# Thread: Problem on product topology/standard topology on R^2.

1. ## Problem on product topology/standard topology on R^2.

Let $\mathbb{R}_{\tau}$ be the set of real numbers with topology
$\tau = \{(-x,x)| x>0\} \cup \{\emptyset, \mathbb{R}\}$
and $\mathbb{R}_{\tau} \times \mathbb{R}_{\tau}$ be the product topology on $\mathbb{R}^2$.

a)Prove that $A = \{(x,y) \in \mathbb{R}^2 | x^2 + y^2 < 1\}$ is open in $\mathbb{R}_{\tau} \times \mathbb{R}_{\tau}$

b)Find $\overline{A}$.Justify your answer.

c) What functions $f: {\mathbb{R}_{\tau}}^2 \rightarrow \mathbb{R}$ are continuous?
Here $\mathbb{R}$ has the standard topology and ${\mathbb{R}_{\tau}}^2 = \mathbb{R}_{\tau} \times \mathbb{R}_{\tau}$ has the product topology.

PICTURE ATTACHED!

2. ## Re: Problem on product topology/standard topology on R^2.

1) Let $(x_0, y_0) \in A.$ Produce an open set in the product topology, which here would be like

$(-c_1, c_1) \times (-c_2, c_2)$, such that $(x_0, y_0) \in (-c_1, c_1) \times (-c_2, c_2) \subset A.$

2) Closed sets are the complement of open sets. Excluding the empty set and all of $\mathbb{R},$ the only

open sets of $\mathbb{R}_{\tau}$ are intervals of the form $(-a, a)$, so, excluding the empty set and all of $\mathbb{R},$

all closed sets are of the form $(-\infty, -a] \cup [a, \infty)$.

Then use that the closure of a set is the intersection of all closed sets containing that set.

3) If $x \in U \in \tau$, then $U = (-a, a)$ for some a > |x|, or $U = \mathbb{R}$.

Thus if $(x,y) \in U \times V \in \tau \times \tau$, you can say something about U and V.

That tells you what the basic open "boxes" in $\mathbb{R}_{\tau}^2$ look like.

So an open set will be an arbitrary union of those, which will put serious restrictions on the kinds of "shapes and fills" they can have.

Take a guess at the symmetries & conditions required to be an open set in $\mathbb{R}_{\tau}^2$,

then prove that a set is open if and only if it has those symmetries & conditions.

(Think about the boundary (a curve), in the usual sense, of an open set of $\mathbb{R}_{\tau}^2$ in the 1st quadrant.)

With the topology of $\mathbb{R}_{\tau}^2$ so characterized, you'll be able to say when the inverse image

of an open interval of $\mathbb{R}$ (i.e. a basis set) is open in $\mathbb{R}_{\tau}^2$,

and so set the condition required for continuity of any $f : \mathbb{R}_{\tau}^2 \rightarrow \mathbb{R}$.

3. ## Re: Problem on product topology/standard topology on R^2.

Are these correct?
1. We consider all open rectangles centered at the origo and are inside of the disk.Clearly, for any (x,y) in A any open rectangle will contin (x,y)
2. It is the origin point
3.we have a union of open filled rectangles centered at the origo

4. ## Re: Problem on product topology/standard topology on R^2.

1) Kinda. Any open (in the traditional sense) rectangle that's inscribed in the boundary circle is a basic open set in the product topology. Since any point (x,y) in A is contained in such an inscribed rectangle, have that any point (x,y) in A is contained in an open set (in this funky tau-topology) that's contained in A, and therefore A is open in this topology.

2) Something about the origin solves this quesiton quickly. Can you find *any* closed set that constains the origin?

To get you thinking about it, here's an example of the smallest closed set consisting of a single point (2, 6):

Closure of {(2, 6)} = [ (-infinity, -2] union [2, infinty) ] x [ (-infinity, -6] union [6, infinty) ].

3) Yes - so what can such unions look like? In particular, think about the symmetries that must result from such unions, and remember that you can include an infinite number of such rectangles, and so can make many shapes with smooth boundaries.Maybe draw some examples, remembering that it's the union.

is b) R^2?

6. ## Re: Problem on product topology/standard topology on R^2.

yep.
Every open set, except the empty set, contains 0. Thus every closed set, except the empty set's complement (= all R), excludes 0. So will get the same effect about the origin in the product topology, since that's what happens in both the coordinates for each basic "box" that contains the origin. Or, seen another way, every open set in the product topology, except the empty set, contains the origin, and so the only closed set which contains the origin is, again, the whole space.

7. ## Re: Problem on product topology/standard topology on R^2.

So,what's the answer for part c? "what can such unions look like?"

8. ## Re: Problem on product topology/standard topology on R^2.

The basic idea is f a continuous (not-strickly) decreasing function on [0, b], then fill it in, and reflect about the x and y axes.
However:
1) Also might have the interval being [0, infinity).
2) Also might have it undefined at x=0, but going to positive infinity.
3) Also might the interval being [a, b] (or [a, infinity)) with a > 0, where on [0, a] the entire vertical strip is filled in.
4) Also might not have a f being a function, because could have vertical lines along which the boundary of the open set "drops straight down" for a time. So the f is really a path that looks like the graph of a decreasing function except where you might have these straight vertical drops.

I wouldn't want to prove that, but that's what it looks like to me. It will be "filled in" (convex) and symmetric about the x and y axes, and have a decreasing boundary "function" in the first quadrant. Those, and the empty set and the whole plane, look to be the only open sets to me. Do you agree?

Other than the constant functions, I'm having a hard time visualizing any continuous function from $\mathbb{R}_{\tau}^2$

to $\mathbb{R}$, although I don't claim to know that - it's just how it seems.