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?

  • 1
    $\begingroup$ 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


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 $d$.


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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.