# A basis for some topology on $\mathbb{R}^2$

I know that the collection $$\mathcal{B}= \{(x,y): a<x\leq b, c<y\leq d\}$$ is not a basis for the standard topology on $\mathbb{R}^2$, but the collection of open rectangles in the plane certainly is. For which topology on $\mathbb{R}^2$ is $\mathcal{B}$ a basis? Would it be called the upper limit product topology? How can this be shown by using the definition of the basis for a topology?

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If you want to show it is a basis for some topology, you just need to check two things. $\bigcup \mathcal B=\Bbb R^2$ and that given a point in $\Bbb R^2$ and two basis elements containing it, there is another basis element contained in the intersection containing this point. And that is not really difficult. – Pedro Tamaroff Aug 30 '12 at 0:12
The way you currently have it written, $\mathcal{B} = (a, b] \times (c, d]$ which is not even a subbasis. – William Aug 30 '12 at 0:13
@William I think the OP means $a,b,c,d$ range through the reals. – Pedro Tamaroff Aug 30 '12 at 0:13

First, what you’ve written isn’t what you really mean. What you want is something like this:

For each $p=\langle a,b\rangle,q=\langle c,d\rangle\in\Bbb R^2$ such that $a<c$ and $b<d$, let $$B(p,q)=\Big\{\langle x,y\rangle\in\Bbb R^2:a<x\le b\text{ and }c<y\le d\Big\}\;,$$ and let $\mathscr{B}$ be the set of all such $B(p,q)$.

That is, you want a collection of many sets of the form $(a,b]\times(c,d]$, not just one such set.

Yes, this $\mathscr{B}$ is a base for a topology $\tau$ on $\Bbb R^2$, and yes, $\tau$ is a product topology. If you call the topology $\tau_u$ on $\Bbb R$ generated by the base $\{(a,b]:a,b\in\Bbb R\text{ and }a<b\}$ the upper-limit topology, as many do, then you could certainly call $\tau$ the product upper-limit topology on $\Bbb R^2$. I usually turn the intervals around and work instead with the Sorgenfrey line, $\Bbb R$ with the Sorgenfrey (lower-limit) topology, and refer to $\Bbb R^2$ with the product Sorgenfrey topology simply as the Sorgenfrey plane.

There are two pretty straightforward ways to see that $\mathscr{B}$ is a base for a topology on $\Bbb R^2$. I’ve already mentioned one. Let $\mathscr{B}_u=\{(a,b]:a,b\in\Bbb R\text{ and }a<b\}$; then $\mathscr{B}_u$ is a base for the topology $\tau_u$, so $\{B_1\times B_2:B_1,B_2\in\mathscr{B}_u\}$ is a base for the product topology on $\Bbb R^2$ as the square of $\langle\Bbb R,\tau_u\rangle$. But $\{B_1\times B_2:B_1,B_2\in\mathscr{B}_u\}=\mathscr{B}$, so $\mathscr{B}$ is a base for this product topology.

The other is simply to show directly that $\mathscr{B}$ satisfies the conditions defining a base for a topology:

1. $\bigcup\mathscr{B}=\Bbb R^2$, and
2. if $B_1,B_2\in\mathscr{B}$, and $p\in B_1\cap B_2$, then there is a $B_3\in\mathscr{B}$ such that $p\in B_3\subseteq B_1\cap B_2$. In fact $\mathscr{B}$ is a very nice base: $B_1\cap B_2\in\mathscr{B}$, so we can simply set $B_3=B_1\cap B_2$.

To verify (1), just note that $\langle x,y\rangle\in(x-1,x]\times(y-1,y]\in\mathscr{B}$.

To verify (2), note that if $B_1=(a_1,b_1]\times(c_1,d_1]$ and $B_2=(a_2,b_2]\times(c_2,d_2]$ have non-empty intersection, then

$$B_1\cap B_2=\Big(\max\{a_1,a_2\},\min\{b_1,b_2\}\Big]\times\Big(\max\{c_1,c_2\},\min\{d_1,d_2\}\Big]\in\mathscr{B}\;.$$

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Why the notation $\langle \cdot, \cdot \rangle$? – Libertron Aug 30 '12 at 1:07
@Sachin: It’s another standard notation for ordered pairs, and I prefer it, especially when I’m using interval notation at the same time. – Brian M. Scott Aug 30 '12 at 1:09
@BrianM.Scott What will you do if you simultaneously face in some problem ordered pairs, intervals and scalar product? :-) – Norbert Aug 30 '12 at 1:14
@Norbert: Given my mathematical interests, I probably don’t have to worry much about it. :-) Seriously, the angle brackets are habitual; I’d have to think about it not to use them. – Brian M. Scott Aug 30 '12 at 1:16