# Product of neighbourhoods

If I define product of two sets $A, B$ such as: $$A*B=\{a*b: a\in A , b\in B\}$$ And then take two neighbourhoods $U_a$ and $U_b$ of points $a,b\in\mathbb{R}$, ($\delta > 0$) $$(a-\delta,a+\delta)\subset U_a$$ $$(b-\delta,b+\delta)\subset U_b$$

Question: What is then their product?

This is perhaps somewhat elementary, but I am slightly confused about the matter and wouldn't want to jump on false conclusions.

Thanks for help!

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It is equivalent to ask, what is the product (defined by you) of two open intervals $(a,b)$ and $(c,d)$. Let me provide you some examples, and let us assume that all numbers are positive. Let $x\in (a,b)$ be any point, then $\{x\}*(c,d) = (cx,dx)$. Due to the fact that everything is positive, and that the multiplication is a continuous functions we obtain $(a,b)*(c,d) = (ac,bd)$.

I guess, you can figure out the general case simply.

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Thanks. I'll try to figure out the general case now. – Dahn Jahn Jan 9 '13 at 10:00
@DahnJahn There are $9$ cases, $3$ for each interval: 1) $a>0$ $b>0$ 2) $a<0$ $b<0$ and 3) $a<0$ but $b>0$, the same cases for $(c,d)$. I described the case $1,1$. For $2,1$ you have $(a,b)*(c,d) = (ad,bc)$ – Ilya Jan 9 '13 at 10:03
The more "interesting" cases are of course when when $0$ is in at least one of the intervals. – Hagen von Eitzen Jan 9 '13 at 10:21
To clarify: For the case $1,3$ it'd be $(bc,bd)$? If that is so, then I suppose I see the logic behind it. Still, I find it slightly baffling; followup question: if I know $a<x<b$, $c<y<d$, $a<0$ and $b,c,d>0$, how do I arrive at $ad<xy<bd$? – Dahn Jahn Jan 9 '13 at 10:21
I shall accept this answer, as it, together with the comments, led me to understanding of the problem. That being said, the "fuzzy numbers" answer is absolutely lovely. – Dahn Jahn Jan 9 '13 at 20:54

With the following setup you are on the safe side:

Introduce "fuzzy numbers" by means of $$(a,\epsilon)\ :=\ [a-\epsilon,a+\epsilon]\qquad(a\in{\mathbb R}, \ \epsilon\geq0)\ .$$ Defining sum and product of such numbers by \eqalign{(a,\epsilon)\oplus(b,\delta)&:=(a+b,\ \epsilon+\delta)\ ,\cr (a,\epsilon)\odot(b,\delta)&:=(a\cdot b,\ |a|\delta +|b|\epsilon+\epsilon\delta)\ \cr} you are guaranteed $$(a,\epsilon)+(b,\delta)\ \subset\ (a,\epsilon)\oplus(b,\delta)$$ and $$(a,\epsilon)\cdot(b,\delta)\ \subset\ (a,\epsilon)\odot(b,\delta)$$

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I love this approach! Thanks. – Dahn Jahn Jan 9 '13 at 11:54