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Show that the countable collection of rectangles

$\{ (a,b)\times (c,d) \mid a<b \text{ and } c<d, \text{ and } a,b,c,d \text{ are rational} \}$

is a topological basis for $\mathbb{R^2}$.

This a question from the book Topology by James Munkres. I am not able figure out. How to go about this question?

Martin Sleziak
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Ramana Venkata
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  • When I google for rational basis topology this is the first result: https://www.math.lsu.edu/~stoltz/Courses/F08/7510/CourseMaterials/HW1ans.pdf - see Problem 6 on p.2 of this document. – Martin Sleziak Oct 30 '11 at 06:59
  • Okay Thanks I'll read it – Ramana Venkata Oct 30 '11 at 07:03
  • What is your definition of a (topological, i presume) basis? Can you prove at least some properties from that definition for your given set? Do you know any bases for $\mathbb R^2$ already? – hmakholm left over Monica Oct 30 '11 at 07:03
  • @HenningMakholm This is first time I am seeing bases of $\mathbb{R^2}$. CAn you explain how $(a,b)\times (c,d)$ is basis element in $\mathbb{R^2}$ ? – Ramana Venkata Oct 30 '11 at 07:10
  • Does the book contain a definition of bases for topological spaces in general? Does it define a topology on $\mathbb R^2$? If so, how is it defined? – hmakholm left over Monica Oct 30 '11 at 07:15
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    Basis is defined for a general topological spaces. I think I misinterpreted the notation I think I got it. I am treating (a,b) as a element of R^2 it should be considered as a subset of R. – Ramana Venkata Oct 30 '11 at 07:20
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    @Ramana: Yes, they must be intervals. Now that you mention it, it is confusing. – hmakholm left over Monica Oct 30 '11 at 08:21
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    @Henning, Ramana: This is a good reason for ordered pairs to be written as $\langle x,y\rangle$ and not as $(x,y)$. – Asaf Karagila Oct 30 '11 at 10:15
  • @Asaf: You denote closed intervals by $[a,b]$? (Many people are used to $\langle a,b \rangle$.) It would not be of much help, if we got just closed intervals instead of open intervals. – Martin Sleziak Oct 30 '11 at 11:28
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    @Martin: There will always be incompatibility in symbols. I come from a place where open intervals are $(x,y)$ and thus I find $]x,y[$ cumbersome. Thus I am left with $\langle x,y\rangle$ for ordered pairs. – Asaf Karagila Oct 30 '11 at 12:21
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    @Asaf: in practice, there is no problem with that overloading of notations. (There are no anecdotes of great mathematicians in the middle of a talk stopping short and saying «ooooooohhhhh, that was an order pair! oooops»...) – Mariano Suárez-Álvarez Nov 01 '11 at 09:01
  • @Mariano: It's almost never the mathematician that confuses that. It's always the students. However I did have a teacher that wrote his $o$ similar to $\theta$, and more than once students were confused about that and asked "What is that $\theta$ over there?", I heard from another student that one time (in a course I did not take with him) he got confused as well. – Asaf Karagila Nov 01 '11 at 09:05
  • @Asaf: I consider that a plus (in a not too subtle to feel less bad about my blackboard handwriting... :) ) in that it keeps students awake or completely asleep. Both of which are good! – Mariano Suárez-Álvarez Nov 01 '11 at 09:24
  • @Mariano: If that is the case, you should start writing complex numbers as $\Xi$. Especially if conjugation is in order. :-) – Asaf Karagila Nov 01 '11 at 09:48

1 Answers1

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Just for the sake of having an answer. (See http://meta.math.stackexchange.com/questions/1148/what-should-one-do-when-ones-question-has-been-answered-in-the-comments or http://meta.math.stackexchange.com/questions/1559/dealing-with-answers-in-comments or http://meta.math.stackexchange.com/questions/3138/unanswered-questions )

If we already know that $$\{(a,b)\times(c,d); a,b,c,d\in\mathbb R,a<b.c<d\}$$ is a basis, then it suffices to show that if $[x,y]\in(a,b)\times(c,d)$ then $x$ is contained in a rectangle with rational endpoints which is a subset of $(a,b)\times (c,d)$. (I am using $[x,y]$ for an ordered pair and $(a,b)$ for an open interval in this answer.)

We have $x\in(a,b)$, $y\in(c,d)$. Then there are rational numbers $a',b',c',d'$ such that $a'\in(a,x)$, $b'\in(b,x)$, $c'\in(c,y)$, $d'\in(y,d)$. Obviously $$[x,y]\in (a',b')\times(c',d')\subset(a,b)\times(c,d).$$

Note that the proof is an easy generalization of the proof of an analogous result for the real line. I've copied here Henno Brandsma's answer from Topology Q+A board.

In reply to "Countable basis of open intervals in R", posted by math layman on Feb 3, 2005:

R is $(-\infty, +\infty)$, how to find a countable basis of open intervals so that for any open point x inside of an open set B, there is an open interval within the basis which contains this point.

The set of intervals with rational endpoints does the trick.
If $x$ is inside an open set $B$, then there is an open interval $x \in (a,b) \subset B$.
But the interval $(a,x)$ contains a rational number $r_1$ and the interval $(x,b)$ contains a rational number $r_2$ and so $x \in (r_1, r_2) \in (a,b) \subset B$ and so there is an interval with rational endpoints inside every open set.
Henno

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Martin Sleziak
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