Let $f$ de multiplication function and complete the proof of the continuity of the function.

Question

Let $f$ de multiplication function and complete the proof of the continuity of the function. Any advice to do in my proof was very helpful

Answer 1

Let $U=(a,b)$, $a<b$, $a,b\in \mathbb{R}$ a open in $\mathbb{R}^2$ we should prove $f^{-1}(U)$ is open in $\mathbb{R}^2$.By definition $f^{-1}(U)=\lbrace a<xy<b \rbrace=L$ geometrically this is the region bounded by two hyperbolas. Fix $P=(p,q)\in L $, by definition $a<pq<b$ and chose $m=Min\lbrace b-pq,pq-a \rbrace$ and pick $\delta>0$ such that $\delta|p|,\delta|q|,3\delta^2<\frac{1}{3}m^2$ without loss of generality suppose thet $m=Min\lbrace b-pq,pq-a\rbrace=b-pq$. \ Consider the square $S=(p-\delta,p+\delta) \times (q-\delta,q+\delta)$ and $l=(l_1,l_2)\in S$ since $l$ is in $S$ $$p-\delta<l_1<p+\delta$$ $$q-\delta<l_2<q+\delta$$ $$(q-\delta)(p-\delta)<l_1 l_2<(p+\delta)(q+\delta)$$ $$pq-\delta p-\delta q+\delta^2<l_1l_2<pq+\delta p+\delta q+\delta^2$$ $$-\delta p-\delta<l_1l_2<\delta p+\delta q$$ since $\delta p< \delta |p|$ and $\delta q< \delta |q|$ $$-\frac{2}{3}m<l_1l_2< \frac{2}{3}m$$ $$-(b-pq)<l_1l_2<b-pq$$ notice that $b-pq<pq-a$ $$-(pq-a)<l_1l_2<b-pq$$ $$-pq+a<l_1l_2<b-pq$$ $$a<l_1 l_2<b$$ therefore $l\in L$ since $l$ was arbitrary point in $S$ then $l$ is open in $\mathbb{R}^2$ and the multiplication function is continuous function.\ Notice that we consider the case where $0<a<b$ in the case where $a<b<0$ is analogue, since our square $S$ is the same, the case where $a<0<b$ not are possibly because the region $L=f^{-1}(U)$ only hcan be in the first and third quadrant