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I'm investigating a topology on that rational plane $\mathbb{Q}^2$, with a subbase which I've had a hard time getting my hands on.

Suppose $X=\{(x,y)\in\mathbb{Q}^2\mid y\geq 0\}$ be the given subset, and let's fix some irrational $j$. Also, I equip $X$ with the topology $\mathcal{T}$ generated by the subbase consisting of sets of form $$ A_\epsilon(x,y)=\{(x,y)\}\cup\{(q,0)\mid q\in\mathbb{Q},\left|q-\left(x+\frac{y}{j}\right)\right|<\epsilon\}\cup\{(q,0)\mid q\in\mathbb{Q},\left|q-\left(x-\frac{y}{j}\right)\right|<\epsilon\}. $$

So I know that a basic set in this topology consists of finite intersections of such sets, and an element $(x,y)$ is in a basic set if it's in at least one of the three types of sets being unioned to form the subbase sets, for each subbase set in the finite intersection.

With this, why is the space $T_2$? The latter two sets used to make the subbase sets are making it difficult for me to find open sets separating two distinct points.

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  • $\begingroup$ See Steen and Seebach, Counterexamples in topology, Example 75, page 93 (Irrational slope topology). $\endgroup$ – t.b. Oct 19 '11 at 0:19
  • $\begingroup$ Thanks @t.b.! I'll look this up. $\endgroup$ – Danielle Intal Oct 19 '11 at 1:07
  • $\begingroup$ You can delete it yourself if you really want to (there's a "delete" button at the bottom of the question), but we'd prefer if you didn't. It's nice to have this site as a repository of questions and answers. You could ask t.b. to post his comment as an answer and then accept it, or you could write your own answer and accept it. $\endgroup$ – Nate Eldredge Oct 19 '11 at 1:15
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For $y_0>0$ look at the lines through $(x_0,y_0)$ with slopes $\pm j$: their equations are $y = jx+y_0-jx_0$ and $y = -jx+y_0+jx_0$, so they have $x$-intercepts $$x=x_0-\frac{y_0}{j}$$ and $$x=x_0+\frac{y_0}{j}.$$ Thus, $A_\epsilon(x_0,y_0)$ is picking up those points of $X$ on the $x$-axis that lie within $\epsilon$ of either of those $x$-intercepts: it’s $(x_0,y_0)$ together with two ‘feet’ on the $x$-axis. (Of course if $y_0=0$ you just get ordinary intervals around $(x_0,0)$ on the $x$-axis.)

If you make a sketch, it’s not hard to see that for sufficiently small $\epsilon$, $A_\epsilon(x_0,y_0)$ and $A_\epsilon(x_1,y_1)$ will be disjoint unless $(x_0,y_0)$ and $(x_1,y_1)$ lie on a straight line of slope $\pm j$; you can even determine how small $\epsilon$ has to be in terms of $x_0,x_1,y_0$, and $y_1$. To finish things off, you just have to show that $(x_0,y_0)$ and $(x_1,y_1)$ cannot lie on a single straight line of slope $\pm j$; here’s where you use the irrationality of $j$ and the rationality of the coordinates.

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  • $\begingroup$ That makes sense, since the slope between any two rational coordinates is again rational. Thank you for the help. $\endgroup$ – Danielle Intal Oct 19 '11 at 20:48

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