Show that the countable collection of rectangles
$\{ (a,b)\times (c,d) \mid a
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?
Show that the countable collection of rectangles
$\{ (a,b)\times (c,d) \mid a
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?
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
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