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In my differential geometry class the professor wrote $$\operatorname{cl}(\Omega) := \Omega \cup\partial \Omega, \tag{1}$$ where $\operatorname{cl}$ is closure. From Topology I recall it was $$\operatorname{cl}(\Omega) := \operatorname{Int}(\Omega) \cup \partial \Omega \tag{2},$$ where $\operatorname{Int}$ is Interior.

Is $(1)$ also correct? If it is not , what is missing in that union? I feel the isolated points may not be taken into consideration in $(1)$.

Furthermore he said that historically the $\partial \Omega$ notation comes from the topological concept of derived set. But I don't think that the boundary is the same as the derived set, is it?


Edit: I am attempting a proof, taking $(2)$ as definition:

$$\operatorname{Int}(\Omega) \subseteq \Omega \implies \operatorname{Int}(\Omega) \cup\partial\Omega \subseteq \Omega\cup \partial\Omega $$

$$\implies \operatorname{cl}(\Omega) \subseteq\Omega\cup \partial\Omega $$

But don't know how to prove the oposite inclusion, any idea?

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    $\begingroup$ I think more often we define the interior and closure before we define the boundary, and we set $$\partial \Omega=\Omega^{\mathsf{C}}\backslash \Omega^{\circ}$$ $\endgroup$ – K.defaoite Dec 24 '20 at 0:49
  • $\begingroup$ Sorry, was the wrong way around. Fixed now. $\endgroup$ – K.defaoite Dec 24 '20 at 0:50
  • $\begingroup$ Definition or not is the equality true? $\endgroup$ – mathlover Dec 24 '20 at 0:50
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Proposition

Let $(X,d)$ be a metric space, let $E$ be a subset of $X$, and let $x_{0}$ be a point of $X$. Then the following statements are logically equivalent

(a) $x_{0}$ is an adherent point of $E$.

(b) $x_{0}$ is either an interior point or a boundary point of $E$.

(c) There exists a sequence $(x_{n})_{n=1}^{\infty}$ in $E$ which converges to $x_{0}$ with respect to $d$.

Proof

Let us start by the implication $(a)\Rightarrow(c)$.

If $x_{0}$ is an adherent point of $E$, then for every $n > 0$ there corresponds a $x_{n}\in B(x_{0},1/n)\cap E$.

Consequently, since $d(x_{n},x_{0}) \leq 1/n$, the sandwich theorem ensures us that $E\ni x_{n}\to x_{0}$, and we are done.

Now, we are going to prove the implication $(c)\Rightarrow(b)$.

In order to do so, we are going to assume that $(c)$ is true and $(b)$ is false.

Since $x_{0}$ is an exterior point of $E$, there exists a $\varepsilon_{0} > 0$ such that $B(x_{0},\varepsilon_{0})\cap E = \varnothing$.

On the other hand, since $x_{n}\in E$ converges to $x_{0}$, for the same value $\varepsilon_{0} > 0$ there is a $n_{\varepsilon_{0}}\in\mathbb{N}$ s.t. \begin{align*} n\geq n_{\varepsilon_{0}} \Rightarrow d(x_{n},x_{0}) \leq \varepsilon_{0} \Rightarrow x_{n}\in B(x_{0},\varepsilon_{0})\cap E \end{align*} which contradicts our assumption. Hence $(c)$ implies $(b)$.

Finally, we are going to prove the implication $(b)\Rightarrow(a)$.

If $x_{0}$ is an interior point of $E$, then there exists an open ball $x_{0}\in B(x_{0},r)\subseteq E$.

Consequently, $x_{0}$ is an adherent point of $E$.

Indeed, for every positive number $r > 0$, one has that $x_{0}\in B(x_{0},r)\cap E$.

On the other hand, if $x_{0}$ is a boundary point of $E$, then it is neither an interior point nor an exterior point.

This means that every open ball $B(x_{0},r)$ is not contained in $E$ nor $E^{c}$.

More precisely, every open ball $B(x_{0},r)$ intersects $E$ and $E^{c}$.

Hence we conclude that $x_{0}$ is an adherent point of $E$, and we are done.

Solution

Since $\overline{\Omega}\supseteq\Omega$ and $\Omega\supseteq\text{int}(\Omega)$, we can conclude based on the previous result that \begin{align*} \overline{\Omega} = \overline{\Omega}\cup\Omega &= \bigl(\text{int}(\Omega)\cup\partial\Omega\bigr)\cup\Omega =\\ &=\bigl(\text{int}(\Omega)\cup\Omega\bigr)\cup\partial\Omega =\\ &=\Omega\cup\partial\Omega \end{align*} and we are done.

Hopefully this helps!

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It's the same thing. If you require $\Omega$ to be an open set (or you use $\operatorname{Int}(\Omega)$ instead) you just end up with a disjoint union, but the same equality holds anyways: $$ \operatorname{cl}(\Omega) = \operatorname{Int}(\Omega)\,\sqcup\,\partial\Omega$$

The derived set (the set of all limit points) is a concept similar to closure, but not the same: the derived set of $\Omega$ doesn't take isolated points into account. In fact, the closure of any set $\Omega$ in a topological space is the disjoint union of limit points AND isolated points: $$ \operatorname{cl}(\Omega)=\operatorname{der}(\Omega)\,\sqcup\,\operatorname{Is(\Omega)}$$ Notice that all isolated points end up to be part of the boundary $\partial\Omega$. In general, you have: $$ \operatorname{Int}(\Omega)\subseteq \operatorname{der}(\Omega) \subseteq \operatorname{cl}(\Omega) $$ $$ \operatorname{Is(\Omega)} \subseteq \partial\Omega \subseteq \operatorname{cl}(\Omega) $$

The difference between $\Omega$ and $\operatorname{Int}(\Omega)$ has not necessarily to do with isolated points.
Take $\Omega=[0,1)\subset\mathbb R$ as an example; $\Omega$ (and its closure, $[0,1]$) have no isolated points, and still $\Omega\smallsetminus\operatorname{Int}(\Omega)\neq\varnothing$. You can "decompose" $\operatorname{cl}(\Omega)$ as $\operatorname{Int}\sqcup\partial=(0,1)\sqcup\,\{0,1\}$, and these are all limit points.

You might try experimenting with $\Omega=[0,1)\cup\{2\}$ to better understand what happens to closure and boundaries when isolated points are included.

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  • $\begingroup$ I feel the isolated points may not be taken into consideration in (1) $\endgroup$ – mathlover Dec 24 '20 at 0:54
  • $\begingroup$ The isolated points of $\Omega$ are in $\Omega$. They’d be in the first part of the identity. $\endgroup$ – Clayton Dec 24 '20 at 0:56
  • $\begingroup$ I'm editing my answer in order to adress the other issues. $\endgroup$ – Ottavio Bartenor Dec 24 '20 at 1:03
  • $\begingroup$ I edited my question to include an incomplete proof, any idea how to complete it? $\endgroup$ – mathlover Dec 24 '20 at 1:06
  • $\begingroup$ It depends on what your definition of boundary set is. Personally, I've always found easier to define closure priorly, and then define $\partial\Omega$ as $\operatorname{cl}\smallsetminus\operatorname{Int}$. Anyway, you just have to show that $\Omega$ is contained in its closure. $\endgroup$ – Ottavio Bartenor Dec 24 '20 at 1:19

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