Let $X$ be a non-empty set and $S$ the collection of all sets $X\setminus\{x\},x\in X$. Prove that $S$ is a sub basis for the finite -closed topology on $X$.
My attempt:
Considering the co-finite topology to be $\tau$, there exists the topological space $X,\tau$. If we have $x_1,x_2\in X$, $\{x_1\},\{x_2\}\notin \tau$, since the singletons are finite. And also considering $x_1\neq x_2$.
$(X\setminus\{x_1\})\cap(X\setminus\{x_2\})=X\setminus\{x_1\}\cup\{x_2\}=X\setminus\{x_1,x_2\}$
By induction we could consider for $n<\infty$
$x_1,x_2...x_n\in X$,so that $(X\setminus\{x_1\})\cap(X\setminus\{x_2\})...(X\setminus\{x_n\})=X\setminus\{x_1\}\cup\{x_2\}\cup...\{x_n\}=X\setminus\{x_1,x_2,...x_n\}$
Since the sets singletons chosen were arbitrary the infinite intersections of different infinite $X\setminus\{x_i\}$ sets generates all the sets whose complement is finite.
Question:
1) Is my proof right? If not how should I proceed?
2) After some search and some thinking I was not able to find a finite-closed topology basis, only mentions of the sub basis? Why? What is a possible basis?