Prove that $\times : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ is continuous I'm trying to solve the first batch of exercises from Topology and Groupoids, but I got stuck with the following one:

Let $C$ be a neighborhood of $c \in \mathbb{R}$, and let $ab = c$. Prove that there are neighborhoods $A \in \mathbb{R}$ of $a$, and $B \in \mathbb{R}$ of $b$, such that $x \in A$ and $y \in B$  implies $xy \in C$.

When $a, b > 0$, I can do the following:


*

*Define $A = \big] a - \delta_1, a + \delta_1 \big[ \subset \mathbb{R}^+$ and $B = \big] b - \delta_2, b + \delta_2 \big[ \subset \mathbb{R}^+$, where $\delta_1$ and $\delta_2$ are yet to be determined.

*If $x \in A$ and $y \in B$, then $xy \in \big] ab - b\delta_1 - a\delta_2 + \delta_1\delta_2, ab + b\delta_1 + a\delta_2 + \delta_1\delta_2 \big[ \subset \big] ab - \delta, ab + \delta \big[$.
Giving the following system of inequations:


*

*$0 < \delta_1 < a$

*$0 < \delta_2 < b$

*$b\delta_1 + a\delta_2 - \delta_1\delta_2 \le \delta$ (subsumed by the next one)

*$b\delta_1 + a\delta_2 + \delta_1\delta_2 \le \delta$


Which has solutions for $\delta_1$ and $\delta_2$.
This solution can be adapted to all other cases where $c \ne 0$ (that is, $a < 0 < b$, $b < 0 < a$, and $a, b < 0$) in a straightforward manner. However:


*

*I have no idea what to do when $c = 0$. What do I do in this case?

*Is there a better solution that doesn't require case-analyzing how $x$ stands in relation to $0$?

 A: Hint: $f:R^2\rightarrow R$ defined by $f(x,y)=xy$ is continuous. Let $U$ be a neighborhood of $c$, $p^{-1}(U)$ is open. Deduce the result from the fact that the projections of $R^2$ onto is factors are open.
Hint: for the continuity Let $(x_0,y_0)\in R^2$, $\mid xy-x_0y_0\mid\leq \mid x\mid\mid y-y_0\mid+\mid y_0\mid\mid x-x_0\mid$.  
A: Using your notation, given $\varepsilon > 0$, we want to find $\delta_1, \delta_2 > 0$ such that if $|x - a| < \delta_1$ and $|y - b| < \delta_2$ then $|xy - ab| < \varepsilon$. Playing with the expression, we have
$$ |xy - ab| = |xy - ay + ay - ab| \leq |x - a||y| + |a||y - b| = |x - a||y - b + b| + |a||y - b|  \\\leq |x - a||y - b| + |b||x - a| + |a||y - b|.$$
Choose $\delta_1, \delta_2 > 0$ such that:


*

*$\delta_1 \cdot |b| < \frac{\varepsilon}{3}$. If $b = 0$, this puts no limitation on $\delta_1$. If $b \neq 0$, we must have $\delta_1 < \frac{\varepsilon}{3|b|}$.

*$\delta_2 \cdot (\delta_1 + |a|) < \frac{2 \varepsilon}{3}$.


Then we will have
$$ |xy - ab| < \delta_1 \delta_2 + |b| \delta_1 + |a| \delta_2 = \delta_2 (\delta_1 + |a|) + \delta_1 \cdot |b| < \frac{\varepsilon}{3} + \frac{2\varepsilon}{3} = \varepsilon. $$
