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For each $x\in \mathbb{R}_{\ell}$, there exist $r_1,r_2$ rational numbers such that $r_1<x<r_2$. Then take $\cal{B}=\{[r_1,r_2):r_1<x<r_2,x\in \mathbb{R}_{\ell}\}$.

Then, for each $x\in\mathbb{R}_{\ell}$, there exists a $B\in\cal{B}$ such that $x\in B$.

Also for $B_1,B_2\in \cal{B}$ (say, $B_1=[r_1,r_2)$ and $B_2=[r_3,r_4)$), if WLOG $r_1<r_2\le r_3<r_4$, then $B_1\cap B_2=\emptyset$ and if $r_1<r_3<r_2<r_4$, then $B_1\cap B_2=[r_3,r_2)$, which is also in $\cal{B}$.

Thus $\cal{B}$ satisfies all the properties of a basis of a topology. Since $\cal{B}$ is countable, $\mathbb{R}_{\ell}$ is second countable.

I cannot understand why it is not a basis for $\mathbb{R}_{\ell}$?

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  • $\begingroup$ What is $\mathbb{R}_\ell$ ? $\endgroup$ – Philippe Gaucher Apr 27 '18 at 7:57
  • $\begingroup$ @PhilippeGaucher It's usually $\Bbb R$ with the "lower limit" topology, i.e., with basis all of the $[a,b)$. $\endgroup$ – Lord Shark the Unknown Apr 27 '18 at 8:01
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A set like $A=[\sqrt2,1+\sqrt2)$ is open in the lower-limit topology, but isn't a union of sets $[a,b)$ with $a$, $b$ rational. So $\mathcal{B}$ isn't a basis for $\Bbb R_l$.

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  • $\begingroup$ Ok, I got it; but in the above answer, $\cal{B}$ satisfies the definition of basis of a topology. Is there something I have done wrong or the definition is wrong? $\endgroup$ – pie Apr 27 '18 at 8:20
  • $\begingroup$ @pie It may be a basis for a topology, but that topology isn't that of $\Bbb R_l$. $\endgroup$ – Lord Shark the Unknown Apr 27 '18 at 8:20
  • $\begingroup$ Oh, ok. It is clear now. Thank you. $\endgroup$ – pie Apr 27 '18 at 8:23
  • $\begingroup$ There's a proof that $\Bbb R_l$ is not second-countable in Munkres's book. $\endgroup$ – Lord Shark the Unknown Apr 27 '18 at 8:24

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