Dtyjlui

Definition of basis for a topology on $X$ given in the Foundation of Topology By Patty given by

Let $(X,mathscr T)$ be a topological space. A basis for $mathscr T$
is a subcollection $mathscr B$ of $mathscr T$ with the property that
if $Uin mathscr T$ then $U=emptyset$ or there is a subcollection
$mathscr B'$ such that $U=bigcup {B:Bin mathscr B'}$

How do I write the definition of basis for the family of closed sets?

My attempt:-

Let $(X,mathscr U)$ be a family of closed sets(closed sets of topology on $X$). A basis for $mathscr
U$ is a subcollection $mathscr D$ of $mathscr U$ with the property
that if $Din mathscr U$ then $D=X$ or there is a subcollection
$mathscr D'$ such that $D=bigcap {D:Din mathscr D'}$

am I correct?

My next aim is to prove the following Result. This result I have seen in the answer given by Brian M Scott.

A family $mathscr Xsubseteqwp(X)$ is a base for the closed sets
of topology on $X$ iff

• 
  $bigcap_{Din mathscr X} D=varnothing$, and


• if $D_0,D_1inmathscr X$ and $xnotin D_0cup D_1$, then there is a $D_2inmathscr X$ such that $xnotin D_2supseteq D_0cup D_1$.

My attempt for the proof:-

Suppose A family $mathscr Xsubseteqwp(X)$ is a base for the closed sets
of topology on $X.$

claim 1:- $bigcap_{Din mathscr X} D=varnothing$

$varnothing in mathscr U$. so by the definition of basis there is a subcollection $mathscr X'$: $varnothing=bigcap {C:Cin mathscr X'}$. Hence,$bigcap_{Din mathscr X} D=varnothing.$

Claim 2:-if $D_0,D_1inmathscr X$ and $xnotin D_0cup D_1$, then there is a $D_2inmathscr X$ such that $xnotin D_2supseteq D_0cup D_1$

if $D_0,D_1inmathscr X implies D_0,D_1in mathscr U implies D_0 cup D_1 in mathscr U implies exists mathscr X'':D_0 cup D_1 =bigcap {C:Cin mathscr X''}. $ If $xnotin D_0 cup D_1 implies exists C'in mathscr X'': x notin C'.$

Conversely suppose family of subsets $mathscr X$ satisfies the following

I. $bigcap_{Din mathscr X} D=varnothing$, and

II. if $D_0,D_1inmathscr X$ and $xnotin D_0cup D_1$, then there is a $D_2inmathscr X$ such that $xnotin D_2supseteq D_0cup D_1$.

Claim:-A family $mathscr Xsubseteqwp(X)$ is a base for the closed sets
of topology on $X$. Let $mathscr
U$ is a subcollection of $wp(X)$ consisting of $X$ and those sets are the intersection of members of $mathscr X$. Our aim is to prove $mathscr U$ is topology of closed sets on $x$.

(a)$Xin mathscr U(because text{ by the definition of} mathscr U$). $emptyset in mathscr U (because text{ by the definition of} mathscr U $ and (I)).

(b)Arbitrary intersection of closed set lie in the $mathscr U$ immediately.

(c)How do I prove the finite union of elements of $mathscr U$ lie in $mathscr U$?

Let $mathscr D$ be the basis for closed sets that we are working with.

If $Acup B$, then there is nothing to prove. So let us assume that $Xsetminus (Acup B)neemptyset$.

Then we have $A=bigcap mathscr D'_0$, $B=bigcap mathscr D'_1$ for some $mathscr D'_{0,1}subseteqmathscr D$.

Let us now take the set
$$S={(x,D_0,D_1,D_2); D_2inmathscr D, xin X, xnotin D_2; D_0cup D_1subseteq D_2, D_0inmathscr D'_0, D_1inmathscr D'_1}.$$
We want to show that
$$Acup B=bigcap {D_2; (x,D_0,D_1,D_2)in Stext{ for some }x, D_0, D_1}=:C.$$

Since for every $D_2$ as above we have $D_2supseteq D_0 supseteq A$, we get $Csupseteq A$. By the same reasoning we have $Csupseteq B$ and, consequently, $Csupseteq Acup B$.

It remains to show that $Csubseteq Acup B$. Assume that, on the contrary, there exists $xin C$ such that $xnotin Acup B$. But since $xnotin Acup B$ we have $xnotin D_0$ and $xnotin D_1$ for some $D_0inmathscr D'_0$, $D_1inmathscr D'_1$. Therefore there exists $D_2inmathscr D$ such that $xnotin D_2$ and $D_0cup D_1subseteq D_2$. But then $(x,D_0,D_1,D_2)in S$, and from $xnotin D_2$ we get $xnotin C$.