# Sub basis for the finite-closed topology

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?

First some theory.

In general every collection $$\mathcal S$$ of subsets of $$X$$ induces a basis $$\mathcal B$$ of a topology of which $$\mathcal S$$ serves as subbasis.

This by:$$\mathcal B:=\{\cap\mathcal T\mid\mathcal T\subseteq\mathcal S\text{ and }\mathcal T\text{ is finite}\}$$

where $$\cap\mathcal T:=\{x\in X\mid\forall T\in\mathcal T\; x\in T\}$$ and the convention that $$X:=\cap\varnothing$$ is practicized (so that $$X\in\mathcal B$$).

So actually the elements of $$\mathcal B$$ are exactly the finite intersections of elements of $$\mathcal S$$ and $$X$$ is defined as empty intersection.

Characteristic for $$\mathcal B$$ is that its elements cover the whole space (simply because $$X\in\mathcal B$$) and secondly that it is closed under finite intersections.

These are exactly the characteristics of a basis of a topology, and this topology is the collection:$$\tau:=\{\cup\mathcal V\mid\mathcal V\subseteq\mathcal B\}$$i.e. its elements are unions of elements of $$\mathcal B$$.

Now let's apply this on $$\mathcal S:=\{\{x\}^{\complement}\mid x\in X\}$$ as in your question.

Then we find the basis: $$\mathcal B=\{X\}\cup\{F^{\complement}\mid F\subseteq X\text{ and }F\text{ finite}\}$$

It is not difficult to see that this collection is closed under unions so that we find for the topology:$$\tau:=\{\cup\mathcal V\mid\mathcal V\subseteq\mathcal B\}=\mathcal B$$

So here we meet the special case that the basis is already is topology itself.

Further it is evidently the cofinite topology on $$X$$ justifying the conclusion that $$\mathcal S$$ is a subbasis for the cofinite topology on $$X$$.

• Thanks for your answer! When you define $\mathscr{B}$ as the basis of co-finite topology are you considering the result of my attempted proof? – Pedro Gomes Oct 7 '18 at 14:31
• Actually I do not define $\mathcal B$ as basis of co-finite topology. I define it as the basis induced by the complements of singletons. Then looking at it I "discover" that this basis is also a topology itself, and that it is exactly the co-finite topology on $X$. Proved is then that $\mathcal S$ is a subbasis for that topology, as is to be shown according to your question. – drhab Oct 7 '18 at 14:37