Prove this is a closed set/Real Analysis

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The discussion focuses on proving that the set {(a1, a2) ∈ R2 : 0 ≤ a1 ≤ 2, 0 ≤ a2 ≤ 4} is a closed set in the Euclidean metric. Participants explore various approaches, including demonstrating that the complement of the set is open by showing that every point in the complement can be surrounded by an open ball that does not intersect the set. A visual representation, likened to a tic-tac-toe board, is suggested to help conceptualize the problem and determine the appropriate radius for the open ball. The discussion emphasizes the importance of carefully selecting the radius based on the position of points in the complement. Ultimately, the goal is to establish that the complement is open, thereby confirming that the original set is closed.
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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?
 
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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.
 
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.
 
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 &gt; 2 and y &gt; 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 &gt; 2 and y &gt; 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
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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