Why is the closedness of the set on which $f = g$ immediate, when proving the Identity Theorem? It's a short proof given on Wikipedia, and I understand the argument for why the set on which $f = g$ must be open, but I'm not sure why closedness of the set is obvious.  Apparently, this comes from the continuity of $f$ and $g$, which are both analytic.
Is it a sequential argument, i.e., all convergent sequences from this set converge to points that are also in the set on which $f = g$? 
Thanks.
 A: The easiest argument uses the topological definition of continuity: $\{ x : f(x)=g(x) \} = \{ x : (f-g)(x)=0 \}$. Since $f-g$ is continuous and $\{ 0 \}$ is a closed subset of $\mathbb{C}$, the preimage of $\{ 0 \}$ under $f-g$ is closed.
A: Suppose $f, g : \mathbb{C} \to \mathbb{C}$ are continuous. Note that
\begin{align*}
\{z \in \mathbb{C} \mid f(z) = g(z)\} &= \{z \in \mathbb{C} \mid f(z) - g(z) = 0\}\\ 
&= \{z \in \mathbb{C} \mid (f - g)(z) = 0\}\\ 
&= (f - g)^{-1}(\{0\}).
\end{align*}
As $f$ and $g$ are continuous, so is $f - g$. As $\{0\}$ is a closed subset of $\mathbb{C}$, and the preimage of a closed set under a continuous map is closed, $(f - g)^{-1}(\{0\}) = \{z \in \mathbb{C} \mid f(z) = g(z)\}$ is closed.
Alternatively, you can use a sequential argument as you suggest. Let $\{z_n\}$ be a sequence in $\{z \in \mathbb{C} \mid f(z) = g(z)\}$ which converges to $a \in \mathbb{C}$. By continuity we have
$$f(a) = f\left(\lim_{n\to\infty}z_n\right) = \lim_{n\to\infty}f(z_n) = \lim_{n\to\infty}g(z_n) = g\left(\lim_{n\to\infty}z_n\right) = g(a)$$
so $a \in \{z \in \mathbb{C} \mid f(z) = g(z)\}$. Therefore, the set is closed.
