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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

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Suppose you don't know what radius $r$ to pick around your point $(y_1,y_2)$, and you've selected a point $(k_1, k_2)$ inside your ball. Still you know for sure $$(k_1-y_1)^2+(k_2-y_2)^2<r^2.\tag1$$ It follows that $$\text{$|k_1-y_1|<r\ $ and $\ |k_2-y_2|<r$}\tag2.$$ (Reason: a single term on the LHS of (1) is less than or equal to the sum of both terms.) Now (2) is the same as $$\text{$y_1-r<k_1<y_1+r\ $ and $\ y_2-r<k_2<y_2+r$}.\tag3$$ Taking just the left side of the inequalities (3), we conclude $$k_1+k_2>y_1+y_2-2r.\tag4$$ All of the above is true for any choice of $r$. So to ensure that the point $(k_1,k_2)$ lies within your original set $U:=\{(x_1,x_2):x_1+x_2>0\}$, what choice of $r$ will guarantee that $k_1+k_2>0$?

What's the motivation behind this argument? If you a draw the picture of (1) and the picture of (2), you've enlarged the circular region defined by (1) into the square region defined by (2). You do this so you can make assertions about the size of each of $|k_1-y_1|$ and $|k_2-y_2|$, which then lead to a statement (4) about the size of $k_1+k_2$. So to ensure that $k_1+k_2>0$ (which implies everything in your ball remains within $U$), it is enough to pick an $r$ that forces the entire square in (2) to remain within $U$.

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The line $x_2 = -x_1$ splits the plane to $2$ half-planes. The set $ A= \{(x_1,x_2) \in \mathbb R^2: x_1 + x_2 >0\}$ is the upper half plane.

We claim that for every point $Y(y_1,y_2)\in A $ we can find a radius $r>0$, such that $B_r(Y)\subset A.$ Setting the radius $r = \frac{y_1+y_2}{2}$ works.

We observe that the line $x_1 +x_2 = 0$ is secant to the ball which has radius $\rho = \frac{y_1+y_2}{\sqrt{2}}.$ Thus, every ball which has radius less than $\frac{y_1+y_2}{\sqrt{2}}$ will have no common points with the line, hence that ball will be exclusively in $A$.

We also observe that $r = \frac{y_1+y_2}{2}< \frac{y_1+y_2}{\sqrt{2}}=\rho$, which implies that $B_r(Y) \in A$.

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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.

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