Can I define a closed subset in a normed vector space as below?

Question

For $A \subset V$, where ($V$, $||.||$) is a normed space,

if $\forall$ $v \in A$ and some $\delta > 0$,
$B$(v,$\delta$) = {x $\in V$ | $||x-v|| \leq \delta$ } and $B$(v,$\delta$) $\subseteq A$

Then A is closed in V


Most books define it as if $A^c$ is open in V then $A$ is closed in $V$. Is there any difference in both ways?

Answer 1

Let $A$ an open set, then for every $x \in A$, we can find some $r>0$ such that $\overset{\circ}B(x,r) \subset A$ (where $\overset{\circ}X$ is the interior of $X$). In the meantime we have $B(x,\frac r 2) \subset A$.

$A$ containing a closed set does not mean $A$ is closed, since, as the little proof above shows, every open set in such space contains a closed ball.

In the definition of a topology, it is said that open sets are stable under finite intersection. An equivalent definition would be that closed sets are stable under finite union (using the fact that $U$ open is the same as $U^c$ closed and De Morgan's law). So a set $A$ is closed iff it can be written as a finite union of closed sets.

The only way to show that a set is closed using finite balls union is then to find a norm $\| \cdot \|$ well suited for the set you are facing. Trying to write a closed ball for the Euclidean norm in $\mathbb{R}^n$ as the finite union of closed ball for the sup norm is ... quite challenging (impossible) and the converse has the same difficulty, but those two sets are clearly closed. I hope I made clear why there is no point in defining closed sets using closed balls.

Answer 2

No, you caannot deduce from that considtion that $A$ is closed. If, for instance $A=B(0,2)$, then $A$ is not closed. However, your condiction holds: just take $v=0$ and $\delta=1$, for instance.

Answer 3

For $x\in V, r>0$, let $D(x,r)=\{v\in V: \|x-v\|\le r\}$, which is the closed ball of radius $r$ centered at $x$ and let $B(x,r)=\{v\in V: \|x-v\|< r\}$ the open ball of radius $r$ centred at $x$

Then the conditions

$$\forall x \in A: \exists r>0: D(x,r) \subseteq A \tag{1}$$

$$\forall x \in A: \exists r>0: B(x,r) \subseteq A \tag{2}$$

are equivalent for all subsets $A$ of $V$.

Proof: Suppose $A$ satisfies $(1)$. We can take the same $r$ at every $x \in A$ to see that $A$ satisfies $(2)$, as $B(x,r) \subseteq D(x,r)$ for all $x,r$. OTOH if $A$ satisfies $(2)$ and we have $x \in A$ and the promised $r$, we see that $D(x,\frac{r}{2}) \subseteq B(x,r)$ (any $s < r$ will do, but I chose $\frac{r}{2}$ for convenience) and so $A$ satsifies $(1)$.

Condition (2) is the definition of open sets so your proposed condition (1) is also another definition of open sets, not of closed sets.

Indeed, a set $A$ is closed by definition if $A^\complement=V\setminus A$ is open, and this can be shown to be equivalent to

$$\forall x \in V: (\forall r>0: B(x,r) \cap A \neq \emptyset) \implies (x \in A)$$ as a more direct way of checking closedness using open balls (we could, by the same reasoning as above, have replaced $B(x,r)$ by $D(x,r)$ again, but using open balls is more common and natural, often).