i need help to prove this problem(functional analysis) show that the annihilator of a set M in an inner product space X is a closed subspace of X.

 A: Here's a way to do it that doesn't explicitly use sequences:
Let the base field be $\Bbb F$, where $\Bbb F = \Bbb R \; \text{or} \; \Bbb C$.
Start with the definition of $M^\bot$:
$M^\bot = \{ x \in X \mid \langle m, x \rangle = 0, \forall m \in M \}; \tag{1}$
next, note that for any $m \in M$, the map
$\phi_m:X \to \Bbb F, \; \phi_m(x) = \langle m, x \rangle \tag{2}$
is a continuous linear functional from $X$ to $\Bbb F$; indeed, it is bounded since, by Cauchy-Schwarz, $\vert \phi_m(x) \vert = \vert \langle m, x \rangle \vert \le \Vert m \Vert \Vert x \Vert$; linearity follows directly from the axiomatic properties of the inner product $\langle \cdot, \cdot \rangle$ on $X$.  Since $\phi_m$ is continuous, $\ker \phi_m$ is a closed subspace of $X$.  Now it is easy to see that
$M^\bot = \bigcap_{m \in M} \ker \phi_m; \tag{3}$
indeed, (3) may be seen as a re-statement of the defintion (1); finish off with the fact that the intersection of a any set of closed subspaces is again a closed subspace.  QED.
Nota Bene:  I think it is worth pointing out that closed here means the general topological notion of closure:  a set $Y$ is closed if and only if its complement is open.  Sequential closure can't be invoked here since don't know that $X$ is Cauchy complete, i.e. that Cauchy sequences actually have limit points, in $X$.  End of Note.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
