Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm making Exercise 9 of paragraph 31 in Munkres, which is a proof that the Sorgenfrey Plane $\mathbb{R}_l^2$ is not normal. I'm having trouble on part c of the question. The full question is:

Let $A$ be the set of all points of $\mathbb{R}_l^2$ of the form $x \times (-x)$, for $x$ rational; let $B$ be the set of all points of this form for $x$ irrational. If $V$ is an open set of $\mathbb{R}_l^2$ containing $B$, show there exists no open set $U$ containing $A$ that is disjoint from $V$, as follows:

(a) Let $K_n$ consist of all irrational numbers $x$ in $[0,1]$ such that $[x,x+1/n) \times [-x,-x+1/n)$ is contained in $V$. Show $[0,1]$ is the union of the sets $K_n$ and countably many one point sets.

(b) User exercise $5$ of paragraph 27 to show that some set $\bar{K}_n$ contains an open interval $(a,b)$ of $\mathbb{R}$.

(c) Show that $V$ contains the open parallelogram consisting of all points of the form $x \times (-x+\epsilon)$ for which $a<x<b$ and $0<\epsilon<1/n$.

(d) Conclude that if $q$ is a rational number with $a < q < b$, then the point $q \times (-q)$ of $\mathbb{R}_l^2$ is a limit point of $V$.

I've done (a) relatively easy I think, simply by proving that for every irrational $x$there must be an $n$ so that $x$ lies in $K_n$, and then adding all rational one-point sets. (b) was something I did in the exercise mentioned before. My problem (at the moment) lies in (c).

First of all, I don't know if they mean only the irrational $x$'s or the rationals too. Second, in (b) we worked with the closure of $K_n$ and not $K_n$ itself. Does the closure of $K_n$ automatically lie in $V$ too? If yes, any hints how to prove this; if no, how do I actually start (c)?

share|cite|improve this question
up vote 1 down vote accepted

They mean all $\langle x,-x+\epsilon\rangle$ such that $x\in(a,b)$ and $0<\epsilon<1/n$; that includes both rational and irrational $x$.

You have a particular $n$ and a non-empty open interval $(a,b)$ such that $(a,b)\subseteq\operatorname{cl}K_n$, where for each $x\in K_n$ we know that $$\left[x,x+\frac1n\right)\times\left[-x,-x+\frac1n\right)\subseteq V\;.$$

Suppose that $x\in(a,b)$ and $0<\epsilon<1/n$. In order to show that $\langle x,-x+\epsilon\rangle\in V$, it suffices to show that there is a $y\in K_n$ such that

$$\langle x,-x+\epsilon\rangle\in\left[y,y+\frac1n\right)\times\left[-y,-y+\frac1n\right)\;.\tag{1}$$

Now $(1)$ holds iff $y\le x<y+\frac1n$ and $-y\le-x+\epsilon<-y+\frac1n$. These can be boiled down to the requirements that $y>x-\frac1n$, $y\ge x-\epsilon$, $y\le x$, and $y<x-\epsilon+\frac1n$. Can you continue from here to show that such a $y$ can always be found? I’ve finished the argument in the spoiler-protected bit below.

Since $\epsilon<\frac1n$, these four inequalities further reduce to $y\ge x-\epsilon$ and $y\le x$, i.e., $x-\epsilon\le y\le x$. Since $\epsilon>0$, the open interval $(x-\epsilon,x)$ is non-empty, and it follows that $(x-\epsilon,x)\cap(a,b)\ne\varnothing$. Finally, $K_n$ is dense in $(a,b)$, and $(x-\epsilon,x)\cap(a,b)$ is a non-empty open subset of $(a,b)$, so $(x-\epsilon,x)\cap(a,b)\cap K_n\ne\varnothing$. Now just choose $y$ in this intersection, and you’re done.

It’s not true that all points of $\operatorname{cl}K_n$ lie in $V$, but the points of $K_n$ are so thickly scattered throughout the interval $(a,b)$ that

$$\{x\}\times\left(-x,-x+\frac1n\right)\subseteq V$$

for each $x\in(a,b)$, even when $x$ itself isn’t in $K_n$. For each $y\in\left(-x,-x+\frac1n\right)$ there is a point $z\in K_n$ to the left of $x$ that is close enough to $x$ that $$\langle x,y\rangle\in\left[x,x+\frac1n\right)\times\left[-x,-x+\frac1n\right)\;,$$ which of course is a subset of $V$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.