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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?

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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 7:20
@Ramana: Yes, they must be intervals. Now that you mention it, it is confusing. – Henning Makholm Oct 30 '11 at 8:21
@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
@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
@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-Alvarez Nov 1 '11 at 9:01
up vote 2 down vote accepted

Just for the sake of having an answer. (See What should one do when one's question has been answered in the comments? or Dealing with answers in comments. or 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.

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