- Apply Lemma 13.2 to show that the countable collection
\[ \mathcal{B} = \ts{(a, b) \mid a < b, \ a \text{ and } b \text{ rational}} \]
is a basis that generates the standard topology on \(\mathbb{R}\).
- Show that the collection
\[ \mathcal{C} = \ts{[a, b) \mid a < b, \ a \text{ and } b \text{ rational}} \]
is a basis that generates a topology different from the lower limit topology on \(\mathbb{R}\).
Solution
The family \[ \mathcal B=\{(a,b):a,b\in\mathbb Q,\ a<b\} \] is countable because \(\mathbb Q^2\) is countable. Given an ordinary open set \(U\) and \(x\in U\), choose an interval \((r,s)\) with \[ x\in(r,s)\subseteq U. \] By density of \(\mathbb Q\), choose rationals \(a,b\) with \[ r<a<x<b<s. \] Then \(x\in(a,b)\subseteq U\). By the basis criterion, \(\mathcal B\) generates exactly the standard topology.
The family \[ \mathcal C=\{[a,b):a,b\in\mathbb Q,\ a<b\} \] covers \(\mathbb R\). If \(x\in[a,b)\cap[c,d)\), put \(r=\max(a,c)\). If \(x=r\), take \(e=r\); otherwise choose rational \(e\) with \(r<e<x\). In either case choose rational \(f\) with \[ x<f<\min(b,d). \] Then \(x\in[e,f)\subseteq[a,b)\cap[c,d)\), so \(\mathcal C\) is a basis.
Every member of \(\mathcal C\) is lower-limit open, so the topology generated by \(\mathcal C\) is no finer than \(\mathbb R_\ell\). It is strictly coarser: for irrational \(\alpha\), the lower-limit open set \([\alpha,\alpha+1)\) is not \(\mathcal C\)-open. If a basic set \([a,b)\in\mathcal C\) contained \(\alpha\) and were contained in \([\alpha,\alpha+1)\), then \(a\le\alpha\) and containment would force \(a\ge\alpha\), hence \(a=\alpha\), impossible because \(a\) is rational.