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

Therefore, a closed ball is a closed set. In summary, we have shown that a closed ball is a closed set by demonstrating that it contains all of its limit points. This can be done by considering the different cases of the size of the closed ball and using the definition of a limit point. This proof does not require knowledge of sequences, making it a simple and elegant solution.
  • #1
Ankit Mishra
6
0

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!
 
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  • #2
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.
 
  • #3
Remember, what is the definition of a limit point? Have you tried using the second definition of a closed set?
 
  • #4
Well I wrote that the boundary points in a closed ball are contained, but i don't know how to prove it?
 
  • #5
The closed about p, of radius r, is defined as [itex]\{ q| d(p, q)\le r\}[/itex]. What are the boundary points of that set?
 
  • #6
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.
 
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1. What is a metric space?

A metric space is a mathematical concept that defines a set of objects (usually points) and a metric, which is a function that measures the distance between any two points in the set. It is a fundamental concept in analysis and topology.

2. What is a closed ball in a metric space?

A closed ball in a metric space is a set of all points that are within a certain distance, called the radius, from a given point, called the center. It is denoted as B(x,r), where x is the center and r is the radius.

3. What does it mean for a set to be closed in a metric space?

A set is considered closed in a metric space if it contains all of its limit points. A limit point is a point that can be approached arbitrarily closely by elements of the set. In simpler terms, a closed set is a set that includes all of its boundary points.

4. Why is a closed ball considered a closed set in a metric space?

A closed ball is considered a closed set in a metric space because it contains all of its limit points. Any point on the boundary of the closed ball can be approached arbitrarily closely by points within the ball, making it a closed set.

5. How can we prove that a closed ball is a closed set in a metric space?

To prove that a closed ball is a closed set in a metric space, we can use the definition of a closed set and show that the closed ball contains all of its limit points. This can be done by considering an arbitrary point on the boundary of the closed ball and showing that it can be approached by points within the ball. This will demonstrate that the closed ball is closed in the metric space.

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