Show that the function $f(x, y) = xy$ is continuous

Question

I know how to show it using the fact, that the product of 2 continuous functions is continuous, but I'd like to prove it simply by using the definition ($f$ is cont. iff preimage of open set is open set).

I got to this: let $U$ be open and let $(a, b) \in f^{-1}(U)$, so $ab = f(a, b) \in U$, since $U$ is open there is some $\delta > 0: (ab - \delta, ab + \delta) \subseteq U$, then $f^{-1}((ab - \delta, ab + \delta)) \subseteq f^{-1}(U)$, hence it lies between lines $y_1 = \frac{ab - \delta}{x}, y_2 = \frac{ab + \delta}{x}$ and then, I guess, if I find $min\{d((a, b), (x, \frac{ab - \delta}{x})), d((a, b), (x, \frac{ab + \delta}{x}))\}$, I would get my open ball around (a, b), but that minimalization problem leads to the 4th degree polynomial equation, which I think shouldn't go that way.

Answer 1

Let $x_n\to x$ and $y_n\to y$ be artitrary sequences. Then $$f(x_n,y_n)=x_ny_n\to xy=f(x,y),$$ since we can simply use product of sequences convergences. You can prove it yourself if you want.

Answer 2

You should not try to find the biggest open set which fits into the preimage. It can also help to divide the problem into more than one to make calculations easier.

For simplicity let's assume $a, b\ge 0$ If $U$ is a neigbbourhood of $ab$ then there is $\varepsilon > 0 $ such that $(ab-\varepsilon, ab+\varepsilon )\subset U$ Now look at $v\in(a, a+\alpha),w\in( b+ \alpha)$ with $\alpha >0$ (the case of $v

Then by monotonicity $$zw <(a+\alpha)(b+ \alpha)= ab + (a+b)\alpha + \alpha^2$$ Now you need to ensure $(a+b)\alpha + \alpha^2<\varepsilon$ to get $zw \in [ab+ \varepsilon)$ The easiest way to do that is to make sure each summand is less than $\varepsilon /2$, which is easy to acchieve.

The case $(ab-\varepsilon, ab]$ can be treated similarly, also the case where $a$, $b$ or both are $< 0$.

(Note that $f^{1}(ab) $ consists of more points than $a, b$, actually it's not a very good idea to )

Answer 3

Hint:

I think it's better consider disk about $(0,0)$ with radius $\delta^2$ where $$|xy|\leq\frac12(x^2+y^2)<\delta^2$$

and for every point $(a,b)$ from $|x-a|^2\leqslant(x-a)^2+(y-b)^2\leqslant\delta^2$ and $|y-b|^2\leqslant(x-a)^2+(y-b)^2\leqslant\delta^2$ $$|xy-ab|=|(x-a)(y-b)+a(y-b)+b(x-a)|\leqslant\delta(\delta+|a|+|b|)$$