Given a straight line in the plane, what topology does this straight line inherit as a subspace of $\mathbb{R_l} \times \mathbb{R}$ and as a subspace of $\mathbb{R_l} \times \mathbb{R_l}$, where $\mathbb{R_l}$ is the lower limit topology?

So trying to figure this out definitely made my brain hurt. I believe that as a subspace of $\mathbb{R_l} \times \mathbb{R}$, all non-vertical straight lines just inherit the lower limit topology $\mathbb{R_l}$, while vertical lines inherit the standard topology on $\mathbb{R}$. As for $\mathbb{R_l} \times \mathbb{R_l}$, as far as I can tell, the only difference is that now all straight lines, including vertical ones, inherit the lower limit topology.

My reasoning was basically that for $\mathbb{R_l} \times \mathbb{R}$, open sets on the line all have an initial left end point, since the x-coordinate of this left end point is always captured by the open sets [a,b) in $\mathbb{R_l}$, and this in turn drags the y-coordinate of the initial left end point along for the ride (so to speak). This is true in all but the vertical line case where there is no shift in the horizontal direction and thus the topology is inherited strictly from $\mathbb{R}$.

As for $\mathbb{R_l} \times \mathbb{R_l}$ it's basically the same argument except now the vertical lines inherit from $\mathbb{R_l}$ as well.

Can someone let me know whether I've reasoned correctly? Thanks.

  • $\begingroup$ They might not have the left point. Just the same as on Sorgenfrey line. In the product of Sorgenfrey lines, the antidiagonal has discrete topology, as do all lines with negative slope. $\endgroup$
    – tomasz
    Jun 24, 2012 at 21:10
  • $\begingroup$ Nollie: As Ross, I'm giving you +1 for showing your work. Your arguments, while not formal, contain the right ideas. $\endgroup$
    – user2093
    Jun 24, 2012 at 21:18
  • $\begingroup$ This Q has appeared before on this site. $\endgroup$ Jun 27, 2018 at 18:16

3 Answers 3


You’re right about $\Bbb R_l\times\Bbb R$, but you can make a better (or at least clearer) argument. Basic open sets in $\Bbb R_l\times\Bbb R$ are of the form $[a,b)\times(c,d)$, so they’re boxes open on all sides except the left. Let $L$ be a non-vertical line, and let $\langle x,y\rangle\in L$. Then the intersection with $L$ of any basic open set $[x,b)\times(c,d)$, where $y\in(c,d)$, is a left-closed, right-open interval with $x$ as left endpoint. It’s not hard to check that these intersections are a base for the subspace topology on $L$, which is therefore the Sorgenfrey (lower-limit) topology. And as you say, the vertical lines are homeomorphic to the second factor, $\Bbb R$.

$\Bbb R_l^2$ is another story, though. Now the topology has a base of boxes of the form $[a,b)\times[c,d)$, open only on the top and righthand edges. If the line $L$ is vertical or horizontal, of course, it’s homeomorphic to one of the factor spaces, i.e., to $\Bbb R_l$. If it has positive slope, you can argue much as in the case of $\Bbb R_l\times\Bbb R$ to conclude that it’s homeomorphic to $\Bbb R_l$. If it has negative slope, however, its topology is discrete: for any $\langle x,y\rangle\in L$, $$L\cap\Big([x,x+1)\times[y,y+1)\Big)=\big\{\langle x,y\rangle\big\}\;.$$

  • $\begingroup$ Excellent, this makes complete sense, thanks. $\endgroup$
    – Set
    Jun 24, 2012 at 21:25
  • 1
    $\begingroup$ If it has negative slope, however, its topology is discrete,,,im not getting this line @Brian M sir can u elaborate little bits $\endgroup$
    – jasmine
    Mar 27, 2018 at 6:26

I wish to respond to @lomber's question to Brian M. Scott's answer, however I don't have enough Math Exchange points to do so directly. With that, I answer it here:

Notice that, if $\langle x,y\rangle\in L $, then

$$\{\langle x,y\rangle\}\stackrel{(1)}{=}L\cap([x,x+1)\times[y,y+1))$$ is an open set in the subspace topology on $ L $, inherited by $ \mathbb{R}_\ell^2 $. This tells us that for every element $ \langle x,y\rangle\in L $, the set $ \{\langle x,y\rangle\} $ is open. Thus, if $ U $ is any subset of $ L $, then $$U=\bigcup_{\langle x,y\rangle\in U}\{\langle x,y\rangle\}$$ is an open set, being a union of open sets. This informs us every subset of $ L $ is open. This is the definition of the discrete topology.

To justify equation (1) (if that was the difficult part), notice if $ L $ has a negative slope and $ \langle x,y\rangle\in L $ (i.e. $y=mx+b$ with $ m<0 $), then $ v>x$ implies $$mv+b<mx+b=y\text{ thus }mv+b\notin[y,y+1) $$ so that $\langle v,mv+b\rangle\in L\backslash\big([x,x+1)\times[y,y+1)\big). $ Otherwise, $ v<x $ clearly gives $ \langle v,mv+b\rangle\in L\backslash\big([x,x+1)\times[y,y+1)\big)$.

Thus if $$\langle v,w\rangle\in L\cap([x,x+1)\times[y,y+1)) $$ and $\langle x,y\rangle\in L $, then $ v=x $ so that $ w=y $, whence $$L\cap([x,x+1)\times[y,y+1))\subset\{\langle x,y\rangle\}.$$ The reverse inclusion comes from $ \langle x,y\rangle\in L $.

  • $\begingroup$ Actually, I was following your answer. But I have some confusion.... So far I have understood that you proved $\{\langle x,y\rangle\}{=}L\cap([x,x+1)\times[y,y+1))$. You are showing that $(x,y)$ is the only point which satisfies the above relation. So, you have considered $v>x$....I have understood the proof for $v>x$. But I did not understand for $v<x$. $\endgroup$
    – user886636
    Jul 12, 2021 at 16:34
  • $\begingroup$ For $v <x$ ...I am trying like ... since $m<0$, we have $mv+b > mx+b = y$, then how to contradict? $\endgroup$
    – user886636
    Jul 12, 2021 at 16:36

Notice that you can only determine the topology up to a regular symmetry of the line, unless you specify a direction of the line. So you might get either lower limit or upper limit topology.

For a more rigorous proof, you have to look at possible traces of basic open sets (so rectangles) on the lines.

For $R_l\times R$ and a nonvertical line, those are indeed intervals with (up to) one endpoint, and for vertical lines, intervals with no endpoints, so you're right with these.

For $R_l\times R_l$ and lines of positive slope, we have the same situation, for vertical lines as well, but for lines of negative slope it is not hard to find that every single point is a trace of a rectangle, so the subspace is discrete.


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