# 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.
$\mathrm{Ext}(A):=(\overline{A})^{c}$ (external of $A$), where $\overline{\cdot}$ is the closure and $\cdot^{c}$ is the complementary.
$\partial A:=\overline{A}\backslash\overset{\circ}{A}$ (boundary of $A$).
Prove that $\partial\mathrm{Ext}(A)=\partial A$.

Thanks.

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

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

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

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

Is that true?

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

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 $A^c$ 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.
$\begin{gathered}
A = (0,1) \cup (1,2)\; \Rightarrow \;\text{Int} (A) = A \hfill \\
\overline A = \left[ {0,2} \right]\; \Rightarrow \;\text{Int} \left( {\overline A } \right) = (0,2) \ne \text{Int} (A) \hfill \\
\end{gathered}$

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.
$\begin{gathered}
A = (0,1) \cup (1,2)\; \Rightarrow \;\text{Int} (A) = A \hfill \\
\overline A = \left[ {0,2} \right]\; \Rightarrow \;\text{Int} \left( {\overline A } \right) = (0,2) \ne \text{Int} (A) \hfill \\
\end{gathered}$
So to continue that example, $\partial A = \{0,1,2\}$, $\text{Ext}(A) = (-\infty,0)\cup(2,\infty)$ and $\partial(\text{Ext}(A)) = \{0,2\} \ne \partial A$.

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 $A:=[0,1]\cap\mathbb{Q}$, then we have $\partial A=[0,1]$ but $\partial\mathrm{Ext}A=\{0,1\}$.
This example also escapes from the gap that $(0,1)=\mathrm{Int}\overline{A}\neq\mathrm{Int}A=\e mptyset$.

Again many thanks for your asistance.