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I'm having a bit of trouble proving this:

The definition of Lower Limit Topology I am working with: $ \{[a, b) \subseteq \mathbb R \ \text s.t \ a < b\}$. The only thing I can think of is that since the rationals are a countable set in the Reals, there will only be countable many intervals and their union can not equal the entire lower limit topology, or in other words there are not enough intervals B in $\mathcal B_{\mathbb Q}$ s.t $\forall x \in$ every open set U of the lower limit topology $\exists B \in \mathcal B_{\mathbb Q} $ s.t $x \in B \in U$. I can't figure out how to rigorously formulate this.

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    $\begingroup$ Your argument will not work: note that the analogous statement for the Euclidean topology on $\mathbb R$ does hold. Hint: consider $[\pi,4)$. $\endgroup$ – Mees de Vries May 18 '16 at 22:49
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    $\begingroup$ It seems like you are trying to formulate something along the loose form of 'the cardinality of the set $B_{\Bbb Q}$ is too low', which would be more formally 'the cardinality of any topology generated by a countable basis will be strictly smaller than the cardinality of the lower limit topology'. However, this statement is false since the ordinary topology is generated by a countable basis, and with some work you can show that the ordinary topology and the lower limit topology have the same cardinality (the continuum). $\endgroup$ – Rolf Hoyer May 18 '16 at 22:57
  • $\begingroup$ Thank you both. @Mees your counter example makes things so much easier. $\endgroup$ – ಠ_ಠ May 18 '16 at 22:57
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What does it mean to say that a collection $\mathcal B$ of open sets is a base? It means that, given any point $a$ and any neighborhood $U$ of $a,$ we can find a set $B\in\mathcal B$ with $a\in B\subseteq U.$

To show that $\mathcal B_\mathbb Q$ is not a base, all we have to do is find a point $a$ and a neighborhood $U$ of $a$ such that there is no set $B\in\mathcal B_\mathbb Q$ with $a\in B\subseteq U.$

Looking at the definition of $\mathcal B_\mathbb Q,$ I think I'd try taking a irrational number for $a,$ say $a=\pi.$

Looking at the definition of the Sorgenfrey topology, oops, I mean the "lower limit topology", I'd try taking a neighborhood of $\pi$ which is not a neighborhood in the usual topology, say, the interval $[\pi,\infty).$

There we have it. Is there a set $B\in\mathcal B_\mathbb Q$ such that $\pi\in B\subseteq[\pi,\infty)?$ If we can prove that there is no such $B,$ then everything is fine.

Hmm. Suppose $B=[p,q)\in\mathcal B_\mathbb Q.$ Now there are three cases: either $p\lt\pi,$ or $p=\pi,$ or $p\gt\pi$ . . .

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