Let $A:D\subset X\to X$ be a linear operator on a Hilbert space. I am trying to understand why the following statement makes sense. Let $a,b\in X$ then, $$\langle a-b,c \rangle = 0 \;\; \forall c\in D(A) \implies \langle a-b,c \rangle = 0 \;\; \forall c\in \overline {D(A)}$$
Since the inner product is a metric, for any limit point $c$, of $D(A)$ the metric should differ only by an infinitesimally small amount, i.e. should be equal.
Is this all there is to it?
No infinitesimals please!
Consider that (fixing $a,b$ for now) that
$$f: X \to \Bbb C, x \to \langle b-a, x \rangle$$ is a continuous function (even Lipschitz, as $|f(x)| \le \|b-a\|\|x\|$) and the condition $\forall x \in D(A): \langle b-a, x \rangle = 0$ just says
$$D(A) \subseteq f^{-1}[\{0\}]$$ where the right hand side is closed as the inverse image of a closed set, and so
$$\overline{D} \subseteq f^{-1}[\{0\}]$$ or $\forall x \in \overline{D(A)}: \langle b-a, x \rangle = 0$
Or more in analysis style, use sequences: if $x \in \overline{D(A)}$ find $x_n$ in $D(A)$ such that $x_n \to x$ and then by continuity of the inner product (you probably covered that?)
$$\langle b-a, x_n \rangle \to \langle b-a,x \rangle$$
and the left hand is a constant sequence of $0$s by assumption and can only converge to $0$.
Both arguments come down to "continuity of inner product", which follows from Cauchy-Schwarz and the way we define the norm and metric etc.