This is a line from this answer by Brian Scott (unfortunately no longer active here). Forcing Question about a sequence of functions from $\aleph_{0}$ into $2 = \{0, 1\}$
I have a sense as to how the separate pieces work, but don't know how thing work in total.
$$F_n:\aleph_0\to\{0,1\}:k\mapsto F(x_n,k)\;$$
Thanks
Read in English: $$\mbox{"$F_n$ is the map from $\aleph_0$ to $\{0, 1\}$ which sends $k$ to $F(x_n, k)$."}$$ One important aspect is the difference between "$\rightarrow$," which is used to describe domain/range, and "$\mapsto$" (whose latex code is "\mapsto"), which describes the value of the function on a given input.
So for example, we could describe the function on the naturals which sends everything to $0$ as $$c: \mathbb{N}\rightarrow\mathbb{N}: n\mapsto 0.$$
Another less trivial example to help:
Consider Cantor's diagonal argument to prove that the set of functions from $\mathbb{N}$ to $\mathbb{N}$ is uncountable.
In natural language, this can be phrased as: "Given a list $(f_i)_{i\in\mathbb{N}}$ of maps $\mathbb{N}\rightarrow\mathbb{N}$, we let $g$ be defined by $g(i)=f_i(i)+1$; then [rest of argument]."
Using the notation above, we can write this instead as: "Suppose $F:\mathbb{N}\rightarrow\mathbb{N}^\mathbb{N}$. ($F$ is just a sequence of functions from naturals to naturals - "$\mathbb{N}^\mathbb{N}$" is the set of functions from naturals to naturals, and a function $h: \mathbb{N}\rightarrow A$ can be thought of as a sequence of elements of $A$ whose $n$th term is $h(n)$.) We let $g$ be $$g:\mathbb{N}\rightarrow\mathbb{N}: i\mapsto f_i(i)+1.$$ Then [rest of argument]."
Depending on your background, the "$\mapsto$" idea may be made more simple if you think in terms of $\lambda$ notation: if we say "$j: A\rightarrow B: a\mapsto [thing]$," we're describing the object "$\lambda x^A. [thing]$." (I'm writing "$x^A$" to denote that $x$ is a variable of type $A$, and I'm being rude and conflating sets and types a bit, but hopefully the point is clear-ish; if not, ignore this part.)