general topology – The annihilator $M^perp$ of a set $M neq emptyset$ in an inner product space X is a closed subspace of X.

I’m trying to prove the following:

Show that the annihilator $M^perp$ of a set $M neq emptyset$ in an inner product space X is a closed subspace of X.

Next is the proof I have done, which indeed only proves that $M^perp$ can’t be open, from what I know, this doesn’t imply it is closed, so I need to ensure $M^perp$ is closed. I would really appreciate any corrections to this argument and if possible any orientation on how to prove the statement with basic concepts of the topic. Thanks.

Proof:

i) By definition
$$
M^perp = {x in X: langle x,y rangle = 0 quad forall y in M}
$$

Suppose $x_1,x_2~in M^perp$ and $alpha$ is a scalar. Then
$$
langle alpha x_1+x_2, y rangle = alpha langle x_1,y rangle +langle x_2, yrangle = 0
$$

This means $~alpha x_1 +x_2 ~in M^perp$, then $~M^perp$ is a subspace of X.

ii) Note that if $M={vec{0}}$, then $M^perp = X$ which is close and open, then in particular $M^perp$ is closed. Suppose then that M contains at least one element $yneq vec{0}$. On the other hand if $M^perp = {vec{0}}$, $M^perp$ is a singleton and therefore is closed. Suppose then that $M^perp$ contains at least one $xneq vec{0}$.

If $M^perp$ is an open subset of X and $x_0 in M^perp$ and $x_0neq vec{0}$, there exist a real number $epsilon>0$ such that the ball with the center in $x_0$ and radius $epsilon$ is contained in $M^perp$

$$
B(x_0, epsilon)={xin X: ||x-x_0||<epsilon} subset M^perp
$$

Now, consider the vector $x’ = x_0+beta y~$ where $yin M$, $y neq vec{0}$ and $beta$ is a scalar, it is true that we can choose $beta$ such that
$$
||x’-x_0||= ||x_0+beta y-x_0||=||beta y||=|beta|~||y||<epsilon longrightarrow x’ in B(x_0,epsilon)
$$

Since $yin M rightarrow langle x_0, yrangle = 0$, and therefore
$$
langle x’,yrangle = langle x_0+beta y, yrangle = langle x_0, yrangle + betalangle y,y rangle = beta langle y,yrangle neq vec{0}. ~Since ~y neq vec{0}.
$$

then $x’ notin M^perp$. In summary, we can always form $x’$ such that $x’ in B(x_0,epsilon)$ but $x’ notin M^perp$. This mean, $B(x_0,epsilon) notsubset M^perp$, which implies $x_0$ is not an interior point of $M^perp$, in consequence $M^perp$ is not open.

Distance between a functional element and the annihilator

To let $ V $ be a normalized vector space and $ W $ a subspace of $ V $, We designate the annihilator of $ W $ by $ W ^ perp $, I have to show that if $ f in V ^ prime $, then $$ d (f, W ^ perp) = | f | _W | _ {W ^ prime}. $$
I've already proven that $ | f | _W | _ {W ^ prime} leqslant d (f, W ^ perp) $ in the following way: for everyone $ f in V ^ prime $ and $ g in W ^ perp $, we have that begin {eqnarray}
| fg | _ {V ^ prime} & = & sup_ {v in V, | v | = 1} | (fg) (v) | = sup_ {v in V, | v | = 1} | f (v) -g (v) | \
& geqslant & sup_ {v in W, | v = 1} | Underbrace {f (v)} _ {= f | _W (v)} – Underbrace {g (v)} _ {= 0, Text {for} v in W} | \
& = & sup_ {v in W, | v | = 1} | f | _W (v) | \
& = & | f | _W | _ {W ^ prime};
end {eqnarray}

and we get that $ | f | _W | _ {W ^ prime} leqslant | f-g | _ {V ^ prime} $for each $ g in W ^ perp $, Therefore $$ | f | _W | _ {W ^ prime} leqslant inf_ {g in W ^ perp} | fg | _ {V ^ prime} = d (f, W ^ perp). $$ But I can't find a way to show the other inequality.

Agal Algebraic Geometry – Annihilator of an element and Jacobson radical

To let $ R $ Be commutative ring with 1. Is there a characterization for an element? $ a $ from $ R $ so that $ ann (1-a) subseteq J (R) $, from where $ ann (x): = {r in r rx = 0 } $ to the $ x in R $ and $ J (R) $ is the Jacobson radical of $ R $ (the intersection of all maximal ideals of $ R $).

I would appreciate comments.