I was reading through general-topology posts, but I couldn't understand the reasoning behind the answer of this part of a post. I'll restate it here:

This concerns the irrational slope topology. That is, let $X = \{(x,y)\in\mathbb{Q}^2:y\geq 0\}$. Fix $\theta\in\mathbb{R}\backslash\mathbb{Q}$. Let $\mathcal{T}$ be the coarsest topology on $X$ containing the sets of the form $$N_\epsilon(x,y)=\{(x,y)\}\cup\{(q,0)\mid q\in\mathbb{Q},\left|q-\left(x+\frac{y}{\theta}\right)\right|<\epsilon\}\cup\{(q,0)\mid q\in\mathbb{Q},\left|q-\left(x-\frac{y}{\theta}\right)\right|<\epsilon\}.$$ for $(x,y)\in X$ and $\epsilon>0$. (The following $B_\epsilon$'s correspond to the latter two sets in the logical manner.)

Claim 1. The closure of each basis neighborhood $N_\epsilon((x,y))$ contains the union of the four strips of slope $\pm\theta$ emanating from $B_\epsilon(x+y/\theta)$ and $B_\epsilon(x−y/\theta)$.

Knowing this, why does it then follow that Claim 2 holds?

Claim 2. The closures of any two open sets intersect (nontrivially).

I think my problems stems largely from the fact that I don't understand the boxed claim to begin with. My interpretation of the claim is that each basis neighborhood must contain all 4 line segments, but this doesn't make much sense:

The basis neighborhoods are merely (open) intervals on the $x$-axis unioned with the point $(x,y)$, whereas each line segment is a line in $\mathbb{R}^2$ with irrational slope (starting at $(x,y)$ and ending at the $x$-intercept). Surely, I'm making a mistake? I appreciate any help.


1 Answer 1


The closure of $N_\epsilon((x,y))$ is the set of all points $p$ such that every neighbourhood of $p$ contains a point of $N_\epsilon((x,y))$. The neighbourhoods of a point $p$ are themselves of the form $N_\delta(p)$ and consist of two intervals on the rational $x$ axis. If $p$ lies within one of the four strips, its projection along one of the two directions lies in an interval $I$ of the intervals forming $N_\epsilon((x,y))$. If so, the interval of the two intervals forming $N_\delta(p)$ that is centred around this projection overlaps with $I$ for any $\delta$. Thus such $p$ is in the closure of $N_\epsilon((x,y))$. Conversely, if $p$ does not lie in one of the four strips, its projection along both directions lies outside both of the intervals forming $N_\epsilon((x,y))$, so we can choose $\delta$ small enough that $N_\delta(p)$ does not overlap with $N_\epsilon((x,y))$.

Hmm -- I just saw that this isn't all that much more than what Brian wrote in his answer to the question you linked to. If this hasn't cleared it up for you, perhaps you could point out more specifically which part of this argument you don't understand?

  • $\begingroup$ Oh, I see now! I guess what made it difficult for me was finding the right picture in my head. I might be missing some 1 or 2-line proof, but how does the next statement follow: that is, that the closure of two open sets have nontrivial intersection? $\endgroup$ Oct 25, 2011 at 9:30
  • $\begingroup$ @Landyn: Both that statement and the one in your question are ungrammatical. I assume that you intended "closure" and "intersects" to be in the plural, that is, you're asking why the closures of two open sets intersect nontrivially. Each of the two open sets contains at least one basis neighbourhood, so its closure contains the closure of that basis neighbourhood. If you look at the picture in the book, it's clear that any two such closures, consisting of four strips each, necessarily overlap. $\endgroup$
    – joriki
    Oct 25, 2011 at 9:44
  • $\begingroup$ Oops, thanks for the correction. That makes perfect sense now! $\endgroup$ Oct 25, 2011 at 10:04

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