Describe the topology $L$ inherits as a subspace of $\mathbb{R}_l \times \mathbb{R}$

Question

$\textbf{Notations:}$ $L$ is a line in plane and $\mathbb{R}_l$ is the lower limit topology.

$\textbf{Question:}$ Describe the topology $L$ inherits as a subspace of $\mathbb{R}_l \times \mathbb{R}$.

First, consider $L$ as a subspace of $\mathbb{R}_l \times \mathbb{R}$. A basis for $\mathbb{R}_l \times \mathbb{R}$ are open sets of the form $[a,b) \times (c,d)$ where $a, b, c, d \in \mathbb{R}$ and $a<b, c<d$. Thus, to find the open sets of $L$, we will find $([a,b) \times (c,d)) \cap L$. Let $([a,b) \times (c,d)) \cap L \neq \emptyset$.

If $L$ be a vertical line, say (x = k), then $([a,b) \times (c,d)) \cap L$ is the collection of points $x \times y \in L$ such that $x \leq x<b$ and $c<dy<d$. Therefore, for the vertical lines $x =k$ we have $([a,b) \times (c,d)) \cap L = \{k\} \times (c,d)$.

Now, I have a $\textbf{question}$.

I am guessing that $([a,b) \times (c,d)) \cap L$ should be an interval of type $(\ )$. But how should I prove it? I have an idea. I know that projections maps $X \times Y \to X$ and $X \times Y \to Y$ are open maps. Now can I project the second factor in $R$? But to do the projection $\{k\} \times (c,d)$ need to be open in $\mathbb{R}_l \times \mathbb{R}$, but $\{k\}$ is not open in $R_l$. Then what to do?

Now, for the horizontal line, say $y = r$, we have $L \cap \left( [a, b) \times (c, d) \right) = [a,b) \times \{r\}$. So, I am guessing that the interval will be of the form $[\ )$.

Now if $L$ has positive or negative slope. Then $$L \cap \left( [a, b) \times (c, d) \right) = \{ \ x\times y \in \mathbb{R} \times \mathbb{R} \ \colon \ a \leq x < b, c < y < d, y = mx+c \ \} $$

Then how should proceed in this case?

Answer 1

If $L$ is a vertical line (say of the form $x=c$), then $L$ is homeomorphic to the vertical axis, so essentially it's $\Bbb R$ in the subspace topology. The map $h: (c,x) \ni L \to x$ from $L$ to $\Bbb R$ is continuous (restriction of the projection), bijective (clear) and open, as $([a,b) \times (a',b')) \cap L$ (a typical basic open subset) has image $\emptyset$ (if $c \notin [a,b)$) or $(a', b')$ (if otherwise) in $\Bbb R$. So $h$ is a homeomorphism.

For any non-vertical line $L$ the restriction of the projection onto the first coordinate from $L$ onto $\Bbb R_l$ is continuous, bijective and open as well, again as basic sets $([a,b) \times (a',b')) \cap L$ map onto $[a,b)$ and so shows it's an open map.

So "all" lines are homeomorphic to $\Bbb R_l$, except the vertical ones, which are homeomorphic to $\Bbb R$.