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I don't really know how to establish the conditions so $AB$ can be open. The problem says:

Let $A$ be an open set in $\Bbb R$ and $B$ any other set. Define: $$AB = \{xy\in\mathbb{R}\,\colon x\in A\text{ and }y\in B\}$$

Is $AB$ open? I believe is not, because if $B= \{0\}$ then $AB= \{0\}$ and is closed

under what conditions of $B$, $AB$ is open?

Stefan Hamcke
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Lore Rodriguez
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    Can you resolve the case where $B={c}$ for $c\neq 0$? If you know that open sets are open under arbitrary union, you can argue that $AB$ is open if $B$ doesn't contain $0$. – Milo Brandt Aug 28 '15 at 03:12
  • what if $B$ does not contain $0$? – user251257 Aug 28 '15 at 03:12
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    But if $0 \not \in B$, then $AB = \bigcup_{b \in B} A{b}$ is a union of... – Stefan Mesken Aug 28 '15 at 03:12
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    Also, be careful: A set being closed does not mean it is not open. The empty set is an example of a set that is both open and closed. However, in $\mathbb{R}$ the only such sets are $\emptyset$ and $\mathbb{R}$. – Ori Aug 28 '15 at 03:17
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    To show that $AB = { 0 }$ is NOT open, use the definition of openness. – Yes Aug 28 '15 at 03:43

1 Answers1

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Assume that both $A$ and $\dot B:=B\setminus\{0\}$ are nonempty. Then $$\dot BA=\bigcup_{b\in\dot B} b\>A$$ is a union of scaled copies of $A$, whence open. If $0\notin B$ then $\dot B=B$, and we are done. If $0\in B\cap A$ then $BA=\dot BA\cup\{0\}=\dot BA$ as well.

There remains the case that $0\in B\setminus A$. Here finer distinctions are necessary: If $A=\ ]0,1[\ $ and $B=[0,1[\ $ then $BA$ is not open. On the other hand, consider the example $A:=\ ]1,2[\ $, $B:=[{-1},1]$. Then $\dot BA=\ ]{-2},0[\ \cup\>]0,2[\ $, and $BA=\ ]{-2},2[$ is open as well.

(Edit: The following conjecture is wrong; see the counterexample by Milo Brandt.)

This leads to the following conjecture concerning the case $0\in B\setminus A$: The set $BA$ is open iff for any $\epsilon>0$ there exist numbers $x$, $x'\in A$ and $y$, $y'\in B$ such that $$-\epsilon<x'y'<0<xy<\epsilon\ .$$

Christian Blatter
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    $B={0}\cup {\frac{1}{2^n}:n\in \mathbb N}$ and $A=(1,2)\cup (-2,-1)$ makes your conjecture false, I believe - $AB$ is then dense in every neighborhood of $0$, but does not contain any neighborhood. – Milo Brandt Aug 28 '15 at 12:58