Showing that a subset of a set of functions is closed. I'm an undergraduate with little to no background in functional analysis and topology. The whole concept of function spaces is quite fuzzy to me, and I'm having a difficult time conceptualizing it. (Things like there being different notions of compactness in general topological spaces is one of many things confusing me because of what I've learned so far. I have learned many things which turn out to be true only in metric spaces, or in $\mathbb{R}^n$ specifically.)
Consider the following situation:
Let $A \subset \mathbb{R}^n$ and $[a, b] \subset \mathbb{R}$. Let $F$ denote the set of all functions from $A$ to $[a, b]$, and $G \subset F$ denote those functions which possess a particular attribute that we are interested in.
In order to finish a larger proof, I'd like to show that $G$ is closed; later on, I want to work with pointwise convergence of functions in $G$. Since we are dealing with a function space, I'm a bit unsure about how to do this, because I'm uncertain about what constitutes a limit point in this space.
I have showed that, for any sequence $(f_n)_1^\infty$ of functions in $G$ converging pointwise to some $f \in F$, $f$ must be in $G$. I believe that this shows that $G$ is closed, and I believe that it has to do with the connection between the product topology and pointwise convergence of functions, but I would really appreciate feedback on this; have I misunderstood what "closed set" means in this context?
Thanks!
EDIT: In my case, $G \subset F$ is the set of all functions $f:A \rightarrow [a, b]$ such that $f(a) + f(b) + f(c) = c(f)$, for any three pair-wise orthogonal unit vectors $a, b, c$, where $c(f)$ is a constant depending only on $f$. 
Second attempt: Take any function $f \in F \setminus G$. There exist two sets of orthogonal unit vectors $(v_1, v_2, v_3)$ and $(v_4, v_5, v_6)$ such that
\begin{equation*}
\Delta = \left| \sum_{i=1}^3 f(v_i) - f(v_{i + 3}) \right| > 0.
\end{equation*}
The set
\begin{equation*}
B_{\Delta/6} = \{ f \in F : \max(|f(v_i) - g(v_i)|) < \Delta / 6, i = 1, 2, \dots 6 \}.
\end{equation*}
is an open neighborhood of $f$. Take any $g \in B_{\Delta/6}$, and let $\delta_i = g(v_i) - f(v_i)$, for $i = 1, 2, \dots 6$, so that $|\delta_i| < \Delta / 6$. We get
\begin{equation*}
\begin{split}
\left| \sum_{i=1}^3 (g(v_i) - g(v_{i+3})) \right| = \left| \sum_{i=1}^3 (f(v_i) + \delta_i - f(v_{i+3}) - \delta_{i+3}) \right| \geq\\
\left| \sum_{i=1}^3 (f(v_i) - f(v_{i+3})) \right| - \left|\sum_{i=1}^3 (\delta_{i+3} - \delta_i) \right|
= \Delta - \left|\sum_{i=1}^3 (\delta_{i+3} - \delta_i) \right| > \Delta - \Delta = 0,
\end{split}
\end{equation*}
using the reverse triangle inequality. Thus $g \in F \setminus G$, so that $B_{ \Delta / 6} \subset F \setminus G$, meaning $F \setminus G$ is open. We conclude that $(F \setminus G)^C = G$ is closed.
Any thoughts?
 A: If you are considering functions from $A \subseteq \mathbb{R}^n$ to $[a,b]$ in the pointwise topology, then for a condition like you describe in the comments: $f(v_1) +f(v_2) + f(v_3) \in C$ where $C$ is closed in $[a,b]$ (like a singleton or a finite set maybe), then the functions that satisfy it is indeed closed. To show this it does not suffice to consider sequences from this set, but nets will do, if you know about them. 
Otherwise let $F = \{f :A \to [a,b]\}$ in the pointwise (=product  ) topology and set $D = \{f: A \to [a,b]: f(v_1) + f(v_2) + f(v_3) \in C\} = (s \circ  p_v)^{-1}[C]$, where $p_v : F \to \mathbb{R}^3, p_v(f) = (f(v_1), f(v_2), f(v_3))$ is continuous as a projection and $s: \mathbb{R}^3 \to \mathbb{R}: s(x,y,z) = x+ y +z$ are both continuous and the inverse image of a closed set under a continuous map is closed. 
A: What you have shown is that $G$ is sequentially closed, which may not imply that $G$ is closed. The two concepts coincide for metric spaces, but not in general. 
Here is a standard counterexample which fits your situation as well: if $A = \mathbb{R}, a=0,b=1$, and $G$ is the set of all functions $f:A\to [0,1]$ such that $f(x) = 0$ for all but countably many $x\in A$.
This set is sequentially closed because the countable union of a countable set is countable. However, it is not closed in $F$ : If $g\in F$ is any function, and $U$ is a basic open neighbourhood in the topology of point-wise convergence, then $\exists x_1,x_2,\ldots, x_n\in A$ and $\epsilon > 0$ such that
$$
U = \{f\in F : |f(x_i) - g(x_i)| < \epsilon \quad\forall 1\leq i\leq n\}
$$
Now simply that $f\in G$ such that $f(x_i) = g(x_i)$ for all $1\leq i\leq n$ and $f(x) = 0$ if $x\notin \{x_1,x_2,\ldots, x_n\}$. Then $f\in U$, so $U\cap G\neq \emptyset$. Hence, $g\in \overline{G}$.
