Prove the intersection of open rectangles/triangles are union of elements of B

Question

I'm proving that the open rectangles and triangles are basis for the euclidean topology on $R^2$, I defined the topology as open rectangles with sides parallel to the axis, others do it with open disc. I need to show that the intersection of any open rectangles (and triangles for the other exercise) is the union of open rectangles with sides paraller to the axis. I don't know how to model those sets (inequalities or something) and solve that problem without visual geometric arguments.

Answer 1

Let U be an open set. Since B is a base, for all x in U
there is some $U_x$ in B with x in $U_x$ subset U.
Show that $\cup\{ U_x : x \in U \} = U.$

Letting U be the intersection of two open sets
will suffice for the triangles and rectangles.
Usually there will infinitely many rectangles that get
smaller and smaller to fit into the open intersection.

Answer 2

Let's prove that rectangles form a basis for Euclidean topology on $\mathbb R^2$.

Let $R((x_1,x_2),a,b)=(x_1-a,x_1+a)\times(x_2-b,x_2+b)$ for $x\in\mathbb R^2$ and $a,b>0$. Let $\mathcal B'=\{R(x,a,b)\;|\;x\in\mathbb R^2, a>0, b>0\}$ and let $\mathcal B$ be the usual basis of open disks for Euclidean topology. We'll make the proof in two steps.

• First step, check $\mathcal B \subset \tau(\mathcal B')$. Here, $\tau(\mathcal B')$ denotes the topology, generated by the basis of rectangles $\mathcal B'$. To see this, it's enough to check that $\forall B\in\mathcal B:\forall y\in B:\exists B'\in \mathcal B': y\in B'\subset \mathcal B'$. What I just wrote means that for every disk in $\mathbb R^2$ and any point inside the disk we can find a rectangle such that the point is inside the rectangle and the rectangle is a subset of the disk. It is obvious that we can achieve this, but let's really derive it. Let $x\in \mathbb R^2$ and $r>0$ and let $y\in D(x,r)$. Denote the distance between the point $x$ and $y$ with $\lVert x-y\rVert$. Then, the distance between the point $y$ and the boundary of the disk is $r-\lVert x-y\rVert=:a(x,y,r)$. It is now obvious that $R(y,\frac{a(x,y,r)}{\sqrt{2}},\frac{a(x,y,r)}{\sqrt 2})$ contains $y$ and lies inside the chosen disk, i.e. $y\in R(y,\frac{a(x,y,r)}{\sqrt{2}},\frac{a(x,y,r)}{\sqrt 2})\subset D(x,r)$. The square root of $2$ comes from Pythagorean theorem; we need to watch out that our rectangle does not fall outside the disk. Draw a picture for intuition.

• Second step, check $\mathcal B'\subset \tau(\mathcal B)$. Similarly, for any open rectangle in $\mathbb R^2$ and any point inside it, we must now find a circle that contains the point and is inside the rectangle. Let $x=(x_1,x_2)\in\mathbb R^2$ and $a,b>0$ and let $y=(y_1,y_2)\in R(x,a,b)$. Now let $c=\min\{a-|x_1-y_1|,b-|x_2-y_2|\}$ be the shortest distance between point $y$ and the rectangle's boundary. Now $y\in D(y,c)\subset R(x,a,b)$.

Why was it enought to check these two things only? Well, now that we've seen this, we can represent any disk as a union of rectangles:

$$ D(x,r)=\cup\{R(y,\frac{a(x,y,r)}{\sqrt{2}},\frac{a(x,y,r)}{\sqrt{2}})\;|\; y\in D(x,r)\} $$

and similarly for the other direction.