Real Analysis - Open Ball

Question

Show that U = {(x, y) ∈ Rⁿ : x² + 4y² < 4} is open by explicitly finding a ball around each point which is contained in U.

Answer 1

I will assume a closed ball around each point in U, but if an open ball is desired then the following explanation will still apply, with just that minor adjustment.

U is an open interior of an ellipse obtained by a vertical compression by a factor of 1/2 of the open ball of radius 2 centered at the origin in ℜ²

Therefore, we can first deal with the open ball of radius 2, define a closed ball around each of its points contained in that open ball, and then compress both the open ball and the closed ball simultaneously by a factor of 1/2, such that our open ball of radius 2 is transformed into U by the compression.

The result of this will be a closed interior of an ellipse around each point in U.

To make these closed interiors of ellipses into closed balls that are still nonetheless guaranteed to be contained in U, we can simply horizontally compress these closed interiors of ellipses by a factor of 1/2.

Let ƒ(R, α) = (R cos(α), R sin(α) / 2)

Then U = { ƒ(R,α) : (R,α) ∈ [0,2) x [0,2π) }

Then for any k ∈ (0,1) the closed ball around the point f(R, α) can be

{ (x,y) ∈ ℜ² : (x - R cos(α))² + (y - R sin(α) / 2)² ≤ k²(2 - R)² / 4 }

For any such k, this closed ball will be contained in U

If we wish for the balls around these points to be open balls, then k is allowed to be 1 as well