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The extended real line can be given the order topology i.e the topology generated by the basis elements $(a,b)=\{a<x<b\},R_a=\{x>a\}$ and $L_a=\{x<a\}$ where $a,b\in \bar R$.

Hence, a countable basis would just be the above sets except with the restriction that $a,b\in \mathbb Q\cup \{+\infty,-\infty\}$

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  • $\begingroup$ Well the real line is second countable? It's just not if you use the lower limit topology. You can check the wikipedia article. en.wikipedia.org/wiki/Second-countable_space $\endgroup$ – 0CT0 Aug 29 '19 at 15:27
  • $\begingroup$ @0CT0 An extension of a second countable space need not always be second countable. So it's a reasonable question. $\endgroup$ – Henno Brandsma Aug 29 '19 at 17:04
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Yes, that's true. Restricting the base elements to rational endpoints (in all 3 types) gives a countable base, as Q is order dense in the extended reals: for any $a < b$ we can find a rational strictly in-between.

As all order topologies are $T_3$ (even $T_5$) Urysohn's metrication theorem already tells us it's metrisable (which is also obvious if you know that $\bar{\Bbb R}$ is homeomorphic to $[-1,1] \subseteq \Bbb R$.)

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  • $\begingroup$ Hi just to clarify the points need only come from $\mathbb Q$ and not necessarily $\mathbb Q\cup \{+\infty,-\infty \}$? $\endgroup$ – Jhon Doe Aug 30 '19 at 10:48
  • $\begingroup$ @JhonDoe no $\Bbb Q$ more than suffices. $\endgroup$ – Henno Brandsma Aug 30 '19 at 10:52

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