Exercise · ZW-BASIS-OF-A-TOPOLOGY-009 · ★★★ Advanced
Let \(\mathcal{B}\) be a basis of \(\mathcal{T}\), \(U\in\mathcal{T}\) and \(x\in U\). Why is there \(B\in\mathcal{B}\) with \(x\in B\subseteq U\)?
Choose one option.
A Because \(U\) is a union of members of \(\mathcal{B}\) and \(x\) lies in one of them B Because \(U\) belongs to \(\mathcal{B}\) C Because \(\mathcal{B}\) contains all the singletons D Because \(x\) is an interior point of \(\mathcal{B}\) Show answer A. Because \(U\) is a union of members of \(\mathcal{B}\) and \(x\) lies in one of them← Back to the question
Explanation Since \(U=\bigcup B_i\) with \(B_i\in\mathcal{B}\), the point \(x\in U\) lies in some \(B_i\), and \(B_i\subseteq U\). Neither \(U\in\mathcal{B}\) nor singletons are needed.
A basis \(\mathcal{B}\) of a topology \(\mathcal{T}\) on \(X\) is a family of open sets such that every open set \(U\) of \(\mathcal{T}\) is a union of members of \(\mathcal{B}\).
\[\forall U\in\mathcal{T}:\ U=\bigcup\{B\in\mathcal{B}:B\subseteq U\}\]
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