For every $n\in \Bbb N$, let $g_{n}:A\mapsto \Bbb C$, where $A\subseteq \Bbb C$. prove from the definition of pointwise convergence, that if the sequence $(g_{n})$ converges pointwise to a function $g:A\mapsto \Bbb C$ then $g$ is unique.
Here is my attempt, if anyone could tell me if its correct, or any detail is missed, and how to correct it.
Proof: Suppose for a contradiction that $(g_{n}) \to g$ and that $(g_{n})\to f$ where $f\ne g$, set $\epsilon=|g-f|>0$ then since $(g_{n})\to g$, $\exists N_{1}\in \Bbb N:|g_{n}-g(x)|<\frac{\epsilon}{2}$, $\forall n>N_{1}$ and since $(g_{n})\to f$ $\exists N_{2}\in \Bbb N:|g_{n}(x)-f(x)|<\frac{\epsilon}{2}, \forall n>N_{2}$ Now set $M=\max\{N_{1},N_{2}\}$. then $\forall n>M$ $$|g-f|=|g-g_{n}(x)+g_{n}(x)-f|\le |g(x)-g_{n}(x)|+|g_n{x}-f(x)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon=|g-f|$$ which is a contradiction. thus if $(g_{n})$ converges pointwise to a function then the function is unique.
Was just wondering if this is correct. (do i need to write $N(\epsilon,x)$ and mention a condition on $x$?
This is mostly correct but there is some confusion about where you are evaluating your functions. What does $|g-f|$ mean? What is this "$x$" you are plugging into some but all of your functions?
To make sense of this, you need to choose some specific $x$ and use it consistently. If $f\neq g$, that means there exists some $x\in A$ such that $f(x)\neq g(x)$. Now fix such an $x$ and go through the rest of your argument, defining $\epsilon=|f(x)-g(x)|$ making sure to evaluate all of your functions at $x$ in all your inequalities.