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

[Ankit Mishra]

Homework Statement

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.

Homework 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.

The Attempt at a Solution

I have no idea, please help!

 

[VeeEight]

I have no idea, please help!

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.

 

[jpjandrade]

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

 

[Ankit Mishra]

Well I wrote that the boundary points in a closed ball are contained, but i don't know how to prove it?

 

[HallsofIvy]

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

 

[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).
This is a contradiction.
In all cases, x' is an element of B[x0,ε].

Since B[x0,ε] contains all of its limit points, it is a closed set.