# Show that this countable collection is a basis for $\mathbb R^2$

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

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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