## 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 $$M^perp$$

$$B(x_0, epsilon)={xin X: ||x-x_0||
Now, consider the vector $$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||
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.

## ra.rings and algebras – Annihilator of an element in a ring

Let $$a$$ and $$b$$ be two elements of a commutative ring $$R$$ with 1 such that $$ann(a) ne ann(b)$$.

is it always possible to find a sequence of elements $$a_1,dots,a_k in R$$ such that $$a in ann(a_1), a_1 in ann(a_2), dots, a_{k-1} in ann(a_k),$$ and $$a_k in ann(b)$$?

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.
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 $$R$$).