A miscellanea of properties of the rational sequence topology on $\mathbb{R}$ For each $x\in \mathbb{R}-\mathbb{Q}$ fix a sequence of rational numbers $(y_i(x))_{i\in \mathbb{N}}$ which converges to $x$. For each irrational point $x$ and each $n \geq 1$ let $M_n(x) = \{x\}\cup \{y_{i}(x): i\geq n\}$. Define a new topology $L$ on $\mathbb{R}$ by taking as a basis the family $$B = \{\{y\}: y\in \mathbb{Q}\}\cup \{M_{n}(x): x\in \mathbb{R}-\mathbb{Q}, n\in \mathbb{N}\}.$$ (This is called the — or a — rational sequence topology on $\mathbb{R}$.)


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*Show that $(\mathbb{R}, L)$ is locally compact and Hausdorff

*Show that $(\mathbb{R}, L)$ is completely regular but not normal.
I spent several hours on this problem but couldn't proceed anywhere due to the complicated definition of the given basis. Can anyone please give some detailed help on this problem?
 A: To show that it is locally compact, you should just try to check that each basis element forms a subspace that is compact.  This is trivial for rational elements, and since the open neighborhoods of $x$ in $M_n(x)$ are all cofinite, we can make the following argument: 
Any open cover $\{U_\alpha\}$ of $M_n(x)$ in the subspace topology has at least one $U_1$ containing $x$, which therefore also contains some $M_k(x)$ for some $k$.  For any such $k$, there are only finitely many points in $M_n(x)$ that are not also in $M_k(x)$.  Choose $U_2, \ldots, U_i$ that contain those points.  Then $U_1, \ldots, U_i$ is a finite subcover, so that $M_n(x)$ is compact.
For Hausdorffness, checking that there are disjoint open sets separating points is nearly trivial if either point is rational, and so you should consider two irrational points.  Given $x \ne y$, you should argue directly that $M_{n}(x)$ and $M_n(y)$ are disjoint for suitably large $n$.
The next thing you should argue is that each basis element is closed in addition to being open, which is only slightly more general than previous arguments.  You should think about why this fact would imply that $(L,R)$ is completely regular.  Remember that for any closed and open set $U$ in a space $X$ there is a continuous function $X \to [0,1]$ sending $U$ to $\{0\}$ and $X \setminus U$ to $\{1\}$.
I think proving that $(L,R)$ is not normal is hard to do by a direct argument.  I found an answer to this question on MO which covered this example, and the best way appears to be to use a cardinality argument with Jones' lemma.
