- #1

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Suppose [tex]X[/tex] is a set, [tex]\mathcal{B}:=\{S\subset{}X:\bigcup{}S=X\}[/tex], \\

[tex]\mathcal{T}:=\{U\subset{}X:U=\bigcup_{\alpha\in{}A}\bigcap_{i\in{}I_\alpha}S_i\}[/tex], where [tex]I_\alpha[/tex] is a finite index set, and [tex]A[/tex] is an arbitrary index set.

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

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

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