(a) If A and B are disjoint closed sets in some

[Jamin2112]

(a) If A and B are disjoint closed sets in some ...

Homework Statement

(a) If A and B are disjoint closed sets in some metric space X, prove that they are separated.

(b) Prove the same for disjoint open sets.

(c) Fix p in X, ∂ >0, define A to be the set of all q in Z for which d(p,q) < ∂, define B similarly, with > in place of <. Prove that A and B are separated.

Homework Equations

Triangle inequality, definitions of open and closed sets

The Attempt at a Solution

Give me some hints to finishing part (c).

 

[mighty2000]

(a) To prove that A and B are separated, we need to show that there exists a distance δ > 0 such that for any two points x ∈ A and y ∈ B, we have d(x,y) > δ.

Since A and B are disjoint closed sets, we know that X \ A and X \ B are open sets. Since X \ A is open, for any point x ∈ A, there exists a δx > 0 such that the open ball B(x,δx) is contained in X \ A. Similarly, for any point y ∈ B, there exists a δy > 0 such that the open ball B(y,δy) is contained in X \ B.

Now, let δ = min{δx, δy}. Since A and B are disjoint, this means that for any x ∈ A and y ∈ B, the open balls B(x,δ) and B(y,δ) are disjoint. Therefore, d(x,y) > δ for all x ∈ A and y ∈ B, which means that A and B are separated.

(b) The proof for (b) is very similar to the proof for (a), except instead of using X \ A and X \ B, we use A^c and B^c, the complements of A and B. Since A and B are disjoint open sets, their complements A^c and B^c are closed. Following the same steps as in (a), we can show that A and B are separated.

(c) To prove that A and B are separated, we need to show that there exists a distance δ > 0 such that for any two points x ∈ A and y ∈ B, we have d(x,y) > δ.

Let x ∈ A and y ∈ B. This means that d(p,x) < ∂ and d(p,y) > ∂. We want to find a δ > 0 that satisfies d(x,y) > δ.

By the triangle inequality, we have d(x,y) ≤ d(x,p) + d(p,y). Since d(p,x) < ∂ and d(p,y) > ∂, we can choose δ = d(p,y) - d(p,x). This gives us d(x,y) ≤ d(x,p) + (d(p,y) - d(p,x)) = d(y,p) < ∂.

Therefore, for any x ∈