Proving a set is compact

[missavvy]

Homework Statement

Prove that K = {(x,y) | y$$\geq$$0, x² + y² $$\leq$$ 4 } is compact.

Homework Equations

The Attempt at a Solution

So a set is compact iff it's closed and bounded.
Closed:
Should I try to show that K^c is open? So that for any point x in the compliment there is r>0 s/t B(x,r) is in K^c.
Or is it better to try contradiction, assuming that there is some sequence in K, {v_n} that converges to v in K^c.
Then v = {(x,y) | y < 0 or x² + y² > 4}.
If I were to do this, how can I come up with a sequence? Because that means this would have to work for all sequences right?

Thanksssss!

 

[mighty2000]

To prove that K is compact, we need to show that it is both closed and bounded. Let's start with showing that K is closed.

To show that K is closed, we can use the fact that the complement of a closed set is open. So, we need to show that Kc is open. Let's take an arbitrary point (x,y) in Kc. This means that either y < 0 or x² + y² > 4. Let's consider the case where y < 0. In this case, we can choose r = -y/2, which is greater than 0. Then, for any point (a,b) in B((x,y),r), we have that b < y + r = -y/2 + y = y/2 < 0. This means that B((x,y),r) is completely contained in Kc, showing that Kc is open. The case where x² + y² > 4 can be shown similarly.

Now, to show that K is bounded, we can use the fact that any closed and bounded set in R² is compact. We can see that K is bounded since the set of points (x,y) satisfying x² + y² ≤ 4 is a disk of radius 2 centered at the origin, which is clearly bounded. Therefore, K is also compact.

I hope this helps. Let me know if you have any further questions.