Is the derived set closed in a metric space?

[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 $B(x,\varepsilon_x) \subseteq X\backslash A'$

 

[quasar987]

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

 

[jdstokes]

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

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

$\overline{A} = A \cup A'$.

Are there any counterexamples where $A' \neq \overline{A}$?

 

[jdstokes]

Ahh yes.

Now that I think about the definition of A', it is not always true that $A'\supseteq A$ 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.

$A' \supseteq A$ 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 $A'$ 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 $$A'\subset \overline{A}$$.

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 $$\overline{A}\subset A'$$.

So it must be that $$A'=\overline{A}$$.

 

[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 $A' \subseteq \overline{A}$ since $\overline{A} = A \cup A'$.

I now think my original post is in error because $x\not\in A' \not\implies x\not\in \overline{A}$.

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 $A' \supseteq A$. When I look at my lecture notes/lectures however, the define

$\overline{A} = A \cup A'$.

Are there any counterexamples where $A' \neq \overline{A}$?

The typical example is:
$$A=\{ \frac{1}{n} \}$$
$$A'=\{ 0 \}$$

 

[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.