$A \times B$ is an open set in $\Bbb R^2 \implies A$ and $B$ are both open in $\Bbb R$; $A,B \neq \emptyset$

I am studying Analysis on my own. Reading The Elements of Real Analysis by Bartle. Came across the above problem and I came up with the following solution but am very unsure about it. Would be very grateful if someone could look through it and verify it for me. Thanks in advance.

My Attempt:

Let $A$, $B \subseteq \Bbb R$ and let $A \times B$ be open in $\Bbb R^2$.

Let $x \in A$ be arbitrary. $\exists b \in B$ since $B$ is non-empty. Since $A \times B$ is open in $\Bbb R^2$, $\;\exists r_{x, \ b} \gt 0$ such that the set $B_{x, \ b} = \{ (z_1, z_2) \ | \ \left|{\left| { (z_1, z_2) - (x, b) } \right |}\right| \lt r_{x, \ b} \}$ is contained in $A \times B$.

Consider the set $A_x = \{ z \ | \ \left|{z - x}\right| \lt r_{x, \ b}\}$. It can be seen that $(z, b) \in B_{x, \ b} \subseteq A \times B \;\; \forall z \in A_x$. This will only be true if $A_x \subseteq A$. As per its definition, $A_x$ is an open ball.

Since $x \in A$ was arbitrary, we have shown that for each point in $A$ there is an open ball entirely contained in $A \implies A$ is an open set.

It can be similarly shown that $B$ is open by taking an arbitrary point in $B$ and a fixed one in $A$.

Q.E.D.

I have the following doubts..

$Q_1$ : Are the above arguments rigorous enough?? Because I solved this purely on geometric imagination thinking of a circle on the plane.

$Q_2$ : According to my proof above it can be shown in a similar way that $A_1 \times A_2 \times ...\times A_p$ is open in $\Bbb R^p \implies$ Each of $A_1, A_2.. A_p$ is open in $\Bbb R$(further worsening my doubts). Is that true in general??

• I think a lot of your doubt comes from the fact your grasp of both the cartesian product and the product topology is still shaky.A big flaw in your arguement is you simply assumed A and B are not the empty set-there's nothing in your arguement that shows that. By definition, if either A or B is the empty set, then A X B is empty! Overall,the basic logic looks ok,but A is an open set only if you can ensure the ball lies entirely on the real line in A for any z in A.Hint:Choose r*= min{Rz,Rb} where Rb is the radius of the open ball in AX B and Rz is the radius of the open ball in A. – Mathemagician1234 Feb 17 '14 at 5:54
• @Mathemagician1234 Well you're right. The result is still true if either $A$ or $B$ is empty. Well I thought I did use the non-void condition in assuming there is an element $b$ in $B$. About the hint: We do not know there is an open ball in $A$ yeah to come up with $R_z$? – Ishfaaq Feb 17 '14 at 6:02
• @Mathemagician1234: If you meant that I had not considered the case where $A$ and $B$ are empty - the question requires me to solve assuming they are both non-void.. – Ishfaaq Feb 17 '14 at 6:08
• The way you worded the question doesn't make that clear. But I think it's worth covering all cases if you're going to write up a full response. Notice since the empty set is open, the implication is true in that case-in effect,this is the trivial case of the answer to your question. Little things like this matter in mathematics. – Mathemagician1234 Feb 17 '14 at 6:36
• As for the hint, for every z,x in A, your definition is clearly an open ball in A. But you made z arbitrary in A. You defined Bx, b={(z1,z2) | ∣∣∣∣(z1,z2)−(x,b)∣∣∣∣<rx, b} is contained in A×B. We know for each ordered pair in A X B, the first coordinate lies in A by definition of the Cartesian product. So now let z=z1 in the ordered pair (z1,z2) in Bx. Hint: Write the norm that defines Bx explicitly in terms of the distance function in R2 by the length of the vectors.Then you'll see the resulting set must lie entirely in A and is indeed an open ball. – Mathemagician1234 Feb 17 '14 at 7:04

I think the proof is, logically, just fine.

Stylistically, I would make a few changes, none of them major or all that significant:

First, it seems weird to use $x$ and $b$ as your chosen elements of $A$ and $B$. I'd use $a$ and $b$.

Second, the notation is subscript heavy. For example, I'd probably just write $r$ instead of $r_{x,b}$ (or $r_{a,b}$). A lot of this depends on the mathematical maturity of you/your intended audience. The notation $r_{x,b}$ is wonderful at reminding you that $r$ depends on the point $(x,b)$, but if you and your audience don't need the reminder, it's kind of cumbersome.

Third, I'd write "consider the open ball $A_x = ...$," as opposed to "consider the set $A_x = ...$. This allows you to completely avoid the last line in that paragraph.

Lastly, rather than write "It can be seen that $(z,b)\in B_{x,b}...$" I would probably write the computation which proves it.

Q2: You're right that a small modification of your proof shows that if $A = A_1 \times \ldots \times A_p\subseteq \mathbb{R}^p$ is open with all of the $A_i$ nonepmty, then every $A_i$ must be open in $\mathbb{R}$.
As an aside, if you consider product topologies in general, then it's easy to see that the projection maps are open on the standard base, and hence open maps. This implies that for any open subset $O \subset \mathbb{R}^2$, $\pi_1[O]$ and $\pi_2[O]$ are open subsets of $\mathbb{R}$. This holds even in any finite or infinite product. So it's not a metric fact, in a way, but more topological. Of course, the relevant fact is that the standard metric on $\mathbb{R}^n$ actually indices the product topology, which is not a priori clear.