Real analysis problem (mathematics)

consider a set S such that S = {(x,y) such that x <1 and y <1}. How do you go about proving that S is an open set?
Here's what I have so far: any point (z,w) that belongs in the e-ball/e-neighborhood of (x,y) will be at a distance less that some small positive number (say e for epsilon) from (x,y). That is essentially the definition of an open set. So if I start there, and go on to prove that (z,w) satisfy the given condition, which is z < 1 and w < 1, would that be enough for proving S is open. If so, how do I prove z < 1 and w < 1

lLet (z,w) be an arbitrary point in S. Then z < 1 and w < 1.

Let e1 = half of 1 - z which is positive

let e2 = half of 1 -w which is positive

let e = the minimum of e1 and e2

now show that z + e (the upper bound of the x coordinate of a point in the ball) < 1

z + e <= z +e1 = z + (1 - z)/2 < z + (1 - z) = 1.

now show that w + e (the upper bound of the y coordinate of a point in the ball) < 1

This shows that any point in the ball you have defined is in S.

The key thing you are missing is to construct an e for an arbitrary point (z,w) that puts the ball inside S.

The idea is:

We can make epsilon as small as we like, so no matter how close the point (x,y) gets to the boundary (ie, x or y very very close to 1) we can always make an epsilon sufficiently small such that the 'e-ball' is also contained within S.

If it was S = {(x,y) such that x =<1 and y =<1}, S would be closed as the point (1,1) would be in S, but on the boundary, so no possible 'e-ball' would exist in S

Hope that helps