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How do I prove that an "open" square, centered in the origin is in fact an open set? I've already have this geometrical argument:

Let $S$ denote the square.

Suppose $(x,y) \in S$. Let $\delta = \min \{1 - |x|, 1 - |y|\}$.

Then, geometrically it is clear that $B_\delta(x,y) \subseteq S$. Hence $S$ is open.

However, How can I write this down and prove it in a formal matter?

Lee David Chung Lin
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Weierstraß Ramirez
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  • Say, for example (x,y) in R2 such that -1<x<1 and -1<y<1. – Weierstraß Ramirez Sep 05 '14 at 21:37
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    Can you be more specific? What topological space are you working with to start with? What tools do you have access too? – Benjamin Sep 05 '14 at 21:39
  • Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Fly by Night Sep 05 '14 at 21:55
  • I added the approach I have been taking with the problem, just need to know how to write it down. – Weierstraß Ramirez Sep 06 '14 at 01:36

3 Answers3

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Let $S$ denote the square you wrote in the comments.

Suppose $(x,y) \in S$. Let $\delta = \min \{1 - |x|, 1 - |y|\}$.

Then $B_\delta(x,y) \subseteq S$. Hence $S$ is open.

William
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  • Actually, I took this approach at first. The thing is I haven't been able to prove that as a matter of fact the elements contained in that ball belong inside the square. – Weierstraß Ramirez Sep 05 '14 at 21:43
  • @WeistrassRamirez: The triangle inequality often helps in that kind of task. – Lee Mosher Sep 05 '14 at 21:44
  • @WeistrassRamirez $B_\delta(x,y)$ is inside the square of side length $2\delta$ centered at $(x,y)$. By choice of $\delta$, the square of side length $2\delta$ centered at $(x,y)$ is inside $S$. I suggest drawing a picture and figure out how to write it up mathematically from there. – William Sep 05 '14 at 22:06
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    The Triangle Inequality can be used to turn geometry into pedantry. – André Nicolas Sep 05 '14 at 23:40
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It is enough to show that every point in the open square is the center of some open ball that is included entirely within the open square. Find the distance from the point in question to the nearest point on the boundary. Then the ball with that radius will serve.

PS in response to comments: Suppose the distance from $(x,y)$ to the nearest side of the square is $\delta=1-x$. If the distance from $(u,v)$ to $(x,y)$ is less than $\delta$ then $(u-x)^2+(v-y)^2<\delta^2$. From that it follows that $(u-x)^2<\delta^2$ so $|u-x|<\delta$, and from that we get $x-\delta<u<x+\delta=x+(1-x)=1$. Since $u>1$, the first component of the pair $(u,v)$ is such that the point must be strictly to the left of the right boundary of the square. And the other boundaries are farther away. Similar arguments handle the cases where one of the three other boundaries is the nearest one.

Michael Hardy
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  • That would be precisely $\delta = \min {1 - |x|, 1 - |y|}$. – Weierstraß Ramirez Sep 06 '14 at 01:52
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    @WeistrassRamirez : I prefer to avoid that detail in this answer because it would distract from the main idea. If I were trying to write a complete answer to an exercise I would of course include that, but I suspect that a person who asked a question like this didn't know precisely the thing that is what my answer says. – Michael Hardy Sep 06 '14 at 01:55
  • My fault, maybe I needed to be more specific in my question (I'm a new user), I've already edited my question. Just having some trouble proving the argument formally. – Weierstraß Ramirez Sep 06 '14 at 01:59
  • It was much more simple than I thought. Thanks a lot, you´ve been very helpful! – Weierstraß Ramirez Sep 06 '14 at 02:48
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It depends on what an open set is. If you mean that every point inside the square is contained in an open ball that is entirely contained within the square, then you can explicitly give the open ball $B_r(x)$ of radius $r$ centered around $x$ that is contained within the square.

davidlowryduda
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