Question

For any set A in a metric space (X,d) show that cl(A)=AUD(A) where cl(A) is the closure of A and D(A) is the set of all limit points of A.

Answer 1

@Alchemista the problem is that i knew previously this is the definition but in a book i saw it is given to prove

Answer 2

You need to tell me what the definition of closure is in your course.

Answer 3

Closure of A is the set of all adherent points points and D(A) is the set of all limit points of A

Answer 4

This is directly from the text or notes, yes?

Answer 5

From the book

Answer 6

definition is from the book

Answer 7

Now, look carefully at the definition of adherent point and limit point.

Adherent point: \(x\) is an adherent point of \(A\) if every nbhd \(U\) containing \(x\) contains points from \(A\), in other words \(U \cap A \neq \varnothing\).

Limit point: \(x\) is a limit point of \(A\) if for every nbhd \(U\) containing \(x\), we have that \(U \setminus \{x\}\) contains points from \(A\). In other words \((U \setminus \{x\}) \cap A \neq \varnothing\).

Answer 8

From the two definitions, try to show that an adherent point is either a point in \(A\) or it is a limit point of \(A\).

Answer 9

The idea is that for a limit point \(x\) of a set \(A\), every neighborhood of \(x\) contains points in \(A\) other than the limit point \(x\).

Answer 10

Suppose \(x\) is an adherent point.

Case 1: \(x \in A\)
There is nothing to show.

Case 2: \(x \not\in A\)
We need to show that \(x\) is a limit point.

Can you do it, or do you need help?

Answer 11

@Alchemista