I am not sure which keywords to use to search for previously asked questions dealing with these problems. I did search for "Topology and Groupoids" but didn't find anything relevant.
Problems:
- Let $C$ be a neighborhood of $c \in \mathbb{R}$, and let $a + b = c$. Prove that there are neighborhoods $A$ of $a$ and $B$ of $b$ such that $x \in A$ and $y \in B$ imply $x + y \in C$.
- Write down and prove a similar result to that of Exercise 6, but with $c = ab$.
Other information:
The definition of a neighborhood used in the book is (from page 1) that $N$ is a neighborhood of $a$ if there is a $\delta > 0$ such that $(a - \delta, a + \delta) \subseteq N$.
Attempt for Exercise 6:
By assumption there exists a $\delta > 0$ such that $(c - \delta, c + \delta) \subseteq C$. We know that $a + b = c$, so we see that
\begin{align*} \Big( a - \frac{\delta}{2} \Big) + \Big( b - \frac{\delta}{2} \Big) &= c - \delta\\ \Big( a + \frac{\delta}{2} \Big) + \Big( b + \frac{\delta}{2} \Big) &= c + \delta. \end{align*}
Therefore if $x \in \big( a - \frac{\delta}{2}, a + \frac{\delta}{2} \big)$ and $y \in \big( b - \frac{\delta}{2}, b + \frac{\delta}{2} \big)$, we see that $x + y \in (a + b - \delta, a + b + \delta) = (c - \delta, c + \delta) \subseteq C$. Let $A = \big( a - \frac{\delta}{2}, a + \frac{\delta}{2} \big)$ and $B = \big( b - \frac{\delta}{2}, b + \frac{\delta}{2} \big)$ and the proof is complete.
Attempt for Exercise 7:
We know there exists a $\delta > 0$ such that $(c - \delta, c + \delta) \subseteq C$. Since $ab = c$, we see that
\begin{align*} (a - \sqrt{\delta})(b + \sqrt{\delta}) &= ab - \delta, \end{align*}
so for $0 < \sqrt{\epsilon} < \sqrt{\delta}$, we have
\begin{align*} ab - \delta < (a - \sqrt{\epsilon})(b + \sqrt{\epsilon}) < ab. \end{align*}
This suggests that $x \in (a - \sqrt{\delta}, a)$ and $y \in (b, b + \sqrt{\delta})$ will work.
The problem is that I'm suspicious my proposed solution will not work if I let $x$ and $y$ vary independently. For example, it's not clear to me if
\begin{align*} ab - \delta < \big( a - \frac{\sqrt{\delta}}{2} \big) \big( b + \frac{\sqrt{\delta}}{4} \big) < ab + \delta \end{align*}
is true or not.
Question:
Can anyone check my proof for Exercise 6, and tell me what's wrong with my attempted proof of Exercise 7? Thank you.
Let $ n$ enough great such that
$$0<a-\frac 1n<x<a+\frac 1n$$ $$0<b-\frac 1n<y<b+\frac 1n$$
then
$$0<ab -\frac 1n(a+b)+\frac{1}{n^2}<xy<ab+\frac 1n(a+b)+\frac{1}{n^2}$$
and $$|ab-xy|<\frac{|a+b|}{n} +\frac{1}{n^2}$$
Since $$\lim_{n\to+\infty}\frac{|a+b|}{n}+\frac{1}{n^2}=0$$ we can choose $ n$ such that $$\frac{|a+b|}{n}+\frac{1}{n^2}<\delta$$