accumulation points

logistic_guy

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Prove that if a set contains each of its accumulation points, then it must be a closed set.

💪:confused:😕
 
Prove that if a set contains each of its accumulation points, then it must be a closed set.

💪:confused:😕

Please define "accumulation points" using a numerical example.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
This problem is a little bit difficult, so I will answer it in steps. If one of my steps doesn't make sense, stop me and I will try to correct it.

For now let us assume that we have a set \(\displaystyle \text{A}\) which contains complex elements.
 
Please define "accumulation points" using a numerical example - as it is presented in your text book.
 
This problem is a little bit difficult, so I will answer it in steps. If one of my steps doesn't make sense, stop me and I will try to correct it.

For now let us assume that we have a set \(\displaystyle \text{A}\) which contains complex elements.
I will assume that set \(\displaystyle \text{A}\) contains all of its accumulation points.
 
Prove that if a set contains each of its accumulation points, then it must be a closed set.
This is an interesting and often used problem. As to the definition that depends upon the textbook in use.
Two major texts are: 1) by Hocking & Young (it in the R.L. Moore school) & 2) the other by R L Kelly.
The first does not even mention accumulation points. Kelly on the other hand
defines an accumulation point, [imath]\alpha[/imath], of a set [imath]A[/imath] to be the same as a limit point or cluster point.
So each open set [imath]O[/imath] that contains [imath]\alpha[/imath] must also contain a point of [imath]A[/imath] distinct from [imath]\alpha[/imath].
 
This is an interesting and often used problem. As to the definition that depends upon the textbook in use.
Two major texts are: 1) by Hocking & Young (it in the R.L. Moore school) & 2) the other by R L Kelly.
The first does not even mention accumulation points. Kelly on the other hand
defines an accumulation point, [imath]\alpha[/imath], of a set [imath]A[/imath] to be the same as a limit point or cluster point.
So each open set [imath]O[/imath] that contains [imath]\alpha[/imath] must also contain a point of [imath]A[/imath] distinct from [imath]\alpha[/imath].
Thank you pka. But the accumulation points in the op context are related to Complex Analysis! Maybe it doesn't matter.

Look for example at these two sets:

\(\displaystyle A = \{ z \in \mathbb{C} : |z| \leq 1 \}\)

\(\displaystyle B = \{ z \in \mathbb{C} : |z| < 1 \}\)

I know that set \(\displaystyle A\) contains all of its accumulation points while set \(\displaystyle B\) doesn't. The idea is simple but how to prove it is very difficult!
 
Let me revise what I did. I assumed that I have a set \(\displaystyle \text{A}\). Its elements are complex and it contains all of its accumulation points. I don't know how to write this with a notation but let me try this:

\(\displaystyle \text{A} = \{ z \in \mathbb{C} \mid z \in \text{A} \text{ and every accumulation point of } \text{A} \text{ is also in } \text{A} \}\)

I am sure that my set \(\displaystyle \text{A}\) must have a boundary. I will let \(\displaystyle \partial \text{A}\) be the boundary set.

I will read more about this topic and if I come with a new idea, I will post it next time!

:alien:👽
 
Prove that if a set contains each of its accumulation points, then it must be a closed set.
\(\displaystyle \textcolor{red}{\bold{Theorem}}\):

If a set \(\displaystyle A \subseteq \mathbb{C} \) contains all of its accumulation points, then \(\displaystyle A \) is a closed set.

\(\displaystyle \textcolor{red}{\bold{Definition}} \rightarrow \textcolor{green}{\bold{Accumulation \ Point}}\)

A point \(\displaystyle a \in \mathbb{C} \) is called an accumulation point (or limit point) of a set \(\displaystyle A \subseteq \mathbb{C} \) if:

\(\displaystyle \forall \epsilon > 0, \quad B_\epsilon(a) \cap (A \setminus {a}) \neq \emptyset\)

That is, every neighborhood of \(\displaystyle a \) contains at least one point of \(\displaystyle A \) different from \(\displaystyle a \).

\(\displaystyle \textcolor{blue}{\bold{My \ proof}}\):

Let \(\displaystyle A \subseteq \mathbb{C} \) be a set that contains all of its accumulation points.
Let \(\displaystyle \partial A \) denote the boundary of \(\displaystyle A \).

We will show that \(\displaystyle \partial A \subseteq A \) which implies that \(\displaystyle A \) is closed.

Let \(\displaystyle a \in \partial A \).
By definition of a boundary point, for every \(\displaystyle \epsilon > 0 \), the open ball:

\(\displaystyle B_\epsilon(a) = { z \in \mathbb{C} : |z - a| < \epsilon }\) contains points of \(\displaystyle A \) and points of \(\displaystyle \mathbb{C} \setminus A \). In particular, every neighborhood \(\displaystyle B_\epsilon(a) \) contains at least one point of \(\displaystyle A \).

We now show that \(\displaystyle a \) is an accumulation point of \(\displaystyle A \).

Let \(\displaystyle \epsilon > 0 \) be arbitrary.
We know that \(\displaystyle B_\epsilon(a) \cap A \neq \emptyset \).

If \(\displaystyle a \notin A \), then any point in \(\displaystyle B_\epsilon(a) \cap A \) is different from \(\displaystyle a \)
If \(\displaystyle a \in A \), we still ask: does \(\displaystyle B_\epsilon(a) \) contain other points of \(\displaystyle A \) besides \(\displaystyle a \)?

Since \(\displaystyle a \in \partial A \), and \(\displaystyle B_\epsilon(a) \) contains points of \(\displaystyle \mathbb{C} \setminus A \) as well, we can find sequences from both inside and outside \(\displaystyle A \) approaching \(\displaystyle a \).

This implies that:

\(\displaystyle B_\epsilon(a) \cap (A \setminus {a}) \neq \emptyset\)

So, for every \(\displaystyle \epsilon > 0 \), the neighborhood \(\displaystyle B_\epsilon(a) \) contains points of \(\displaystyle A \) other than \(\displaystyle a \).

Then, \(\displaystyle a \) is an accumulation point of \(\displaystyle A \).

But by the assumption, \(\displaystyle A \) contains all of its accumulation points,

then \(\displaystyle a \in A \).

This shows that \(\displaystyle \partial A \subseteq A \), meaning that \(\displaystyle A \) contains all of its boundary points.

Since \(\displaystyle A \) contains all its boundary points, then \(\displaystyle \textcolor{blue}{\bold{it \ is \ a \ closed \ set}}\).
 
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