# Subbasis help

1. Sep 7, 2009

### littleHilbert

Hello, could you please check if the reasoning is correct. This is not a homework, just a part of an exercise in a book I'm reading at the moment.

Suppose $$X$$ is a set, $$\mathcal{B}:=\{S\subset{}X:\bigcup{}S=X\}$$, \\
$$\mathcal{T}:=\{U\subset{}X:U=\bigcup_{\alpha\in{}A}\bigcap_{i\in{}I_\alpha}S_i\}$$, where $$I_\alpha$$ is a finite index set, and $$A$$ is an arbitrary index set.

(b) It is asked to prove that $$\mathcal{T}$$ is the smallest topology for which all the sets in $$\mathcal{B}$$ are open, i.e. $$\mathcal{T}=\bigcap_{\tau\supset\mathcal{B}}\tau$$ is the intersection of all topologies containing $$\mathcal{B}$$.

Let $$U$$ be an open set, contained in every topology, where $$S\in\mathcal{B}$$ is among open sets. Then being a subset of $$X$$ the set $$U\subset\bigcup_{S\in\mathcal{B}}S$$, so every $$x\in{}U$$ is in some $$S_{\beta}\in\mathcal{B}$$. By assumption every finite intersection of sets in $$\mathcal{B}$$ is open. Intersecting $$S_\beta$$ with finitely many elements in $$\mathcal{B}$$ that contain $$x$$ will give us the open subset $$\bigcap_{S\in\mathcal{B}'\subset\mathcal{B}}S\subseteq{}S_\beta$$. Taking the union $$\bigcup_{x\in{}U}\bigcap_{S\in\mathcal{B}'\subset\mathcal{B}}S$$ we thus obtain an open (in each $$\tau$$) set, which is of the form given by $$\mathcal{T}$$.

Conversely, let $$U\in\mathcal{T}$$ be open. We need to show that $$U$$ is open in every topology $$\tau$$, containing $$\mathcal{B}$$. For some $$\alpha\in{}A$$ an element $$x\in{}U$$ is contained in $$V:=\bigcap_{i\in{}I_\alpha}S_i$$. But on the other hand $$V$$ is open in $$\tau$$ and is a subset of $$U$$. This means that every point $$x\in{}U$$ is contained in a subset of $$U$$ that is open in $$\tau$$. Hence $$U$$ is a union sets open in $$\tau$$ and is thus open in $$\tau$$.