A topology is second countable if there exist a countable basis. For the Sorgenfrey line, each basis are of the form [a,b) where a and b are both real numbers. This is uncountable, but if I restrict them to only rationals, they it would be countable right?
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1The collection $[p, q)$ with $p, q$ rational is not a basis for the Sorgenfrey line. For e.g. there is no way to produce $[\sqrt{2}, 3)$ with that collection. – balddraz Jan 13 '21 at 07:05
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A base element could also have the form $(a,b)$ (Also open). The standard base from the definition are sets of the form $[a,b)$, but a general base can have many open sets of other forms too. – Henno Brandsma Jan 13 '21 at 14:41
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If $\mathcal{B} $ is a basis for the Sorgenfrey line, then for any $b \in \mathbb R$, since $b \in [b,\infty)$ and the latter set is open, we can find $V_b \in \mathcal B$ such that $b \in V_b \subseteq [b,\infty)$. Notice that $\inf V_b =b$. Thus, for $b_1 \neq b_2$ we get that $V_{b_1}\neq V_{b_2} $. In other words, the map $b \mapsto V_b$ is injective and $\mathcal{B}$ has cardinality at least $\mathfrak{c}=|\Bbb R|$.
Henno Brandsma
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Evangelopoulos Foivos
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