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

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In summary, the problem is to show that the square {(x,y) : |x|< 1, |y|< 1} is open in the metric space (R^2, d) with d being the standard euclidean metric. This can be done by showing that every point in the square is an interior point, meaning that an open ball can be constructed around the point that is still contained within the square. To find such a ball, take a typical point (x,y) and show that a small enough ball can be constructed to fit into the square.
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Pyroadept
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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:
 

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

What is an open set?

An open set is a set of points where every point has a neighborhood (a region around it) that is completely contained within the set. In other words, none of the points on the boundary of the set are included in the set itself.

How do you show that a square is an open set?

To show that a square is an open set, we need to prove that for any point within the square, there exists a small enough distance around that point that is also contained within the square. This can be done using the definition of an open set and by showing that the distance between any point and the boundary of the square is greater than zero.

3. Can a square be both an open set and a closed set?

No, a square cannot be both an open set and a closed set. A closed set is a set that includes all of its boundary points, while an open set does not include any of its boundary points. Since a square includes some of its boundary points (its corners), it cannot be an open set.

4. Are all squares open sets?

No, not all squares are open sets. A square is only considered an open set if it does not include any of its boundary points. If a square includes any of its boundary points, it is considered a closed set.

5. How does the concept of open sets relate to topology?

In topology, open sets are used to define the properties of a topological space. A topological space is a set of points with a defined notion of "nearness" between them. Open sets are used to describe the neighborhoods of points in a topological space, and this allows for the study of properties such as continuity and connectedness.

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