# Thread: A simple topology question

1. ## [SOLVED: a counter example is given] A simple topology question

Hi all again.

I have a very simple question that I cannot find a clue.
$\displaystyle \mathrm{Ext}(A):=(\overline{A})^{c}$ (external of $\displaystyle A$), where $\displaystyle \overline{\cdot}$ is the closure and $\displaystyle \cdot^{c}$ is the complementary.
$\displaystyle \partial A:=\overline{A}\backslash\overset{\circ}{A}$ (boundary of $\displaystyle A$).
Prove that $\displaystyle \partial\mathrm{Ext}(A)=\partial A$.

Thanks.

2. A point $\displaystyle x \in \partial (A)$ if and only if each open set that contains $\displaystyle x$ contains a point in $\displaystyle A$ and a point in $\displaystyle A^c$.
Read that definition carefully several times.
It follows from that definition that any such point is a boundary point of both $\displaystyle A$ and $\displaystyle A^c$, recall that $\displaystyle \left[ {A^c } \right]^c = A$.
If $\displaystyle X$ is the whole space then $\displaystyle X = \text{Int} (A) \cup \partial (A) \cup \text{Ext} (A)$.

3. Thanks Plato, but I guess I need more help.

For instance, do we have $\displaystyle \mathrm{Int}(\overline{A})=\mathrm{Int}(A)$?
I could not find a counter example.

If so, I have a solution as follows.
$\displaystyle \partial(\mathrm{Ext}(A))=\partial((\overline{A})^ {c})$
.............._$\displaystyle =\overline{(\overline{A})^{c}}\backslash\mathrm{In t}((\overline{A})^{c})$
.............._$\displaystyle =(\mathrm{Int}(\overline{A}))^{c}\backslash(\overl ine{(\overline{A})})^{c}$
.............._$\displaystyle =(\mathrm{Int}(\overline{A}))^{c}\backslash(\overl ine{A})^{c}$
.............._$\displaystyle \stackrel{?}{=}(\mathrm{Int}(A))^{c}\backslash(\ov erline{A})^{c}$
.............._$\displaystyle =\overline{A^{c}}\backslash\mathrm{Int}(A^{c})$
.............._$\displaystyle =\partial(A^{c})$
.............._$\displaystyle =\partial(A).$

Is that true?

Note. We work on $\displaystyle \mathbb{R}$.

4. A slight variation on what Plato said: if you define the interior points of a set, A, as you do in metric topology- points, p, such that some neighborhood of p is a subset of A- then you can define the exterior of A as the set of all interior points of the complement of A and the boundary points of A as the set of all points in the space that are neither interior points nor exterior points of A. Because "complement of complement of A" is A, The interior points of A are the exterior points of $\displaystyle A^c$ and vice versa. What does that tell you about the boundary points of A.

5. I don't know why in the world you are making it so complicated.
In fact, I do not follow some of those steps.
$\displaystyle \begin{gathered} A = (0,1) \cup (1,2)\; \Rightarrow \;\text{Int} (A) = A \hfill \\ \overline A = \left[ {0,2} \right]\; \Rightarrow \;\text{Int} \left( {\overline A } \right) = (0,2) \ne \text{Int} (A) \hfill \\ \end{gathered}$

6. May be this is just because of my field (applied mathematics), I always wish to see some computations instead of words.

7. Originally Posted by Plato
I don't know why in the world you are making it so complicated.
In fact, I do not follow some of those steps.
$\displaystyle \begin{gathered} A = (0,1) \cup (1,2)\; \Rightarrow \;\text{Int} (A) = A \hfill \\ \overline A = \left[ {0,2} \right]\; \Rightarrow \;\text{Int} \left( {\overline A } \right) = (0,2) \ne \text{Int} (A) \hfill \\ \end{gathered}$
So to continue that example, $\displaystyle \partial A = \{0,1,2\}$, $\displaystyle \text{Ext}(A) = (-\infty,0)\cup(2,\infty)$ and $\displaystyle \partial(\text{Ext}(A)) = \{0,2\} \ne \partial A$.

In fact, as Plato showed, a set and its complement always have the same boundary. But a set and its "external" need not have the same boundary.

8. Thank you both Plato and Opalg, I really appreciate this help.
I could never find that counter example.

9. ## [Solved: A counter example is given]

Okay, I have a counter example too.
I cannot belive how I could not see it earlier.
Let $\displaystyle A:=[0,1]\cap\mathbb{Q}$, then we have $\displaystyle \partial A=[0,1]$ but $\displaystyle \partial\mathrm{Ext}A=\{0,1\}$.
This example also escapes from the gap that $\displaystyle (0,1)=\mathrm{Int}\overline{A}\neq\mathrm{Int}A=\e mptyset$.

Again many thanks for your asistance.