# Real Analysis: Local Metric Topology

## Interior, Boundary, and Closure

We will now formalize the notion of the three closely related concepts of the interior, boundary, and closure of a set, plus the exterior. For all discussions here, we assume that $(X, d)$ is a metric space.

**Interior**

Let $A$ be a subset of $X$. A point $x$ is called an **interior point** of $A$ if there exists an open set $S$ such that $x \in S \subseteq A.$ The **interior** of $A$ is the set of all of its interior points and is denoted by $A^{\circ}$. Formally,

$$A^{\circ} = \{ x \in A : \text{ there exists an open set } S \text{ such that } x \in S \subseteq A \}.$$

An equivalent and more useful definition of interior point says that $x$ is an interior point of $A$ if there exists an $r > 0$ such that $B_{r}(x) \subseteq A.$ Formally,

$$A^{\circ} = \{ x \in A : \text{ there exists an } r > 0 \text{ such that } B_{r}(x) \subseteq A \}.$$

This second definition is directly implied by the definition of open set, since every open set containing a point $x$ must contain as a subset an open ball centered at $x.$ This definition is more useful for computing interior points for particular sets, as it gives us direct access to the underlying metric to compute the distance between $x$ and other points of interest. In contrast, the open set definition requires that we first introduce an open ball containing $x$ by using the definition of open set each time. A third and equivalent definition of interior point says that $x$ is an interior point of $A$ if $A$ is a neighborhood of $x.$ This definition is easier to read aloud, but also hides the open ball behind an additional layer of abstraction.

In either case, the rigorous definition of the interior of a set lines up with our geometric intuition. For example, consider the set $(0, 1] \subseteq \mathbb{R}$. Neither $0$ nor $1$ is an interior point, since $0$ is not an element of $S$, and every open set containing containing $1$ contains points greater than $1$ and thus outside $(0, 1].$ On the other hand, the point 0.5 is an interior point, since $B_{0.2}(0.5) \subseteq (0, 1]$.

The interior of a set is easiest to visualize in $\mathbb{R}^2$ using the standard Eucliean metric. Consider the set $X$ as the square region of $\mathbb{R}^2$ in the picture below, and let $S$ be the subset of $X$ colored blue. The point $x$ is an element of $S$, and there exists an open ball $B_{r}(x)$ with a positive radius $r$ around $x$, which is colored red. Since $B_{r}(x)$ is an open subset of $S$, we can say that $x$ is an interior point of $S$.

**Boundary**

A point $x$ is called a **boundary point** of $S$ if every open set containing $x$ contains at least one point in $S$ and one other point not in $S$. Likewise, the **boundary** of $S$ is the set of all its boundary points, and is denoted $\partial S$. Formally,

$$\partial S = \{ x \in X : \text{ for all open sets } A \subseteq X \text{ containing } x, \text{ there exists a } y, z \in A \text{ such that } y \in S \text{ and } z \notin S \}.$$

As with the definition of interior points, we may make an equivalent definition in terms of open balls:

$$\partial S = \{ x \in X : \text{ for all } r > 0, \text{ there exists a } y, z \in B_{r}(x) \text{ such that } y \in S \text{ and } z \notin S \}.$$

This formalization of the notion of boundary also lines up with our geometric intuition; the points $0$ and $1$ are boundary points of $(0, 1]$, but no other points are. Note that boundary points need not be elements of the set they form the boundary of!

A direct corollary of the definition is that $\partial S = \partial \left(S^c\right)$. This also lines up with our intuition - the boundary between a set and what isn't the set is the same as the boundary between what isn't the set and the set itself. Less formally, the wall between your living room and the outside world is the same as the wall between the outside world and your living room.

In the diagram below, the point $x$ is a boundary point of $S$, as there is a point $y$ in $B_{r}(x)$ that is inside $S$ and a point $z$ in $B_{r}(x)$ that is not in $S$. Be careful, though! Points only need to be contained in *one* open set that is a subset of $S$ to qualify as interior points, but they need *every* open set containing them to have a point in $S$ and a point outside of $S$ to qualify as boundary points.

To distinguish it from the set complement and the set interior, the boundary of $S$ is drawn as a black line.

**Closure**

The **closure** of $S$ is the union of $S$ with its boundary and is denoted $\overline{S}$. Formally, $\overline{S} = S \cup \partial S$. For example, the boundary of $(0, 1]$ is the two-element set $\{0, 1\}$, so its closure is $\overline{(0,1]} = (0, 1] \cup \{0, 1\} = [0, 1]$.

**Exterior**

The **exterior** of $S$ is the complement of its closure and is denoted $\text{Ext}(S)$. Formally, $\text{Ext}(S) = \left(\overline{S}\right)^c$. A point in the exterior is called an **exterior point**. As shown below, an equivalent definition of the exterior states that the exterior is the interior of the complement. The interior and the exterior are thus similar in that they are both defined as unions of open sets and are separated by the boundary. To continue with the above example, the exterior of $(0, 1]$ is $(-\infty, 0) \cup (1, \infty)$.

## Problems

Show that $S^{\circ} \cap \partial S = \varnothing.$

Let $x \in S^{\circ}.$ Then there exists some $r > 0$ such that $B_{r}(x) \subseteq S.$ But this means that no point of $B_{r}(x)$ is outside of $S$, so $x \notin \partial S.$ Therefore $S^{\circ} \cap \partial S = \varnothing.$

Alternatively: Let $x \in \partial S.$ Then every open set centered containing $x$ contains a point outside of $S.$ Therefore no open set containing $x$ is a subset of $S$, so $x$ is not an interior point of $S.$ Therefore $S^{\circ} \cap \partial S = \varnothing.$

Alternative definition of interior: Show that $S^{\circ} = S - \partial S$.

First, assume $S^{\circ}$ is nonempty, and let $x \in S^{\circ}$. Then $x \in S$. Since $x$ is an interior point, there exists an open set $A$ such that $x \in A \subseteq S.$ However, because $A \subseteq S$, it does not contain any points outside of $S$, so $x$ is not a boundary point of $S$. Therefore $x \notin \partial S$, so $x \in S - \partial S$. If instead $S^{\circ}$ is empty, then there is no element $x \in S$ that is contained within an open set $A$ that is a subset of $S.$ Thus if there is an open set $A$ containing $x,$ it must contain an element outside of $S.$ Therefore all elements of $S$ are boundary points, and so $S - \partial S$ is also empty.

Conversely, assume $S - \partial S$ is nonempty, and let $x \in S - \partial S$. Then $x \in S.$ Since $x \notin \partial S$, it is not the case that every open set containing $x$ contains a point outside of $S$. Thus there is some open set containing $x$ that contains no points outside of $S$ and is therefore a subset of $S$. Therefore $x \in S^{\circ}$. If instead $S - \partial S$ is empty, then for all $x \in S,$ it is the case that any open set $A$ containing $x$ is not a subset of $S$. Therefore $x$ is not an interior point, and so $S^{\circ}$ is also empty.

Alternative definition of closure: Show that $\overline{S} = S^{\circ} \cup \partial S$.

Let $x \in S^{\circ} \cup \partial S$. Since $S^{\circ} \subseteq S$, it follows that $S^{\circ} \cup \partial S \subseteq S \cup \partial S$.

Conversely, assume $x \in \overline{S}$. Then $x \in S \cup \partial S$. If $x \in S - \partial S$, then $x \in S^{\circ}$. Likewise, if $x \notin S - \partial S$, then $x \in \partial S$. Therefore $x \in S^{\circ} \cup \partial S$.

Alternative definition of exterior: Show that $\text{Ext}(S) = (S^c)^{\circ}$.

Corollary: $x$ is an exterior point of $S$ if there exists an open set $A$ such that $x \in A \subseteq S^c.$

Assume $x \in (S^c)^{\circ}$. Then $x \in S^c$, so $x \notin S$. Likewise, because $\partial \left(S^c\right) \cap \left(S^c\right)^{\circ} = \varnothing$, it follows that $x \notin \partial \left(S^c\right)$. Therefore $x \notin S \cup \partial S = \overline{S}$, thus $x \in \left(\overline{S}\right)^c = \text{Ext}(S)$.

Conversely, assume $x \in \text{Ext}(S)$. Then $x \in \left(\overline{S}\right)^{c}$. Therefore $x \notin \overline{S}$. Since $S \subseteq \overline{S}$, it follows that $x \notin S$, therefore $x \in S^c$. Likewise, since $\partial S = \partial (S^c)$, and $\partial (S^c) \subseteq \overline{S^c}$ it follows that $x \notin \partial (S^c)$. Therefore $x \in S^c - \partial (S^c) = (S^c)^{\circ}$.

Show that $\partial S \cap \text{Ext}(S) = \varnothing$.

$ \partial S \cap \text{Ext}(S) = \left(\partial \left(S^c\right)\right) \cap \left(\left(S^c\right)^{\circ}\right) = \varnothing.$

Let $(X, d)$ be a metric space, and let $S \subseteq X$. Show that $X = S^{\circ} \cup \partial S \cup \text{Ext}(S)$.

$ S^{\circ} \cup \partial S \cup \text{Ext}(S) = \overline{S} \cup \text{Ext}(S) \\ S^{\circ} \cup \partial S \cup \text{Ext}(S) = \overline{S} \cup \left(\overline{S}\right)^{c} \\ S^{\circ} \cup \partial S \cup \text{Ext}(S) = X$

Let $(X, d)$ be a metric space. Calculate $X^{\circ}$, $\partial X$, and $\overline{X}$.

$X^{\circ} = X$, as every open set containing every point $x \in X$ is a subset of $X$. $\partial X = \varnothing$, since no open set containing any $x \in X$ contains any points outside of $X$. Therefore $\overline{X} = X$.

Consider $\mathbb{Q}$ as a subset of $\mathbb{R}$. Calculate $\mathbb{Q}^{\circ}$, $\partial \, \mathbb{Q}$, and $\overline{\mathbb{Q}}$.

Let $x \in \mathbb{R}$, and consider some open ball $B_{r}(x)$. Then $B_{r}(x)$ contains the points $x$ and $x + \dfrac{r}{2}$. Between these two points are at least one other rational number and one other irrational number, neither of which is $x$, and both of which are in $B_{\delta}(x)$. Therefore $x \in \partial \, \mathbb{Q}$, and likewise $x \notin \mathbb{Q}^{\circ}$. Therefore $\mathbb{Q}^{\circ}$ is empty, and $\partial \, \mathbb{Q} = \mathbb{R}$. It follows that $\overline{\mathbb{Q}} = \mathbb{R}$.

Consider the open interval $I = (a, b) \in \mathbb{R}$ where $a < b.$ Calculate $I^{\circ},$ $\partial I,$ and $\overline{I}.$

### Interior

Let $x \in I,$ and let $r= \dfrac{1}{2}\text{min}(x - a, b - x).$ Then $B_{r}(x) \subset I,$ so $x$ is an interior point of $I.$ Therefore $I^{\circ} = I.$

### Boundary

Next, consider some distance $r < \dfrac{b-a}{2}.$ The open ball $B_{r}(a)$ contains the point $p = a + \dfrac{r}{2},$ where $a < p < \dfrac{b+3a}{4} \in I,$ as well as the point $q = a - \dfrac{r}{2},$ where $a > q > \dfrac{5a-b}{4} \notin I.$ Furthermore, every open ball centered at $a$ with a radius greater than $r$ contains $B_{r}(a)$ as a subset, so $a \in \partial I.$ Likewise, the open ball $B_{r}(b)$ contains the points $m = b - \dfrac{r}{2},$ where $b > m > \dfrac{3b+a}{4} \in I,$ as well as the point $n = b + \dfrac{r}{2},$ where $b < n < \dfrac{5b-a}{4} \notin I.$ Furthermore, every open ball centered at $b$ of radius greater than $r$ contains $B_{r}(b)$ as a subset, so $b \in \partial I.$

### Exterior

Finally, consider the point $x \in (-\infty, a),$ and let $r = \dfrac{a-x}{2}.$ Then $B_{r}(x) \subset (-\infty, a).$ Similarly, consider the point $x \in (b, \infty)$ and let $r = \dfrac{x - b}{2}.$ Then $B_{r}(x) \subset (b, \infty),$ so $x \in \text{Ext}(I).$ Thus $(-\infty, a) \cup (b, \infty) \subseteq \text{Ext}(I).$

### Closure

By construction, the three regions above are mutually disjoint. Therefore there are no other points in $\partial I$ or $\text{Ext}(I).$ Thus $\partial I = \{a, b\}$ and $\overline{I} = [a, b].$

Consider the open unit disc $D_1 = \{ \textbf{x} \in \mathbb{R}^2 : d(\textbf{0}, \textbf{x}) < 1 \}$. Calculate $D_1^{\circ}$, $\partial D_1$, and $\overline{D_1}$.

### Interior

Let $\textbf{x} \in D_1$. Then $d(\textbf{0}, \textbf{x}) < 1$. Set $r = \dfrac{1 - d(\textbf{0}, \textbf{x})}{2}$. It follows that $d(\textbf{0}, \textbf{x}) + r < 1$. Consider the open ball $B_{r}(\textbf{x}) = \{ \textbf{n} : d(\textbf{n}, \textbf{x}) < r \}$. For any $\textbf{n} \in B_{r}(\textbf{x})$, it follows from the triangle inequality that $d(\textbf{0}, \textbf{n}) \leq d(\textbf{0}, \textbf{x}) + d(\textbf{x}, \textbf{n}) < d(\textbf{0}, \textbf{x}) + r < 1$. Thus $\textbf{x}$ is an interior point of $D_1$, so $D_1^{\circ} = D_1$.

### Exterior

Next, consider the set $E = \{ \textbf{e} : d(\textbf{0}, \textbf{e}) > 1 \}$. For any $\textbf{e} \in E$, set $r = \dfrac{d(\textbf{0}, \textbf{e}) - 1}{2}$. It follows that $d(\textbf{0}, \textbf{e}) - r > 1$. Consider the open ball $B_{r}(\textbf{e}) = \{ \textbf{n} : d(\textbf{n}, \textbf{e}) < r \}$. Then for any $\textbf{n}$ in $B_{r}(\textbf{e})$ we can see that $d(\textbf{0}, \textbf{e}) - d(\textbf{e}, \textbf{n})$ is positive, as $d(\textbf{0}, \textbf{e}) - d(\textbf{e}, \textbf{n}) > d(\textbf{0}, \textbf{e}) - r$. Therefore by the alternate triangle inequality, $d(\textbf{0}, \textbf{n}) \geq |d(\textbf{0}, \textbf{e}) - d(\textbf{e}, \textbf{n})| = d(\textbf{0}, \textbf{e}) - d(\textbf{e}, \textbf{n}) > d(\textbf{0}, \textbf{e}) - r > 1$. Therefore every $\textbf{e} \in E$ is an exterior point, so $E \subseteq \text{Ext}(D_1)$.

### Boundary

Finally, consider the unit circle $C = \{ \textbf{c} : d(\textbf{0}, \textbf{c}) = 1 \}$, and consider an open ball $B_{r}(\textbf{c})$ for any $\textbf{c} \in T$ where $r < 1$. Consider the point $\textbf{m} = \left(1 - \dfrac{r}{2}\right)\textbf{c}$. Then

$ d(\textbf{c}, \textbf{m}) = d\left(\textbf{c}, \left(1 - \dfrac{r}{2}\right)\textbf{c}\right) \\ d(\textbf{c}, \textbf{m}) = \sqrt{\sum\limits_{i=1}^2 \left(t_i - \left(1 - \dfrac{r}{2}\right)t_i\right)^2} \\ d(\textbf{c}, \textbf{m}) = \sqrt{\sum\limits_{i=1}^2 \left(t_i - t_i + \dfrac{r}{2}t_i\right)^2} \\ d(\textbf{c}, \textbf{m}) = \sqrt{\sum\limits_{i=1}^2 \left(\dfrac{r}{2}t_i\right)^2} \\ d(\textbf{c}, \textbf{m}) = \sqrt{\sum\limits_{i=1}^2 \dfrac{r^2}{4}t_i^2} \\ d(\textbf{c}, \textbf{m}) = \dfrac{r}{2}\sqrt{\sum\limits_{i=1}^2 t_i^2} \\ d(\textbf{c}, \textbf{m}) = \dfrac{r}{2}d(\textbf{0}, \textbf{c}) \\ d(\textbf{c}, \textbf{m}) = \dfrac{r}{2} \\ d(\textbf{c}, \textbf{m}) < r $

Therefore $\textbf{m} \in B_{r}(\textbf{c})$. Next observe that

$ d(\textbf{0}, \textbf{m}) = d\left(\textbf{0}, \left(1 - \dfrac{r}{2}\right)\textbf{c}\right) \\ d(\textbf{0}, \textbf{m}) = \sqrt{\sum\limits_{i=1}^2 \left(0 - \left(1 - \dfrac{r}{2}\right)t_i\right)^2} \\ d(\textbf{0}, \textbf{m}) = \sqrt{\sum\limits_{i=1}^2 \left(1 - \dfrac{r}{2}\right)^2t_i^2} \\ d(\textbf{0}, \textbf{m}) = \left(1 - \dfrac{r}{2}\right)\sqrt{\sum\limits_{i=1}^2 t_i^2} \\ d(\textbf{0}, \textbf{m}) = \left(1 - \dfrac{r}{2}\right)d(\textbf{0}, \textbf{c}) \\ d(\textbf{0}, \textbf{m}) = 1 - \dfrac{r}{2} \\ d(\textbf{0}, \textbf{m}) < 1 $

Therefore $\textbf{m} \in D_1$. Likewise, every open ball centered at $\textbf{c}$ with radius greater or equal to $1$ is a superset of $B_{r}(\textbf{c})$, so therefore contains $\textbf{m}$ as well. Therefore every open ball centered at $\textbf{c}$ contains an interior point of $D_1$.

Next, consider the point $\textbf{n} = \left(1 + \dfrac{r}{2}\right)\textbf{c}$. Then

$ d(\textbf{c}, \textbf{n}) = d\left(\textbf{c}, \left(1 + \dfrac{r}{2}\right)\textbf{c}\right) \\ d(\textbf{c}, \textbf{n}) = \sqrt{\sum\limits_{i=1}^2 \left(t_i - \left(1 + \dfrac{r}{2}\right)t_i\right)^2} \\ d(\textbf{c}, \textbf{n}) = \sqrt{\sum\limits_{i=1}^2 \left(t_i - t_i - \dfrac{r}{2}t_i\right)^2} \\ d(\textbf{c}, \textbf{n}) = \sqrt{\sum\limits_{i=1}^2 \left(-\dfrac{r}{2}t_i\right)^2} \\ d(\textbf{c}, \textbf{n}) = \sqrt{\sum\limits_{i=1}^2 \dfrac{r^2}{4}t_i^2} \\ d(\textbf{c}, \textbf{n}) = \dfrac{r}{2}\sqrt{\sum\limits_{i=1}^2 t_i^2} \\ d(\textbf{c}, \textbf{n}) = \dfrac{r}{2}d(\textbf{0}, \textbf{c}) \\ d(\textbf{c}, \textbf{n}) = \dfrac{r}{2} \\ d(\textbf{c}, \textbf{n}) < r $

Therefore $\textbf{n} \in B_{r}(\textbf{c})$. Next observe that

$ d(\textbf{0}, \textbf{n}) = d\left(\textbf{0}, \left(1 + \dfrac{r}{2}\right)\textbf{c}\right) \\ d(\textbf{0}, \textbf{n}) = \sqrt{\sum\limits_{i=1}^2 \left(0 - \left(1 + \dfrac{r}{2}\right)t_i\right)^2} \\ d(\textbf{0}, \textbf{n}) = \sqrt{\sum\limits_{i=1}^2 \left(1 + \dfrac{r}{2}\right)^2t_i^2} \\ d(\textbf{0}, \textbf{n}) = \left(1 + \dfrac{r}{2}\right)\sqrt{\sum\limits_{i=1}^2 t_i^2} \\ d(\textbf{0}, \textbf{n}) = \left(1 + \dfrac{r}{2}\right)d(\textbf{0}, \textbf{c}) \\ d(\textbf{0}, \textbf{n}) = 1 + \dfrac{r}{2} \\ d(\textbf{0}, \textbf{n}) > 1 $

Therefore $\textbf{n} \notin D_1$. Likewise, every open ball centered at $\textbf{c}$ with radius greater or equal to $1$ is a superset of $B_{r}(\textbf{c})$, so therefore contains $\textbf{n}$ as well. Therefore every open ball centered at $\textbf{c}$ contains an exterior point of $D_1$, and so $T \subseteq \partial D_1$.

### Closure

Note that $\mathbb{R}^2 = D_1 \cup E \cup T$ and $D_1$, $E$, and $T$ are all mutually disjoint. Thus there are no other elements of $E$ or $T$, so $\text{Ext}(D_1) = E$ and $\partial D_1 = T$. Therefore $\overline{D_1} = D_1 \cup T = \{ \textbf{x} : d(\textbf{0},\textbf{x}) \leq 1 \}$.

Show that if $S \subseteq T,$ then $\overline{S} \subseteq \overline{T}.$

Let $x \in \overline{S}.$ Then either $x \in S$ or $x \in \partial S - S.$ If $x \in S,$ then $x \in T \subseteq \overline{T}.$ If $x \in \partial S - S,$ then $x \notin S$ and every neighborhood of $x$ contains a point in $S$ and therefore contains a point in $T.$ If $x \in T - S,$ then $x \in T \subseteq \overline{T}.$ If $x \notin T,$ then $x \in \partial T \subseteq \overline{T}.$

Let $(X, d)$ be a metric space and let $S \subseteq X$. Show that $\left(S^{\circ}\right)^c = \overline{S^c}$.

Assume $x \in \left(S^{\circ}\right)^c$. Then $x \notin S^{\circ}$. Therefore $x \in \partial S \cup \text{Ext}(S)$. By definition of boundary, $\partial S = \partial (S^c)$. Likewise, $\text{Ext}(S) = (S^c)^{\circ}$. Therefore $x \in \partial S^c \cup (S^c)^{\circ} = \overline{S^c}$.

Conversely, assume $x \in \overline{S^c}$. Then $x \in \partial (S^c) \cup (S^c)^{\circ}$. By definition of boundary, $\partial (S^c) = \partial S$. Likewise, $(S^c)^{\circ} = \text{Ext}(S)$. Therefore $x \in \partial S \cup \text{Ext}(S)$, so $x \notin S^{\circ}$. Therefore $x \in \left(S^{\circ}\right)^c$.

Let $S$ be a subset of a metric space (X, d). Show that $x \in \overline{S}$ if and only if every open set containing $x$ contains a point in $S$.

Assume $x \in \overline{S}$. Then either $x \in S^{\circ}$, in which case every open set containing of $x$ contains a point in $S^{\circ}$, namely $x$ itself, or $x \in \partial S$, in which case every open set containing $x$ contains a point in $S$ by definition.

Conversely assume every open set containing $x$ contains a point in $S$. If $x \in S$, then $x \in \overline{S}$ by definition. If $x \notin S$, then $x \in \partial S$, as every open set containing of $x$ contains a point outside of $S$, namely $x$ itself, and a point in $S$ by assumption. Therefore $x \in \partial S \subseteq \overline{S}$.

Consider two subsets $S$ and $T$. Show that following $S^{\circ} \cup T^{\circ} \subseteq (S \cup T)^{\circ}$.

Let $x \in S^{\circ} \cup T^{\circ}$. Then there exists an $r > 0$ such that either $B_{r}(x) \subseteq S$ or $B_{r}(x) \subseteq T$. Since $S \subseteq S \cup T$ and $T \subseteq S \cup T$, it follows that $B_{r}(x) \subseteq S \cup T$. Therefore $x \in (S \cup T)^{\circ}$.

Counterexample: Show that $(S \cup T)^{\circ} \not\subseteq S^{\circ} \cup T^{\circ}$.

Let $S = D_1 - \{(0,0)\}$, i.e. the open unit disk without the origin, and let $T = \{(0,0)\}$. Then $S^{\circ} = S$ and $T^{\circ} = \varnothing$, so $S^{\circ} \cup T^{\circ} = S$. However, $S \cup T = D_1$, and $D_1^{\circ} = D_1$, which is a superset of $S$.

Counterexample: Show that $\left(\overline{S}\right)^{\circ} \neq S^{\circ}$.

Consider the set $S = (0, 1) \cup (1, 2) \subset \mathbb{R}$. We see that $\overline{S} = [0, 2]$, so $\left(\overline{S}\right)^{\circ} = (0, 2)$, but $S^{\circ} = (0, 1) \cup (1, 2)$.

The set $D_1 - \{(0,0)\}$ from the preceding proof is another such counterexample.

Let $S$ be a subset of a metric space $(X, d)$. Show that $\partial \left(\overline{S}\right) \subseteq \overline{S}$.

Recall that $\text{Ext}(S) = \left(S^c\right)^{\circ}$. Therefore $\partial \, \text{Ext}(S) \cap \text{Ext}(S) = \varnothing$, so $\partial \, \text{Ext}(S) \subseteq \left(\text{Ext}(S)\right)^c$. Therefore $\partial \left(\overline{S}\right) = \partial \left(\text{Ext}(S)^c\right) = \partial \, \text{Ext}(S) \subseteq \left(\text{Ext}(S)\right)^c = \overline{S}$.

Idempotence: Show that $\overline{S} = \overline{\left(\overline{S}\right)}$. In other words, show that the closure of the closure is the closure.

If $x \in \overline{S}$, then $x \in \overline{\left(\overline{S}\right)}$ by definition.

Conversely, let $x \in \overline{\left(\overline{S}\right)}$. Then either $x \in \left(\overline{S}\right)^{\circ}$ or $x \in \partial \left(\overline{S} \right)$. If $x \in \left(\overline{S}\right)^{\circ}$, then $x \in \overline{S}$ and the proof is complete. If instead $x \in \partial\left(\overline{S}\right)$, then $x \in \overline{S}$ by the preceding proof. Therefore $\overline{\left(\overline{S}\right)} \subseteq \overline{S}$. Therefore $\overline{S} = \overline{\left(\overline{S}\right)}$.

Show that the closure of a set is closed.

Let $S$ be a subset of some metric space $(X, d)$. Since $\text{Ext}(S) = \left(S^c\right)^{\circ}$, and the interior of a set is open by definition, it follows that complement of the exterior $\overline{S}$ is closed.

Equivalent definition of open: Show that a subset $S$ is open if and only if $S \cap \partial S = \varnothing$.

Assume $S$ is open. Then $S \cap \partial S = S^{\circ} \cap \partial S = \varnothing$. Conversely, assume $S \cap \partial S = \varnothing$. Then $S^{\circ} = S - \partial S = S$.

Equivalent definition of closed: Show that a subset $S$ is closed if and only if it $\partial S \subseteq S$.

Assume $S$ is closed. Then $S^c$ is open and $(S^c)^{\circ} = S^c$. Then $\partial (S^c) \cap S^c = \varnothing$. Therefore $\partial S = \partial (S^c) \subseteq (S^c)^c = S$.

Conversely, assume $\partial S \subseteq S$. Then $\partial S = \partial (S^c) \not\subseteq S^c$, and so $(S^c)^{\circ} = S^c - \partial (S^c) = S^c$. Therefore $S^c$ is open, so $S$ is closed.

Show that a set is clopen if and only if it has an empty boundary.

Let $(X, d)$ be a metric space.

First, let $U$ be a clopen subset of $X.$ Because $U$ is open, it follows that $U \cap \partial U = \varnothing.$ Because $U$ is closed, it follows that $\overline{U} = U,$ and so $\partial U \subseteq U.$ Therefore $\partial U = U \cap \partial U = \varnothing.$

Conversely, let $U$ be a set with an empty boundary. It follows that $U \cap \partial U = U \cap \varnothing = \varnothing,$ and so $U$ is open. Likewise, $\overline{U} = U \cup \partial U = U \cup \varnothing = U.$ Therefore $U$ is closed.

Show that every element of a finite subset of $\mathbb{R}$ is a boundary point, and therefore that finite subsets of $\mathbb{R}$ are closed.

Let $A = \{a_1, \ldots, a_n\}$ be a subset of $\mathbb{R}.$ For each $a_i \in A$, consider $r_i = \dfrac{1}{2} \min\limits_{j \neq i}(d(a_i, a_j))$, or half the distance to the other nearest point in $A$. Then $B_{r_i}(a_i)$ is an open set that contains $a_i$ and infinitely many other points of $\mathbb{R}$ that are not in $A$, and so $a_i$ is a boundary point. Therefore $\partial A = A$, so $\partial A \subseteq A$, so $A$ is closed.

Counterexample: Show that the closure of an open ball is not always the closed ball of the same radius.

Let $(X, d)$ be a metric space with at least two elements under the discrete metric. Then $B_1(x) = \{x\}$ for all $x \in X$. From this same fact it follows that $\partial B_1(x) = \varnothing$, so $\overline{B_1(x)} = \{x\}$. However, $\overline{B}_1(x) = X$.

Determine whether $\mathbb{Q}$ is relatively open, relatively closed, or neither with respect to $\mathbb{R}$ by analyzing its interior, boundary, closure, and exterior.

Recall that $\partial \, \mathbb{Q} = \mathbb{R}$ and $\mathbb{Q}^{\circ} = \varnothing$, so $\mathbb{Q}$ is neither open nor closed.

Show for every $S \subseteq \mathbb{R}$ that if $\inf(S)$ exists, then $\inf(S) \in \partial S.$

Let $S \subseteq \mathbb{R},$ and assume that $\inf(S)$ exists. Consider for any $r > 0$ the ball $B_{r}(\inf(S)).$ Note that $\inf(S) - \frac{r}{2} \in B_{r}(\inf(S))$ and that $\inf(S) - \frac{r}{2} \in S^c.$ Next, consider any point in the interval $(\inf(S), \inf(S) + \frac{r}{2}).$ The intersection of this interval with $S$ cannot be empty, as otherwise it would contain a value greater than $\inf(S)$ that was a lower bound for $S,$ which is a contradiction. Thus $B_{r}(\inf(S))$ always contains a point outside $S$ and a point inside $S.$ Since every open set containing $\inf(S)$ must contain such a ball, it follows that $\inf(S) \in \partial S.$

Show for every $S \subseteq \mathbb{R}$ that if $\sup(S)$ exists, then $\sup(S) \in \partial S.$

Let $S \subseteq \mathbb{R},$ and assume that $\sup(S)$ exists. Consider for any $r > 0$ the ball $B_{r}(\sup(S)).$ Note that $\sup(S) + \frac{r}{2} \in B_{r}(\sup(S))$ and that $\sup(S) + \frac{r}{2} \in S^c.$ Next, consider any point in the interval $(\sup(S) - \frac{r}{2}, \sup(S)).$ The intersection of this interval with $S$ cannot be empty, as otherwise it would contain a value less than $\sup(S)$ that was an upper bound for $S,$ which is a contradiction. Thus $B_{r}(\sup(S))$ always contains a point outside $S$ and a point inside $S.$ Since every open set containing $\sup(S)$ must contain such a ball, it follows that $\sup(S) \in \partial S.$