Show a set is open using open balls The set is $ \{ (x_1 , x_2)  :  x_1 + x_2 > 0 \}$ 
I wanted to solve this using open balls, so I said let $y = (y_1, y_2)$ be in the stated set. Then create an open ball $ B_r (y)$ around this point with radius $r = \frac {y_1 + y_2}{2} $. (My professor suggested this radius, but I understand why it'll work) Next, pick some point $k = (k_1, k_2)$ such that $k$ is in this ball. We must show that $k$ must also be in the original set. 
So then I though, ok, since k is in the ball then:
$$ \sqrt { (k_1 - y_1)^2 + (k_2 - y_2)^2 } < \frac {y_1 + y_2}{2}    $$
which implies that 
$$  (k_1 - y_1)^2 + (k_2 - y_2)^2 < (y_1 + y_2)^2  $$
I tried playing around with this, but couldn't arrive at the needed result of $k_1 + k_2 > 0$. It's frustrating, because it's a problem we did long ago in class, but it never clicked :(
Thank you for your help :D

Actually there is a general result: 
If function $f:\mathbb{R}^n\to \mathbb{R}$ is continuous on $\mathbb{R}^n,$ then for every real number $a,$ the set $\{x\in\mathbb{R}^n\mid f(x)>a\}$ is open in $\mathbb{R}^n$ (with respect to the natural Euclidean topology).
The proof of this result is elementary. For convenience, I give its proof here. 
Proof. Let $f$ be a real-valued function which is continuous on $\mathbb{R}^n,$ and $a$ be a real number. Let $E=\{x\in\mathbb{R}^n\mid f(x)>a\}.$ Suppose $x\in E.$ Then $f(x)>a$ and $f$ is continuous at $x.$  Assume $a<c<f(x).$  Put $\epsilon=f(x)-c.$ Then $\epsilon>0.$ By definition of continuity of $f$ at $x,$ there exists some $\delta>0$ such that for every $y$ in the ball $\mathbb{B}(x,\delta)$ of center $x$ with radian $\delta,$ we have $|f(y)-f(x)|<\epsilon.$ Thus we deduce that
$f(y)>f(x)-\epsilon=f(x)-\big(f(x)-c\big)=c>a,$ which implies that $y\in E,$ and so $\mathbb{B}(x,\delta)\subset E.$ Since $x$ is chosen arbitrarily, we have proved that $E$ is open.
Finally, what you need to do is to let $n=2, a=0,$ and $f(x_1,x_2)=x_1+x_2,$ which is surely continuous on $\mathbb{R}^n,$ and then, the desired result follows. 
