# Metric Space, closed ball is a closed set. prove this

1. Nov 19, 2009

### Ankit Mishra

1. The problem statement, all variables and given/known data

Let (X, d) be a metric space. The set Y in X , d(x; y) less than equal to r is called a closed set with radius r centred at point X.

Show that a closed ball is a closed set.

2. Relevant equations

In a topological space, a set is closed if and only if it coincides with its closure. Equivalently, a set is closed if and only if it contains all of its limit points.

3. The attempt at a solution

2. Nov 19, 2009

### VeeEight

What have you tried? There are a few ways to show a set is closed. You can try showing that it contains all of its limit points, or try showing that the complement is open.

3. Nov 19, 2009

Remember, what is the definition of a limit point? Have you tried using the second definition of a closed set?

4. Nov 19, 2009

### Ankit Mishra

Well I wrote that the boundary points in a closed ball are contained, but i dont know how to prove it?

5. Nov 20, 2009

### HallsofIvy

Staff Emeritus
The closed about p, of radius r, is defined as $\{ q| d(p, q)\le r\}$. What are the boundary points of that set?

6. Oct 1, 2011

### ndwork

Here is my proof. I would appreciate any comments. Assuming it's correct, I like this proof because it doesn't require knowledge of sequences.

Let B[x0,ε] be the closed ball.
We will show that this set contains all of its limit points.
Case 1: |B[x0,ε]| < ∞
A finite set contains all of its limit points.
Case 2: Otherwise
Let x' be a limit point of B[x0,ε]. W.t.s. x' is in B[x0,ε].
Suppose x' is not in B[x0,ε]. Then d(x0,x') > ε.
Let γ = 1/2 * (d(x0,x')-ε). Then γ > 0.
Consider the open set B(x',γ). Since x' is a limit point of B[x0,ε], (B(x',γ) intersect B[x0,ε]) - {x'} ≠ null set.
Let a be an element of (B(x',γ) intersect B[x0,ε]) - {x'}.
By definition, d(x0,x') = ε + 2γ.
d(x0,a) + d(x',a) < ε + γ < ε + 2γ = d(x0,x').
By the triangle inequality, d(x0,x') < d(x0,a) + d(x',a).