Dear Anesh,

You want to show that a certain set, which I''ll call the unit square for convenience'' sake, is open. You stated your end goal correctly: Given an arbitrary point (x,y), you must produce some circle that 1) has (x,y) at its center, and 2) is entirely contained within the unit square. This is a fine goal, but it would help to refine it further — to have a guess about what circle will meet these two criteria.

If it''s not obvious how to proceed at this point, you should stop and make a diagram. Draw a unit square, draw a point somewhere inside the unit square, and draw a circle around your point that stays within the unit square. Do that now.

OK, you''ve drawn the diagram and have come back? Great, now describe the circle you drew: It''s entirely to the right of the square''s left side, to the left of the square''s right side, above the square''s bottom, and below the square''s top. What''s more, the point (x,y) is distance x away from the left side of the unit square, (1-x) away from the right side, y away from the bottom, and (1-y) away from the top. So, you know that the circle that you drew has a radius r that''s less than the minimum of these four numbers.

Now that you know the essential properties of the ball that will serve to demonstrate the openness of the unit square, your goal has become more precise: Given an arbitrary point (x,y) within the unit circle, and a radius r that is less than x, y, 1-x, and 1-y, you want to show that the open disc D of radius r around (x,y) lies entirely within the unit square. Now, proving that any object lies within the unit square requires you to show that the object lies to the right of the y-axis, above the x-axis, to the left of the line x=1, and below the line y=1. So let''s prove each of those four things for the open disc D…

Proof that D lies to the right of the y-axis.
Let''s use (p,q) to denote an arbitrary point in D, meaning that . We need to show that (p,q) lies to the right of the y-axis; in other words, that p>0. Suppose, with a view towards contradiction, thatp<=0 instead. Then we would have dist((x,y),(p,q)) = sqrt{(x-p)^2 + (y-q)^2} >= sqrt{(x-0)^2 + (y-q)^2} = sqrt{x^2 + (y-q)^2} >= sqrt{x^2} = x > r, which contradicts the statement that . Thus, p>0.

The proof that D lies above the x-axis is identical, but with x and p changed for y andq.

Proof the D lies to the left of the line x=1.
We again use (p,q) to denote an arbitrary point in D, meaning that . We need to show that (p,q) lies to the left of x=1; in other words, that p<1. Suppose, with a view towards contradiction, that p>=1 instead. Then we would have dist((x,y),(p,q)) = sqrt{(x-p)^2 + (y-q)^2} >= sqrt{(x-1)^2 + (y-q)^2} = sqrt{(x-1)^2} = abs{x-1} = 1-x > r, which contradicts the statement that . Thus, p<1.

And the proof that D lies below the line y=1 is again identical, but with x and pchanged for y and q.

So, given an arbitrary point (x,y) within the unit square, we''ve now demonstrated how to construct a disc/ball that has (x,y) at its center and is itself contained within the unit square. Therefore, the unit square is open.

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