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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?

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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 $X$.

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  • $\begingroup$ 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? $\endgroup$ – Pedro Gomes Oct 7 '18 at 14:31
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    $\begingroup$ 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. $\endgroup$ – drhab Oct 7 '18 at 14:37

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