} $. If I consider the functions $g: \mathbb{N} \to X$ given by $g(n) = \\{ n \\} $ then can we conclude that $g$ is bijective ?

Frank has already covered the answer quite sufficiently: I'll address your second paragraph.

There is no reason to believe that $g: n \mapsto \\{n\\}$ is even a proper function from $\mathbb{N}$ to $X$, much less a bijection. If we take $A=\\{\text{Banana}\\}$, then $X = \\{\\{\text{Banana}\\}\\}$ does not even intersect the range of $g$. On the other hand, if we take $A= \mathbb{Z}$ (the set of all integers), then $X$ is denumerable, but $g$ fails to surject onto $X$ (since, in particular, $\\{-1\\} \
ot\in \operatorname{Range}(g)$).

The definition of $g$ you want* is $g(n) = \\{f(n)\\}$. This follows from the fact that, for any $x \in X$, we can write $x=\\{a\\}$ for a unique $a \in A$, and we can write $a=f(n)$ for a unique $n\in\mathbb{N}$ because $f$ is a bijection.

* * *

*: It can actually be shown that all bijections $g: \mathbb{N} \to X$ can be written as $g(n) = \\{f(n)\\}$ for some bijection $f:\mathbb{N} \to A$: I'll leave the proof as an exercise.