Exercise · ZW-BASIS-OF-A-TOPOLOGY-007 · ★★☆ Intermediate
In \(\mathbb{R}\) with the usual topology , let \(\mathcal{B}=\{(p,q):p<q\ \text{rational}\}\). Which statement is correct?
Choose one option.
A It is not a basis: intervals with irrational endpoints are missing B It is not a basis: it is countable and bases must be uncountable C It is a basis: around each point of an open set there fits an interval with rational endpoints inside the set D It is a basis only for the subspace \(\mathbb{Q}\) Show answer C. It is a basis: around each point of an open set there fits an interval with rational endpoints inside the set← Back to the question
Explanation If \(x\in U\) is open, there is \(\varepsilon>0\) with \((x-\varepsilon,x+\varepsilon)\subseteq U\), and by density of \(\mathbb{Q}\) there is \((p,q)\) with rational endpoints such that \(x\in(p,q)\subseteq U\). So \(U\) is a union of members of \(\mathcal{B}\).
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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