Is the Square {(x,y) : |x|< 1, |y|< 1} an Open Set in the Metric Space (R^2, d)?

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The discussion focuses on proving that the square defined by {(x,y) : |x|< 1, |y|< 1} is an open set in the metric space (R^2, d) using the standard Euclidean metric. A set is considered open if every point within it is an interior point, meaning an open ball can be constructed around each point that remains entirely within the set. Participants suggest starting with a typical point (x,y) and determining a sufficiently small radius for the open ball that fits within the square. The conversation emphasizes the need for a rigorous proof rather than intuitive understanding or visual representation. The conclusion is that constructing the proof involves demonstrating the existence of such open balls for all points in the square.
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Homework Statement


In the metric space (R^2, d), where d is the standard euclidean metric, show that the square {(x,y) : |x|< 1, |y|< 1} is open.


Homework Equations


A set is closed if and only if its complement is open.

A set is open if every point in the set is an interior point, that is, you can construct an open ball around the point such that the open ball is contained in the set.



The Attempt at a Solution


Hi everyone,

I'm really stuck. Intuitively I can of course see it's right and if I draw a picture it looks right. My problem is with constructing the rigid proof; I don't have a clue where to begin. If I draw a unit circle within the square I could show that it's an open set as I know how to show an open ball is an open set. But does the proof for the square work along a similar principle? But how do I apply it?
I know you guys can't give me the full solution, but if you could even point me in the right direction I would really appreciate it!

Thanks in advance for any help. :)
 
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Hi Pyroadept! :smile:

Just take a typical point (x,y), and find a ball round it small enough to fit into the square. :wink:
 
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