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(a) Show that A = {$(x,y) \in \mathbb R^2 : x + y > 5$} is open.

(b) Show that B = {$(x,y) \in \mathbb R^2 : x.y \ge 5$} is closed. I couldnt solve in my homework.How should I start this proof? Should I take any arbitary pair satisfying the condition?

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What was your definition of an open/closed set? – Stefan Oct 15 '12 at 19:45

$A$ is the set of points lying strictly above the line $L$ whose equation is $x+y=5$. Suppose that $p=\langle a,b\rangle\in A$. The slope of the line $x+y=5$ is $-1$, so the slope of the perpendicular from $p$ to $L$ is $1$. The line of slope $1$ through $p$ has the equation $y-b=x-a$, or $y=x+b-a$. Solve the system

$$\left\{\begin{align*} &y=5-x\\ &y=x+b-a \end{align*}\right.$$

for the point of intersection of these two lines: $5-x=x+b-a$, $2x=5+a-b$, $=\frac12(5+a-b)$, and $y=5-x=\frac12(5-a+b)$. The point of intersection, $q=\left\langle\frac12(5+a-b),\frac12(5-a+b)\right\rangle$, is the point of $L$ closest to $p$. If $r$ is the distance from $p$ to $q$, the open ball of radius $r$ centred at $p$ lies entirely within $A$, and of course you can calculate $r$ from the coordinates for $p$ and $q$. This shows that $A$ is open.

To show that $B$ is closed, I think that it’s easiest to show that $\Bbb R^2\setminus B$ is open. Since $$\Bbb R^2\setminus B=\big\{\langle x,y\rangle:\in\Bbb R^2:xy<5\big\}\;,$$ your first step should be to draw the hyperbola $xy=5$, which has branches in the first and third quadrants, and figure out which part of the plane is $\Bbb R^2\setminus B$. This clearly includes the axes, where $xy=0$, and the second and fourth quadrants, where $xy<0$. A little thought will show you that in the first quadrant it’s the region under the curve, and in the third quadrant it’s the region above the curve. Now proceed much as in the first question: pick an arbitrary point $p=\langle a,b\rangle\in\Bbb R^2\setminus B$, and find an open ball around it lying entirely inside $\Bbb R^2\setminus B$.

You will probably want to consider at least a couple of cases. It’s very easy to do if $p$ is in the second or fourth quadrant, and almost as easy if $p$ is on one of the coordinate axes. Symmetry ensures that if you can do it for the part of $\Bbb R^2\setminus B$ in the first quadrant, you can do it for the part in the third quadrant, too. The only slightly hard work, then, is to show that if $0\le a,b$ and $ab<5$, there’s an open ball centred at $p$ and lying entirely under the righthand branch of the hyperbola. Don’t try to be too fancy: it’s not necessary to find the point of the hyperbola closest to $p$, as I did with the straight line $L$ in the first problem. Just pick a radius $r$ for $B(p,r)$ that you can prove is small enough for $B(p,r)$ to miss the hyperbola.

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