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I find it hard to prove that R with lower limit topology is not second countable.

Let's construct a collection B consists of half-open interval [a,b) where a,b $\in$ Q, so there's a mapping from B to Q x Q which is a countable set

Consider an arbitrary open set [x,y), then there's an open set [q1,q2) $\subset$ [a,b) such that q1,q2 $\in$ Q which means B is a countable basis for lower limit topology which means R with lower limit topology is 2nd countable

P/S: I've read some of other posts on this but nothing gave me a satisfied answer

JokingBear
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  • Your question was closed as a duplicate of this one. I suppose that is one of the other posts you've read. Can you tell us what is unsatisfactory about the answer I posted there? – bof May 11 '17 at 08:46
  • Apparently you think that the collection $\mathcal B$ of half-open intervals $[q_1,q_2)$ with rational endpoints is a base for the lower limit topology? It's true that for every (basic) open set $[a,b)$ there is a rational interval $B\in\mathcal B$ such that $B\subset[a,b).$ But that's not enough for a base. If $\mathcal B$ is a base, then $[a,b)$ must be the union of some subcollection of $\mathcal B.$ In other words, for each $x\in[a,b)$ there must be some $B\in\mathcal B$ such that $x\in B\subset[a,b).$ In particular, we must have $a\in B\subset[a,b)$ for some $B\in\mathcal B.$ – bof May 11 '17 at 08:53
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    $[\sqrt2,2)$ is an open set in the lower limit topology, but there is no half-open interval $[q_1,q_2)$ with rational endpoints such that $\sqrt2\in[q_1,q_2)\subset[\sqrt2,2).$ This shows that the set of those rational intervals is not a base for the topology. – bof May 11 '17 at 08:57

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