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This question is based on Munkers 16.8, which is asked here. My question is spurred by the answer given. First, Munkers Ex 16.8 asks:

if $L$ is a straight line in the plane, describe the topology $L$ inherets as a subspace of $\mathbb{R}_l \times \mathbb{R}$ and as a subspace of $\mathbb{R}_l \times \mathbb{R}_l$.

The answer that was given at the linked question does make sense to me, but raises a question. For the purposes of my question, assume that $L$ is neither vertical nor horizontal. Now, the basis for the order topology on $\mathbb{R}_l \times \mathbb{R}$ is $\mathscr{B}$: $$ \mathscr{B} = \{[a_1, b_1) \times (a_2,b_2): a_1<b_1, a_2<b_2, \text{ and } a_1,a_2,b_1,b_2 \in \mathbb{R}\} $$

The basis we inheret is contains sets of the form $L \cap [a_1, b_1) \times (a_2, b_2)$ whatever that turns out to be. To paraphrase the original answer, let $U$ be a general element from $\mathscr{B}$. For any $U$, either $U$ intersects $L$ or else $L \cap U = \emptyset$. But if $L$ intersects $U$ we get two cases. Either $L$ intersects on the solid left edge, or it does not. When it does, we get:

$$ L \cap U = \{\langle x, mx + c \rangle : x \in [a,b) \text{ and } a,b \in \mathbb{R}\} $$

for some $a$ and $b$. But if $L$ intersects only the open edges we get:

$$ L \cap U = \{\langle x, mx + c \rangle : x \in (a,b) \text{ and } a,b \in \mathbb{R}\} $$ Here $m$ and $c$ are just constants from $\mathbb{R}$ defining the line. Now both kinds of intersection contribute elements to the basis of the topology of $L$. My question: isn't the topology for $\mathbb{R}_l$ supposed to have the basis $\{[a,b):a<b \text{ and } a,b \in \mathbb{R}\}$ and not $$ \{[a,b): a<b\} \cup \{(a,b): a<b\}, \quad a,b \in \mathbb{R}$$

If we really do inherit a topology like $\mathbb{R}_l$ please show me how one can get open intervals like $(a,b)$ in $\mathbb{R}_l$ using only half-open intervals $[a,b)$.

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Why $\bigcup_{0<a<1}[a,1) = (0,1)$. –  Jonathan Y. Aug 29 '13 at 20:55
    
@Janathan I see. And that could almost be an answer if you explained it. So it seems that sets in the topology can be unions of infinitely many elements from the basis, yet only intersections of finitely many elements from the topology are required to be in the topology. I'm sure it's obvious -- if you already know topology, my apology. –  Edward Newell Aug 29 '13 at 21:01
    
You are completely right. I supposed, seeing as how you've been reviewing $\mathbb{R}_l$ topology and bases and product topology, that the question reflected a familiarity/comfort with the basic notions of point-set topology. My intention was to point you in the right direction, not to lecture or imply no explanation was required (hence I didn't feel my hint was appropriate as a full answer). Perhaps I should've said, the finer the topology, the less sequences are converging, so taking a basis for a finer topology and adding to it other open sets changes nothing. –  Jonathan Y. Aug 29 '13 at 21:54

2 Answers 2

up vote 2 down vote accepted

As for the general question, it's as you think:

If the line crosses the "solid" edge of any open set, then the inherited topology will be $\mathbb{R}_\ell$.

As for your final question, is you're reading Munkres' book, the one I'm also using to study, you should have had to use it in some other exercises, but it's this way:

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

I will leave it to you to prove that's actually correct

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Argh face-palm. Of course, I did use it to prove that $\mathbb{R}_l$ is finer than $\mathbb{R}$! Thanks for closing the circuit! –  Edward Newell Aug 29 '13 at 21:18

Consider the union of open sets $[\frac12,1)\cup[\frac14,1)\cup[\frac18,1)\cup\cdots=(0,1)$. Does that answer your question?

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