My try on an exercise of Alexandrov topological spaces

Question

I did this one but I don't know if my way to attack this kind of problems is the right one:

Let $X$ be an $A$-space (Alexandrov space, a space where arbitrary intersections of open sets are open. Given a subset $A\subseteq X$, we define the open cover $U(A)$ as the intersection of all open sets containing $A$. Prove the following:

$i)$ If $A\subset V$, where $V$ is an open set, then $U(A)\subset V$. So $U(A)$ is the smallest open set containing $A$.

$ii$)Open sets $U_x=U(\{x\})$ define a basis for the topology of $X$.

Here's my try (a bit loose, but I want to know if that sketch is correct):

$i)$ If $V$ is the only open set such that $A\subset V$, then $V$ is the open cover of $A$ and we have finished.

If we have more than one set, we intersect them and that intersection will be open and will contain $A$ ($X$ is Alexandrov). Thus $\cap V_i$, with $V_i$ containing $A$ for all $i\in I$, is the open cover we're looking for.

So $U(A)\subset V$.

$ii$) We have to see now that open sets $$U(\{x\})=\bigcap_{i\in J} V_i, \ \ x\in V_i \ \ \forall i$$

are a basis.

So we have to see that every point on $X$ has an element of the basis (it's obvious, $z\in X$ has the "basis" (we cannot call it basis yet) element $U(\{z\})$).

And that the points $z$ living on the intersection of two "basis" elements $U_x$ and $U_y$ are contained in a $U_z\subset U_x\cap U_y$.

If $U_x\cap U_y$ it's empty, it's straightforward.

If $U_x\cap U_y$ contains an element, say $z$, the intersection $U_z=U_x\cap U_y$ is an open set containing $z$ (using $X$ is Alexandrov). We can find that this is the open cover, or the open cover is smaller so it's contained there. Anyway, we found that $U_z$.

Is my solution correct?

Thanks for your time.

Answer 1

You are somewhat on the right track, you just need formal precision. So lets start with

$i)$ If $A\subset V$, where $V$ is an open set, then $U(A)\subset V$. So $U(A)$ is the smallest open set containing $A$.

Let's do this formally. In a given poset $(P,\leq)$ an element $x\in P$ is the smallest element if $x\leq y$ for all $y\in P$. Such element is unique if it exists. So our poset is

$$P=\{U\ |\ U\text{ is open and contains }A\}$$

and the $\leq$ ordering is simply the inclusion $\subseteq$. So lets apply the definition to $U(A)$.

Let $V$ be any open subset containing $A$. Then $U(A)\subseteq V$ because $U(A)$ is the intersection of all open subsets containing $A$, in particular $V$ is a part of that intersection as well.

Since $V$ was chosen arbitrarly then $U(A)$ is the smallest element.

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As for the second part. First of all

So we have to see that every point on $X$ has an element of the basis

is not a correct definition of basis. That would imply that $\{X\}$ is a basis of any topology. The correct definition is: "a collection of open subsets is a basis if every open subset can be written as a union of elements from that collection".

So lets get to the point:

$ii$)Open sets $U_x=U(\{x\})$ define a basis for the topology of $X$.

So let $V\subseteq X$ be an open subset. Then

$$V=\bigcup_{x\in V}\{x\}\subseteq \bigcup_{x\in V}U(\{x\})$$

because $x\in U(\{x\})$. On the other hand each $U(\{x\})\subseteq V$ for $x\in V$, meaning that the "$\supseteq$" inclusion holds as well. Therefore

$$V=\bigcup_{x\in V}U(\{x\})$$

which completes the proof.