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I am reading topology from Munkres book. While reading the countability and Separation axioms, I came across several references to Lower limit topology ($\mathbb{R}_l$) which essentially comprises of basis elements of the form $\{[a,b)\mid a<b, a,b\in \mathbb{R}\}$.

These are some of the properties for $\mathbb{R}_l$ mentioned in the book.

1) It is Hausdorff

2) It is first countable but not second countable

3) It is normal

Here is my understanding of the proofs for these properties.

1) $\mathbb{R}_l$ is Hausdorff as if I consider two points $x$ and $y$ with $x<y$, I can find disjoint open sets $[x,y)$ and $[y,y+\delta)$ containing $x$ and $y$, respectively.

2) It is first countable as every point $x$ in $\mathbb{R}_l$ has a countable basis, but I don't understand why it is not second countable, as it should have countable basis for this topology. Is it because if I consider an open set for the standard topology on $R$, say $(a,b)$, this set is an union of uncountable open sets in $\mathbb{R}_l$

$(a,b)=\cup_{a<x<b}[x,b)$

3) I understand the proof given in the book for $\mathbb{R}_l$ to be normal.

Please tell me if my understanding of these concepts is correct. I would appreciate if someone can provide more deeper insight about these concepts.

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  • $\begingroup$ Assume you have a countable base $[a_n,b_n)$. Now take a point $x$ distinct from any $a_n$. Can you find a basic neighborhood of $x$ within $[x,\infty)$ ? $\endgroup$ – Stefan Hamcke Jul 3 '15 at 17:25
  • $\begingroup$ I suppose I can. It should be of the form $[x,y)$ where $y<b_n$. Is that right? And I don't know how it proves it non second countable. $\endgroup$ – Khushboo Jul 3 '15 at 17:28
  • $\begingroup$ But this set $[x,y)$ is not in the countable base since $x$ was chosen to be distinct from any $a_n$. $\endgroup$ – Stefan Hamcke Jul 3 '15 at 17:30
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    $\begingroup$ For $\Bbb R$ with standard topology it works. The difference is that given an $a\in\Bbb R$, you don't have set of the form $[a,b)$. So you don't have an open set which is "as small as possible to the left of $a$". Every neighborhood of $a$ must contain some interval $(c,d)$ for $c<a$, and then you can find a rational number $q$ with $c<q<a$. $\endgroup$ – Stefan Hamcke Jul 3 '15 at 17:40
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    $\begingroup$ Right you can take as a base the intervals $(a,b)$ with rational $a,b$. Note that $\Bbb Q$ with the lower limit topology is second countable. $\endgroup$ – Stefan Hamcke Jul 3 '15 at 18:16
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Assume $X=(\Bbb R,\tau_l)$ were second-countable. One can prove that in a second countable space, every base $\cal B$ contains a countable base, so we can assume that the countable base $\cal C$ has only sets of the form $[a,b)$ for $a<b$, and we can enumerate them as $[a_n,b_n)$. Now take a point $x$ distinct from any $a_n$. Then there is no neighborhood of $x$ in $\cal C$ contained in the neighborhood $[x,\infty)$. That means $\Bbb R$ is not second countable.

For the Hausdorffness, you can give a direct argument, as you did here, or you can use the fact that the lower limit topology is finer that the usual topology on $\Bbb R$ which is Hausdorff. For normality, however, this doesn't work as a finer topology can cease to be normal.

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