Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
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
1  
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
show 3 more comments

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 $$(a,\epsilon)\cdot(b,\delta)\ \subset\ (a,\epsilon)\odot(b,\delta)$$

share|improve this answer
    
I love this approach! Thanks. –  Dahn Jahn Jan 9 '13 at 11:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.