Does $A = \{T\in B(R \ ^{d}) : T(x_0) = 0\} $ is open?

Question

Let $(X,||||)$ be a norm space.

for $T\in Hom(X)$ define $||T||_{op} = sup\{||Tx|| : x \in X , ||x||=1 \}$

Let $B(X) = \{T\in Hom(X) : ||T||_{op}< \infty \}$

I have proved that $(B(X) , ||||_{op})$ is a metric space.

Now I have the following question:for $(B(R \ ^{d}) , ||||_{op})$ with euclidean norm fix $x_0 \in R \ ^ d $ and let $A = \{T\in B(R \ ^{d}) : T(x_0) = 0\} $

Does A closed , open , both or none ?

I managed to prove that A is closed.

I'm not sure how to show that A is open, or maybe it is not?

Thanks for helping

Answer 1

No, if $x_0\neq 0$, $A$ is not open. If $A$ is open, and $T \in A$, $\exists \epsilon>0$ such that $B(T, 2\epsilon) \subset A$.

$\|\epsilon I_d\|_{op} = \epsilon$, so $T'=T+\epsilon I_d \in A$. But $T'(x_0)=\epsilon x_0 \neq 0$. Contradiction.

So $A$ contains no open ball around $T$.

If $x_0=0$ , $A=B(R^d)$, so $A$ is a closed and open set.

Answer 2

As for closedness:

I'll assume here that $d$ is some positive integer. So $\mathbb R^d$.

$B(\mathbb R^d)$ is a linear subspace of $\mathrm{End}(\mathbb R^d)$, which is finite-dimensional, so is itself finite-dimensional (we even have $B(\mathbb R^d) = \mathrm{End}(\mathbb R^d)$ since any linear map that originates from a finite dimensional space is bounded).

Note that $L:T\in B(\mathbb R^d)\longmapsto T(x_0)\in \mathbb R^d$ is a linear map from $B(\mathbb R^d)$ to $\mathbb R^d$, and its kernel is $A$.

Thus $A$ is a linear subspace of the finite-dimensional vector space $B(\mathbb R^d)$, so is closed.