Topical webpage title: Is the Derived Set Closed? A Proof and Discussion

[jdstokes]

Hi all,

Would anyone be able to comment if my proof is sound or can be simplified.

Let (X,d) be a metric space, A a subset of X. The derived set A' of A is the set of all limit points of sequences in A.

Proposition: A' is closed.

Proof: Show that its
complement is open. Let x\in X\backslash A' Then x\not\in \overline{A} so there is a
non-zero \varepsilon_x s.t. B(x,\varepsilon_x) \cap A = \emptyset<br /> \Rightarrow B(x,\varepsilon_x) \subseteq X \backslash A. Suppose
\exists y \in B(x,\varepsilon_x) which in addition belongs to
A'. Then there's an open ball B(y,\varepsilon_y) \subseteq<br /> B(x,\varepsilon_x) about y which intersects A. Then
B(x,\varepsilon_x) \cap A \neq \emptyset. This contradiction shows
that B(x,\varepsilon_x) \subseteq X\backslash A&#039;

 

[quasar987]

Please correct me if I'm wrong, but as far as I can see, A&#039;=\overline{A} !

 

[jdstokes]

Please correct me if I'm wrong, but as far as I can see, A&#039;=\overline{A} !

Now that I think about it, it seems like you are correct. Since surely A&#039; \supseteq A. When I look at my lecture notes/lectures however, the define

\overline{A} = A \cup A&#039;.

Are there any counterexamples where A&#039; \neq \overline{A}?

 

[jdstokes]

Ahh yes.

Now that I think about the definition of A', it is not always true that A&#039;\supseteq A since the definition of A' requires that for each x in A there by a point, a ,NOT EQUAL to x s.t. d(x,a) < epsilon for all epsilon > 0.

A&#039; \supseteq A will fail for any discrete space.

 

[quasar987]

I don't understand your last two posts.

Is it true or not that

The derived set A' of A is the set of all limit points of sequences in A.

?

If so, then we can show that every point in A&#039; is in \overline{A} and vice versa. Firstly, observe that \overline{A}=A\cup \mbox{Acc}(A) (Acc(A) being the set of all accumulation points of A).

Let x be in A'. If x is in A, then x is in the closure also. If not, then x is obviously an accumulation point of A, so x is in the closure again. This shows that A&#039;\subset \overline{A}.

Now let x be in \overline{A}. Then x is in A or in Acc(A). If x is in A, then define the constant sequence x_n=x. So x is in A'. If x is in Acc(A), then it is also easy to find a sequence of elements of A converging to x, so that x is in A' again. This shows that \overline{A}\subset A&#039;.

So it must be that A&#039;=\overline{A}.

 

[jdstokes]

The word limit point and accumulation point are equivalent in the course I'm taking. You cannot claim that x is a limit point because it is the limit of the constant sequence {x,x,...}.

If you take the discrete space consisting of just two points, say. Then no point in the space is an accumulation point/limit point.

It is of course true that A&#039; \subseteq \overline{A} since \overline{A} = A \cup A&#039;.

I now think my original post is in error because x\not\in A&#039; \not\implies x\not\in \overline{A}.

Do you have any suggestions on how to correct this?

thanks

James

 

[quasar987]

To show Acc(A) is closed, show that a sequence of points in Acc(A) lies in Acc(A).

 

[Jimmy Snyder]

To show Acc(A) is closed, show that a sequence of points in Acc(A) lies in Acc(A).

I'm sure you meant to write:

To show Acc(A) is closed, show that if a sequence of points in Acc(A) converges to a limit, then the limit lies in Acc(A).

 

[NateTG]

Now that I think about it, it seems like you are correct. Since surely A&#039; \supseteq A. When I look at my lecture notes/lectures however, the define

\overline{A} = A \cup A&#039;.

Are there any counterexamples where A&#039; \neq \overline{A}?

The typical example is:
A=\{ \frac{1}{n} \}
A&#039;=\{ 0 \}

 

[quasar987]

I'm sure you meant to write:

To show Acc(A) is closed, show that if a sequence of points in Acc(A) converges to a limit, then the limit lies in Acc(A).

Yes.