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

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.
