Prove this is a closed set/Real Analysis

[amanda_ou812]

Homework Statement

Prove that {(a1, a2) ∈ R2 : 0 ≤ a1 ≤ 2, 0 ≤ a2 ≤ 4} is a closed set in the Euclidean metric.

Homework Equations

Not sure

The Attempt at a Solution

How do I approach this problem? Do I prove there is a closed ball in my set? Or, do I prove there is an open ball in the complement of my set which makes my complement open which makes my set closed? Or, is there a better way to do it using union/intersections of open/closed sets?

 

[gauss^2]

Draw a picture, pick a point in the complement, and from that see how small your open ball centered at that point should be to ensure the open ball stays in the complement. Once you have that, then you have showed every point in the complement is an interior point of the complement, so the complement is open.

 

[amanda_ou812]

So my complement would be (-∞ < a1 < 0) U (2 < a1< ∞); (-∞ < a2< 0 ) U (4 <a2 < ∞)

And I have a ball centered at point x (x is in the complement) with radius r. So this ball is in the complement. What should I make my radius? I have a hard time choosing the radius. And then I chose an arbitrary point in my ball and show that it is in the complement.

 

[gauss^2]

Let's set $A = \{(a1,a2) \in \mathbb{R}^2: 0 \leq a1 \leq 2, 0 \leq a2 \leq 4\}$ and consider the case that $x > 2$ and $y > 4$ holds so that $(x,y)$ is in the complement of $A$. Then take $r$ to be the minimum of $a1 - 2$ and $y-4$. Then the ball of radius $r$ centered at $(x,y)$ doesn't intersect $A$ since $a1 - r \geq a1 - (2-a1) = 2$ and $a2 - r \geq a2 - (4-a2) = 4$. Hence, $(x,y)$ in an interior point of the complement of $A$ when $x > 2$ and $y > 4$ hold.

If you divide up the plane $\mathbb{R}^2$ like a tic-tac-toe board with your set $A$ in the middle, then you can apply similar arguments for the other cases. Just draw the tic-tac-toe board out and pick a point not in the center block (e.g., in $A$), and it should be easier to see how to choose $r$. Of course, you also have to consider the lines making up the grid too (other than the ones forming the boundary of $A$).

[PLAIN]http://img148.imageshack.us/img148/1112/drawingx.png