# Real Analysis: 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.

### Neighborhoods

The set $N_r(x) = \{ a \in X : d(a, x) < r \}$ is called the **neighborhood** of radius $r$ at $x$ and consists of all points less than $r$ away from $x$ according to the metric $d$. The collection of sets $\mathfrak{N}(x)$ is the set of all neighborhoods of $x$ and is called the **complete system of neighborhoods** of $x$. For example, the open interval $(3, 5)$ is the neighborhood of radius $1$ around the point $4 \in \mathbb{R}$. When $d$ is the $2$-norm, $N_3([2, 4])$ is the set of points within the circle of radius $3$ centered at $[2, 4] \in \mathbb{R}^2$. The set $\overline{N}_r(x) = \{ a \in X : d(a, x) \leq r \}$ is called a **closed neighborhood** of radius $r$ at $x$. The choice of the word "closed" will become clear as we explore the concept of closure below.

### Interior

Let $S$ be a subset of $X$. A point $x$ is called an **interior point** of $S$ if there exists a $r > 0$ such that $N_{r}(x) \subseteq S$. In other words, $x$ is an interior point of $S$ if there is a neighborhood of $x$ that is also inside $S$. The **interior** of $S$ is the set of all of its interior points and is denoted by $S^{\circ}$. Formally,

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

This rigorous definition of interior 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 neighborhood of $1$ contains points greater than $1$ and thus outside $(0, 1]$, but the point 0.5 is an interior point, since $N_{0.2}(0.5) \subseteq (0, 1]$.

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 a neighborhood $N_{r}(x)$ with a positive radius $r$ around $x$, which is colored red. Since $N_{r}(x)$ is a subset of $S$, we can say that $x$ is an interior point of $S$.

**Note on Topological Diagrams: **Generic sets like $S$ are usually drawn as blobs rather than recognizable shapes to indicate the fact that they are arbitrarily chosen. Likewise, drawing them as blobs distinguishes them from those sets, such as $N_{r}(x)$, that are *not* arbitrary and have particular geometric qualities to them. In this case, neighborhoods like $N_{r}(x)$ are indeed necessarily circles (when using the standard Euclidean metric).

### Boundary

A point $x$ is called a **boundary point** of $S$ if every neighborhood of $x$ 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,

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

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 $N_{r}(x)$ that is inside $S$ and a point $z$ in $N_{r}(x)$ that is not in $S$. Be careful, though! Points only need to have *one* neighborhood that is a subset of $S$ to qualify as interior points, but they need *every* neighborhood to have a point in $S$ and a point outside of $S$ to qualify as boundary points. The entire 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**. 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 $N_{r}(x) \subseteq S$. But this means that no point of $N_{r}(x)$ is outside of $S$, so $x \notin \partial S$. Therefore $S^{\circ} \cap \partial S = \varnothing$.

Alternatively: Let $x \in \partial S$. Then every neighborhood of $x$ contains a point outside of $S$. Therefore no neighborhood of $x$ 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, let $x \in S^{\circ}$. Then $x \in S$. Since $x$ is an interior point, there exists some positive $r \in \mathbb{R}$ such that $N_{r}(x) \subseteq S$. However, because $N_{r}(x)$ is a subset of $S$, it does not contain any points outside of $S$, so $x$ is not a boundary point of $S$. Therefore $x \notin \partial S$, so $x \in S - \partial S$.

Conversely, let $x \in S - \partial S$. Since $x \notin \partial S$, it is not the case that every neighborhood of $x$ contains a point outside of $S$. Thus there is some neighborhood of $x$ that contains no points outside of $S$ and is therefore a subset of $S$. Therefore $x \in S^{\circ}$.

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 $\text{Ext}(S) = (S^c)^{\circ}$.

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 neighborhood of every point $x \in X$ is a subset of $X$. $\partial X = \varnothing$, since no neighborhood of 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 neighborhood $N_{\delta}(x)$. Then $N_{\delta}(x)$ contains the points $x$ and $x + \dfrac{\delta}{2}$. 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 $N_{\delta}(x)$. 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 $N_{r}(x) \subset I,$ so $x$ is an interior point of $I.$ Therefore $I^{\circ} = I.$

### Boundary

Next, consider some distance $r < \dfrac{b-a}{2}.$ The neighborhood $N_{r}(a)$ 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 neighborhood with a radius greater than $r$ contains $N_{r}(a)$ as a subset, so $a \in \partial I.$ Likewise, the neighborhood $N_{r}(b)$ 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 neighborhood of radius greater than $r$ contains $N_{r}(b)$ 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 $N_{r}(x) \subset (-\infty, a).$ Likewise consider the point $x \in (b, \infty)$ and let $r = \dfrac{x - b}{2}.$ Then $N_{r}(x) \subset (b, \infty),$ 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 neighborhood $N_{r}(\textbf{x}) = \{ \textbf{n} : d(\textbf{n}, \textbf{x}) < r \}$. For any $\textbf{n} \in N_{r}(\textbf{x})$, 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 neighborhood $N_{r}(\textbf{e}) = \{ \textbf{n} : d(\textbf{n}, \textbf{e}) < r \}$. Then for any $\textbf{n}$ in $N_{r}(\textbf{e})$ 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 a neighborhood $N_{r}(\textbf{c})$ 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 $\textbf{m} \in N_{r}(\textbf{c})$. 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 neighborhood of $\textbf{c}$ with radius greater or equal to $1$ is a superset of $N_{r}(\textbf{c})$, so therefore contains $\textbf{m}$ as well. Therefore every neighborhood of $\textbf{c}$ 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 $\textbf{n} \in N_{r}(\textbf{c})$. 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 neighborhood of $\textbf{c}$ with radius greater or equal to $1$ is a superset of $N_{r}(\textbf{c})$, so therefore contains $\textbf{n}$ as well. Therefore every neighborhood of $\textbf{c}$ 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 $\overline{D_1} = D_1 \cup T = \{ \textbf{x} : d(\textbf{0},\textbf{x}) \leq 1 \}$.

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 neighborhood of $x$ contains a point in $S$.

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

Conversely assume every neighborhood of $x$ 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 neighborhood 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 $N_{r}(x) \subseteq S$ or $N_{r}(x) \subseteq T$. Since $S \subseteq S \cup T$ and $T \subseteq S \cup T$, it follows that $N_{r}(x) \subseteq S \cup T$. 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 $\overline{S} = \overline{\left(\overline{S}\right)}$.