I'm looking for some topologies on $\mathbb{R}$ that do not have a countable basis. Specifically, I want something where some of the open sets can not be written as a countable union of basis elements. I'm trying to figure out why the definition of a topology states that the union of arbitrarily many open sets is open and I haven't found any examples that require more than a countable union.
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2Welcome to Mathematics Stack Exchange. How about the discrete topology? – J. W. Tanner Jan 26 '21 at 21:37
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Give an uncountable space the particular point topology. (This example is also connected.) – Randall Jan 26 '21 at 21:41
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Some fairly familiar topologies on $\Bbb R$ that do not have countable bases are the discrete topology, the cofinite topology, the co-countable topology, and the Sorgenfrey topology (also called the lower limit topology). One that has some importance as a counterexample is the Michael line, and another is the rational sequence topology.
Brian M. Scott
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Take the discrete topology on $\Bbb R$ as a very easy example.
The Sorgenfrey line is another classic: this has as basic open sets all sets of the form $[x,y)$ where $x < y$. This is separable ($\Bbb Q$ is dense) but does not have a countable base.
The Michael line topology on $\Bbb R$ is the usual topology on $\Bbb R$ where we add all irrational points as isolated points (so it has subbase $\mathcal{T}_e \cup \mathscr{P}(\Bbb P)$ e.g.)
Henno Brandsma
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