Hello everyone!
I'm having a hard time figuring out how to do this problem.
Any help would be highly appreciated!
Show that U = {(x, y) ∈ Rn : x2 + 4y2 < 4} is open by explicitly finding a ball around each point which is contained in U.
As for what understand, since we have x2 + 4y2 < 4, we can conclude that,
For each (x_0, y_0) contained in U, we have,
((x_0)^2, 4*(y_0)^2) = 4 - epsilon , for some epsilon>0
From there onwards, I would appreciate if I can have any of your help on this.
Sure! I can help you with that.
To show that U = {(x, y) ∈ R² : x² + 4y² < 4} is open, we need to demonstrate that for every point (x₀, y₀) in U, there exists a ball around that point which is entirely contained in U.
Let's start by taking an arbitrary point (x₀, y₀) in U. Since (x₀, y₀) is in U, it must satisfy the inequality x₀² + 4y₀² < 4.
To find a ball around (x₀, y₀), we need to find a positive value ε such that every point (x, y) within a distance of ε from (x₀, y₀) is also in U. In other words, we need to find a positive ε for which x² + 4y² < 4 for all (x, y) that satisfy |x - x₀| < ε and |y - y₀| < ε.
Now, let's analyze the equation x² + 4y² < 4. We can start by rearranging it to obtain x² < 4 - 4y², which implies |x| < √(4 - 4y²). This means that the x-coordinate of any point (x, y) in U must fall within the interval (-√(4 - 4y²), √(4 - 4y²)).
Similarly, for the y-coordinate, we have 4y² < 4 - x², which implies |y| < √((4 - x²) / 4). So, the y-coordinate of any point (x, y) in U must fall within the interval (-√((4 - x²) / 4), √((4 - x²) / 4)).
Now, let's choose ε to be the smaller of (√(4 - 4y₀²), √((4 - x₀²) / 4)). This ensures that any point (x, y) within a distance of ε from (x₀, y₀) will satisfy both the x and y conditions.
Therefore, the ball centered at (x₀, y₀) with radius ε is entirely contained in U. It can be written as {(x, y) ∈ R² : |x - x₀| < ε, |y - y₀| < ε}.
By explicitly finding this ball around each point (x₀, y₀) in U, we have shown that U is open.