the Sorgenfrey topology is characterised by the base $\mathcal{B} = \{[a,b) \:{:}\: a,b \in \mathbb{R}\}$. However, I would like to know if there exists any other base for the Sorgenfrey topology. My intuition says that there can't be any other as this base defines the Sorgenfrey topology. Thank you all in advance.
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4I guess $ {[a,b):;a\in\mathbb{R},b\in\mathbb{Q}}$ would do just as well. – Ivan Neretin Mar 11 '19 at 07:02
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2A base $\mathscr{B} \subsetneq \mathscr{T}$ for a topology $\mathscr{T}$ is never unique since $\mathscr{T}$ is a base for itself. – parsiad Mar 11 '19 at 07:06
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I suspect there are $2^{c}$ many bases for the Sorgenfrey topology, and also I suspect this can be proved similarly to how I proved here (see the Question paragraph) that there are $2^c$ many bases for the usual topology on ${\mathbb R}.$ – Dave L. Renfro Mar 11 '19 at 07:29
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Thank you all for the generous helps !! – James Mar 21 '19 at 18:27
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If $D$ is a dense subset of $\mathbb{R}$ (usual topology), like $\mathbb{Q}$ or the irrationals, then
$$\mathcal{B} = \{[a,b): a \in \mathbb{R}, d \in D\}$$
is also a base for the Sorgenfrey topology, as can easily be checked.
Or take all standard countable local bases together:
$$\{[a,a+\frac{1}{n}): a \in \mathbb{R}, n \in \mathbb{N}^+\}$$
etc.
Henno Brandsma
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