# Is it correct to define $\operatorname{cl}(\Omega) := \Omega \cup\partial \Omega$?

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?

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

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!

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.

• I feel the isolated points may not be taken into consideration in (1) – mathlover Dec 24 '20 at 0:54
• The isolated points of $\Omega$ are in $\Omega$. They’d be in the first part of the identity. – Clayton Dec 24 '20 at 0:56
• I'm editing my answer in order to adress the other issues. – Ottavio Bartenor Dec 24 '20 at 1:03
• I edited my question to include an incomplete proof, any idea how to complete it? – mathlover Dec 24 '20 at 1:06
• 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. – Ottavio Bartenor Dec 24 '20 at 1:19