Does $\{A={\rm core}(A)\}$ form a topology? I've the following question: If $V$ be an arbitrary (real) vector space and if for any subset $A\subset V$ we denote by $ {\rm core}(A)$ the set $\{a\in A\> |\> \forall v\in V \exists T>0 \>\forall \,|t|<T: a+tv \in A  \}$. Then $ {\rm core}(A)$ is the set of all $a\in A$ such that in every direction $v$ there is a environment of $a$ on the line $a+tv$ which also belongs to $A$ (I hope this set is really call the core of $A$...).
Now it seems tempting to call set $A$ open if $A={\rm core}(A)$.
Does the family $\{A={\rm core}(A)\}$ form a topology on $V$?
(I think I once read that it does not... But arbitrary unions of sets of this form as well as finite intersections seem to belong this this family again...)
Thanks a lot in advance for any suggestions
 A: Let us denote
$$\mathcal{C}=\{A\subseteq V\,:\,A=\mathrm{core}(A)\}$$

Claim: $\mathcal{C}$ is a topology on $V$.

Proof: Clearly, $\emptyset,V\in\mathcal{C}$.
Let $(U_i)_{i\in I}\subset \mathcal{C}$ be a collection in $\mathcal{C}$. Then also $U=\bigcup_{i\in I} U_i\in \mathcal{C}$: we only need to show $U\subseteq \mathrm{core}(U)$. Pick $a\in U$. Then there exists $i_0\in I$ such that $a\in U_{i_0}$. Since $U_{i_0}\in\mathcal{S}$ we have 
$$\forall v\in V\exists T>0 \forall |t|<T:a+tv\in U_{i_0}\subseteq U$$
This implies $a\in \mathrm{core}(U)$, so $U\in\mathcal{C}$.
Let $(U_i)_{i=1,\dots,n}\subset \mathcal{C}$ be a finite collection in $\mathcal{C}$. Then also $U=\bigcap_{i=1}^n U_i\in\mathcal{C}$: again let $a\in U$. Then $a\in U$ for all $i=1,\dots,n$. Let $v\in V$ be arbitrary. Then there exist $T_1,\dots,T_i>0$ such that for all $|t|<T_i$ we have $a+tv\in U_i$. Now set $T:=\min_{i=1,\dots,n} T_i$. Then, for $|t|<T\le T_i$ we have $a+tv\in U_i$ for all $i=1,\dots,n$, so $a+tv\in U$. That is, $U\in\mathcal{C}$. $\square$
Addendum:
We can say a bit more about this topology:
Note that, if $V$ is equipped with a norm $\|\cdot\|$, then $\mathcal{C}$ contains all open balls, i.e. sets of the form
$$B(x,r)=\{y\in V\,:\,\|x-y\|<r \}$$
This corresponds to the case where the parameter $T$ doesn't depend on the direction $v$.
But there are also sets in $\mathcal{C}$ which are not open with respect to the norm topology generated by $\|\cdot\|$.
Consider for example $V=\mathbb{R}^2$ with the Euclidean norm and define for $x\in\mathbb{R}^2$, the argument of $x\not=0$ to be the unique number $\varphi=\mathrm{arg}(x)\in [0,2\pi)$ such that
$$x=(\|x\| \cos(\varphi),\|x\|\sin(\varphi))$$
Also set $\mathrm{arg}(0)=0$.
Then notice that for all $x\in V$ we have $2\pi - \mathrm{arg}(x)>0$. Therefore, the set
$$M=\{x\in\mathbb{R}^2\,:\,|x|<2\pi -\mathrm{arg}(x)\}$$
is in $\mathcal{C}$. But it is not open with respect to $\|\cdot\|$ because $0\in M$, but there is no open ball around $0$ which is contained in $M$. You can understand $M$ as a sort of distorted ball where the radius depends on the direction.
