How does Munkres prove that lower limit topology is finer than the standard topology on $\mathbb{R}$

Question

The proof in question (from the book "Topology" by Munkres):

Let $\mathcal T$ and $\mathcal T_\mathscr{l}$ be the standard and lower limit topology on $\mathbb R$ respectively. Given a basis element $(a,b)$ for $\mathcal T$ and a point $x\in (a,b)$, the basis element $[x,b)$ for $\mathcal T_\mathscr{l}$ contains $x$ and lies in $(a,b)$. Conversely, given the basis element $[x,d)$ for $\mathcal T_\mathscr{l}$, there is no open interval in $\mathcal T$ that contains $x$ and lies in $[x,d)$. Thus $\mathcal T_\mathscr l$ is strictly finer than $\mathcal T$.

There was a question on it before, but his actual proof was never addressed. I understand the proof involving the union of an infinite collection of sets. However, I don't get how the above proof proves that all elements of the latter topology are in the former, which as I understand is the definition of a finer topology.

Answer 1

Of $O\in\mathcal{T}$, then $O$ can be written as the union of intervals of the type $(a,b)$. Each interval $(a,b)$, in turn, can be written as$$\bigcup_{x\in(a,b)}[x,b),$$which belongs to $\mathcal{T}_1$. Therefore, $O\in\mathcal{T}_1$.

Answer 2

What Munkres is doing:

If $T_1,T_2$ are topologies on a set $R,$ and $B_1,B_2$ are bases for $T_1,T_2$ respectively, then to show that $T_1\subset T_2,$ it suffices to show that whenever $x\in b_1\in B_1$ there exists $b_2\in B_2$ with $x\in b_2\subset b_1.$

That implies that every $b_1\in B_1$ is a union of members of $B_2,$ so $b_1\in T_2$. So $B_1\subset T_2,$ so any $t\in T_1,$ being a union of members of $B_1,$ is a union of members of $T_2,$ so $t\in T_2.$

In this case we have $R=\Bbb R$ and $T_1=\mathcal T$ and $T_2=\mathcal {T_l}$ and $B_1=\{(a,b): a,b\in \Bbb R\}$ and $B_2=\{[x,b):x,b\in \Bbb R\}.$

Therefore $\mathcal T\subset \mathcal T_l.$

(And, obviously, they are not equal, because $[0,1)\in \mathcal T_l$ \ $\mathcal T.$)