Write $(a,\ b)$ as a union of basis elements in $(\mathbb{R},\ \mathcal{T}_l)$

Question

Here $\mathcal{T}_l$ is the lower limit topology.

I have proved that $(a,\ b)\in\mathcal{T}_l$ as follows:

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Let $x\in(a,\ b)\implies a<x<b$. $[x,\ b)$ is a basis element of $(\mathbb{R},\ \mathcal{T}_l)$. $x\in[x,\ b)$ and $[x,\ b)\subset(a,\ b)$.

Therefore, $(a,\ b)\in\mathcal{T}_l$

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Now how do I write $(a,\ b)$ as a union of basis elements in $(\mathbb{R},\ \mathcal{T}_l)$?

Answer 1

Let $A=\{x|a<x<b, x\in \mathbb{Q}\}=\{x_{\alpha}\}_{\alpha\in J}$. Clearly, $A$ is countable. In fact, we may represent $(a,b)$ as a countable union of $[x,b)$, where $x\in A$, i.e., $$(a,b)=...\cup[x_i,b)\cup[x_{i+1},b)\cup...=\bigcup_{x_i\in A}[x_i,b)$$

To prove so is easy. Let $y\in \bigcup_{x_i\in A}[x_i,b)$. Therefore, $a<y<b$, so that $y\in(a,b)$. As a consequence, $\bigcup_{x_i\in A}[x_i,b)\subseteq (a,b)$. Now, let $y\in(a,b)$ and $y'=x_n$ a rational such that $y'>y$. Indeed, $y\in \bigcup_{i =1}^{n}[x_i,b)\subseteq\bigcup_{x_i\in A}[x_i,b)$. Because of this, $(a,b)\subseteq\bigcup_{x_i\in A}[x_i,b)$, so that we proved what we wanted.