$X$ is an Alexandroff space iff each point in $X$ has a minimal open neighborhood. $X$ is an Alexandroﬀ space iﬀ each point in $X$ has a minimal
open neighborhood.
 A: If $\{U_\alpha\}_{\alpha\in A}$ is a collection of open neighborhoods and $x\in \bigcap \limits_{\alpha \in A}U_\alpha$, $x$ has a minimal neighborhood $U(x)$. Now use this to prove that $x$ is an interior point of $\bigcap \limits_{\alpha \in A}U_\alpha$.
A: If $X$ is an Alexandroff space, for each $x \in X$, define the open set $M_x = \cap\{O: O \text{ open}, x \in O\}$, which is well-defined (as the intersection is of a non-empty collection, as $X$ itself is open and contains $x$, and arbitrary intersections of open sets are open in Alexandroff spaces. If $O$ is then any open set containing $x$, it appears as one of the sets we intersect so trivially $M_x \subset O$, which exactly means it is a minimal neighbourhood of $x$.
Now suppose every $x\in X$ has such a minimal open neighbourhood $M_x$, and let $O_i, i \in I$ be any collection of open sets. If $O = \cap_i O_i$ is empty, it is certainly open, so let $x$ be any point in $O$. Then for all $i$, $x \in O_i$, so $M_x \subset O_i$ by minimality, also for all $i$. This means that $M_x \subset O$, so any point of $O$ is an interior point of $O$, which means $O$ is open. In a formula: $\cap_i O_i = \cup \{M_x: x \in \cap_i O_i\}$, which is open.
A: Proof. 
  ⇒) Suppose $X$ is an Alexandroﬀ space with $x ∈ X$. Let $$O(x) = \{U ⊂
X : U\text{ is an open neighborhood of}\ \  x\}$$. Take $S(x)=\bigcap
U$ for $U ∈ O(x)$, then
$S(x)$ is an open neighborhood of $x$ because $X$ is Alexandroﬀ. And from the
deﬁnition of $S(x)$, it is clear that $S(x)$ is a minimal open neighborhood of $x$.\
⇐) Suppose each $x \in X$ has a minimal open neighborhood $S(x)$. Consider
an arbitrary intersection of open sets, $V =\bigcap  _{α∈A} U_α$, where each $U_α$ is open
in $X$. If $V = \phi$, then $V$ is open and we are done. If $V \neq\phi$  then pick $x ∈ V$
and we have $ x \in U_α$ for all $α \in A$. Hence, $S(x) ⊂ U_α$ for all α because $S(x)$
is the minimal open neighborhood of  $x$. Therefore, $S(x) ⊂ V$ . Hence, $ V$ is
open because it contains an open set around each of it’s points. 
am I right?
