Find the Set of Accumulation Points

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Mathyes

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The Problem: Let M be a set with the discrete metric and A \(\displaystyle \subset \) M be any subset. Find the set of accumulation points of A.

My Question:

Now I know that in discrete metric, d(x,y) = 1 if x \(\displaystyle \ne \) y and d(x,y) = 0 if x = y. So is this mean I have to consider two cases separately to find the set of accumulation points of A?
 
Mathyes said:
The Problem: Let M be a set with the discrete metric and A \(\displaystyle \subset \) M be any subset. Find the set of accumulation points of A. discrete metric, d(x,y) = 1 if x \(\displaystyle \ne \) y and d(x,y) = 0 if x = y.
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Suppose that \(\displaystyle t\in M\). Consider the ball, open set, \(\displaystyle \frak{B}(t;0.5)\). What is that set?
Can \(\displaystyle t\) be an accumulation point of any set?
 
To answer that question, is it the set {t} which is itself? But I am sorry that I still do not know how you come to choose \(\displaystyle {\rm B}(t,\frac{1}{2})\) instead of \(\displaystyle {\rm B}(t,5)\) or something. I guess that I am trying to understand which one shall I use, B(t, 1/2) or B(t,3), or B(t,5), etc. Would you please explain?
 
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Mathyes said:
To answer that question, is it the set {t} which is itself? But I am sorry that I still do not know why you choose \(\displaystyle {\rm B}(t,\frac{1}{2})\) instead of \(\displaystyle {\rm B}(t,5)\) or something. I guess that I am trying to understand which one shall I use, B(t, 1/2) or B(t,3), or B(t,5).
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You are correct: \(\displaystyle \frak{B}(t;\frac{1}{2})=\{t\}\). So that shows that in the discrete metric every singleton set is open.

Have you forgotten the definition of accumulation point?
If \(\displaystyle t\) is an accumulation point of a set \(\displaystyle A\) then every open set that contains \(\displaystyle t\) must also contain a point of \(\displaystyle A\setminus\{t\}\).

Is that possible here?
 
Thank you for the explanation,

Yes, it is possible here. So the set of accumulation point of A is t.

I hope I get it. As I read the problem again, it states that M is a set of discrete metric and A is a subset of M. This mean if I consider B(t,5) then every one in M get to be the accumulation points which is not an open set, unless I choose B(t, 1/2). This way t is an accumulation point because B is an open set which contain and t and contain point 1/2 of A.
 
Mathyes said:
Yes, it is possible here. So the set of accumulation point of A is t. NO!

I hope I get it. As I read the problem again, it states that M is a set of discrete metric and A is a subset of M. This mean if I consider B(t,5) then every one in M get to be the accumulation points which is not an open set, unless I choose B(t, 1/2). This way t is an accumulation point because B is an open set which contain and t and contain point 1/2 of A.
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You simply do not understand the definition of accumulation point.

The whole point of this exercise is to show you that in a discrete metric space there are no accumulation points.
 
Thanks again for the corrections,

Definition of the accumulation point: A point t in a metric space M is called an accumulation point of a set \(\displaystyle A \subset M\) if every open set V containing t contains some point of A other than t.

The condition is not met here, so there are no accumulation points in a discrete metric space.

Thank you so much for your help,
 
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