My friend asked me to show continuity of $f(x,y) = xy$ at all points in $\Bbb R^2$. I started
We need to show $|xy-ab| \lt \epsilon$ whenever $d((a,b), (x,y)) < \delta$. So we have $|x-a| < \delta$ and $|y-b| < \delta$ then $$|xy-ab | = |xy-ab-bx+bx-ay+ay-xy+xy-ab| \\ = |x(y-b) + y(x-a) -(x-a)(y-b)| \\ < |x(y-b)| + |y(x-a)|+|(x-a)(y-b)|$$
Now we have $|x-a| < \delta$ or $a-\delta<x<a+\delta$ or $|x| < \max (|a+\delta|, |a-\delta|)$ and similarly $|y| < \max(|b-\delta|, |b+\delta|)$
So we obtain:
$$|xy-ab| < \delta (\max (|a+\delta|, |a-\delta))+\max(|b-\delta|, |b+\delta|)) \delta ^2 < \epsilon$$
Now how to conclude from here?? Do we take four cases?
Thanks a lot!
Edit
As Holo said, i mistakenly didnt write modulus $|a-\delta|$, now I have edited.
There is no need to add an subtract so much$$|xy-ab|=|xy-xb+xb-ab|\le|xy-xb|+|xb-ab|=|x||y-b|+|b||x-a|,$$ you can bound $|x|$ using delta and go on from there
In your case you have a little mistake: $a-\delta<x<a+\delta\implies |x| < \max (|a+\delta|, |a-\delta|)$, the $|\cdot|$ is important. From there you have $$|xy-ab| < \delta (\max (|a+\delta|, |a-\delta|)+\max(|b-\delta|, |b+\delta|)) \delta ^2 < \epsilon$$
Now you say that $\delta$ is at most $1$, so $\max (|a+\delta|, |a-\delta|),\max(|b-\delta|, |b+\delta|)<\max (|a+1|, |a-1|),\max(|b-1|, |b+1|)$, those $2$ are just regular old positive numbers, let's call them $A,B$ respectively.
Thus $$|xy-ab| < \delta (\max (|a+\delta|, |a-\delta|)+\max(|b-\delta|, |b+\delta|)) \delta ^2 < A\delta +B\delta ^2 =\epsilon$$
Now, because $Ax+Bx^2$ is bijective on $(0,\infty)$ it has inverse: $g(x)$(Just the positive side of the Quadratic Formula) so set $\delta=\min\{1,g(\epsilon)\}$