Proving that a set is open using epsilons. I'm trying to prove that the set $$A=\{x=(x_{1},x_{2})\in\mathbb{R}^2:x_{1}^{2}+x_{2}^{2}>1\}$$ is open in $\mathbb{R}^2$ with the usual norm is open with the definition of "epsilons". 
My attempt is based in taking $\epsilon=x_{1}^{2}+x_{2}^{2}-1,$ but simply don't achieve to prove that $B(x,\epsilon)\subset A.$ I tried too using another norm because of the equivalence among norms, but it's useless. 
Another form to attack the problem is taking complement over A, so $A^c$ is closed that's why is an closed ball. However I'd know the path using epsilon definition.
Anyone could help me?
Thanks in advance.
 A: Draw a picture of the circle that has centre the origin $O$, and radius $1$.
Let $P=(a,b)$ be such that $(OP)^2=a^2+b^2=r^2\gt 1$.  It is obvious that there is an open disk $D$ with centre $P$ such that $D$ lies entirely within $A$.  Now it's over.
But if we want to be explicit, the disk $D$ with radius $\epsilon=\frac{r-1}{2}$ will do the job. For suppose to the contrary that there is an $Q=(x,y)$ such that $x^2+y^2\le 1$ and the distance $PQ$ is less than $\frac{r-1}{2}$.  Then by the Triangle Inequality, we have
$$OP\le OQ +QP.$$
Thus $r\le 1+\frac{r-1}{2}=\frac{1+r}{2}$. This contradicts the fact that $r\gt 1$.
A: Let $\epsilon = d(x, 0) - 1$. Thus $d(x, 0) - \epsilon = 1$ where $d$ is the standard metric. 
Then if $y \in B_\epsilon(x)$ (open ball of radius $\epsilon$ centered at $x$), we want to show $d(y, 0) > 1$ (since $d(y, 0) > 1$ if and only if $y_1^2 + y_2^2 > 1$). 
By the triangle inequality 
$$d(y, 0) + \epsilon > d(y, 0) + d(x, y) \ge d(x, 0) > 1$$
Thus $$d(y, 0) > d(x, 0) - \epsilon = 1.$$
A: Your guess at using $\epsilon = x_1^2 + x_2^2 - 1$ is not correct, but perhaps it is going in the correct definition. Instead, you should use $\epsilon = \sqrt{x_1^2 + x_2^2} - 1$. 
The geometric/intuitive reason for this choice of $\epsilon$ is that it is equal to the shortest distance from $x$ to the set $A^c=\overline B(O,1)$: draw the segment $\overline{Ox}$ whose length equals $\sqrt{x_1^2+x_2^2}$, then subtract the radius of $\overline B(O,1)$ which equals $1$.
Once you have intuited the correct formula for $\epsilon$, the formal proof using the triangle inequality goes just like in the other answers.
