# Show the set is open

I've been seriously trying to solve this for really lots of time. It's pretty basic, which furthers the frustration...

Let there be a set $A=\{x=(x_1,x_2)\in \mathbb R^2 :|x_1|+|x_2|<1\}$. Show it is an open set.

So first of all I say it's all the points within a square in the plane on $(1,0), (0,-1), (-1,0)$ and $(0,1)$.

I took the following approach - and if you could, I would like you to show me or hint me how to solve in this approach: Let there be $P = (x_1,x_2) \in A$, and lets show $P$ is an interior point.

Now that's what I got stuck with. I want to define a circle $B((x_1,x_2),r)$ for some radius $r$, and finding $r$ is what I got problems with.

My attempt: Lets assume $|x_1| \leq |x_2|$, and so we'll choose $r=1-|x_2|$, and define $B((x_1,x_2),r)$. And lets take $(y_1,y_2) \in B$, and I need to prove that $(y_1,y_2)\in A$. Just couldn't make it...

Thanks for any help in advance!

• Just take $r=K(1-|x_1|-|x_2|)$ for an appropriately chosen $K$. (For instance, $K=1/10$ clearly works, but isn't the largest value that does.) Commented Mar 3, 2014 at 21:51
• can you perhaps show me how to algeribcally prove it's good?
– Jim
Commented Mar 4, 2014 at 13:17

There are several ways to show it:

Let the function $f: \mathbb R^2\to\mathbb R$, with $$f(x_1,x_2)=\lvert x_1\rvert+\lvert x_2\rvert.$$ Clearly $f$ is continuous and as $(-\infty,1)$ is open in $\mathbb R$, so is its inverse image: $$f^{-1}(-\infty,1)=\big\{(x_1,x_2)\in\mathbb R^2: \lvert x_1\rvert+\lvert x_2\rvert<1 \big\}.$$

You can do it geometrically. Take a point in your set, and suppose it is in the first quadrant. Drop lines parallel to the axes to the line $x+y=1$, and mark half the distance to the intersection. Then the circle with that radius is entirely contained in the square, because the diagonal is strictly longer that the radius.

As you said let $P=(x_1,x_2)$ then set $r_0=|x_1|+|x_2|$ then it is on the square $|x|+|y|=r_0$ and $r_0<1$.

Then choose $r=(1-r_0)/\sqrt 2$ then $B(P,r)$ is insede of the big square.

Why did I choose $r$ in that pattern, it is completley geometric,Just drow the squares!

The complement of $A$ contains its boundary. Hence, $A^C$ is closed, so $A$ is open.

• This almost begs the question! You could've easily said "$A$ is disjoint from $\partial A$".
– Pedro
Commented Mar 4, 2014 at 2:16
• @PedroTamaroff ;) Commented Mar 4, 2014 at 2:49
• But I am serious here. =)
– Pedro
Commented Mar 4, 2014 at 2:56
• Yes, well, the definition of an open set is a set whose complement is closed. An equivalent definition for a closed subset of $\mathbb{R}^n$ is a subset that contains its boundary. $A^C$ obviously contains its boundary, although demonstrating that could be done with a bit of work, if so desired. Commented Mar 4, 2014 at 3:06
• Yes, the set in the post is obviously open too. That's my point.
– Pedro
Commented Mar 4, 2014 at 3:14