
Interior points, boundary points, open and closed sets

Let $(X,d)$ be a metric space with distance $d\colon X \times X \to [0,\infty)$.

• A point $x_0 \in D \subset X$ is called an interior point in D if there is a small ball centered at $x_0$ that lies entirely in $D$,

$x_0 \text{ interior point } \defarrow \exists\: \varepsilon > 0; \qquad B_\varepsilon(x_0) \subset D.$

• A point $x_0 \in X$ is called a boundary point of D if any small ball centered at $x_0$ has non-empty intersections with both $D$ and its complement,

$x_0 \text{ boundary point } \defarrow \forall\: \varepsilon > 0 \quad \exists\: x,y \in B_\varepsilon(x_0); \quad x \in D,\: y \in X \setminus D.$

• The set of interior points in D constitutes its interior, $\mathrm{int}(D)$, and the set of boundary points its boundary, $\partial D$. $D$ is said to be open if any point in $D$ is an interior point and it is closed if its boundary $\partial D$ is contained in $D$; the closure of D is the union of $D$ and its boundary:

$\overline D := D \cup \partial D.$ $\qquad$Alternative notations for the closue of $D$ in $X$ include $\overline{{D\,}^X}$, $\mathrm{clos}(D)$ and $\mathrm{clos}(D;X)$.1)

Ex.
• In $\R$ with the usual distance $d(x,y) = |x-y|$, the interval $(0,1)$ is open, $[0,1)$ neither open nor closed, and $[0,1]$ closed.2)
• The set $D := \{(x,y) \in \R^2 \colon x > 0, y \geq 0\}$ is neither closed nor open in Euclidean space $\R^2$ (metric coming from a norm, e.g., $d(x,y) = \|x-y\|_{l_2} = ((x_1-y_1)^2 + (x_2-y_2)^2)^{1/2}$), since its boundary contains both points $(x,0)$, $x > 0$, in $D$ and points $(0,y)$, $y \geq 0$, not in $D$. The closure of D is

$\overline D = \{(x,y) \in \R^2 \colon x \geq 0, y \geq 0\}.$

• An entire metric space is both open and closed (its boundary is empty).
• In $l_\infty$, $B_1 \not\ni (1/2,2/3,3/4,\ldots) \in \overline{B_1}.$
• For a general metric space, the closed ball $\tilde B_r(x_0) := \{ x \in X\colon d(x,x_0) \leq r\}$ may be larger than the closure of a ball, $\overline{B_r(x_0)}$. If we let $X$ be a space with the discrete metric, $\begin{cases} d(x,x) &= 0,\\ d(x,y) &= 1, \quad x\neq y. \end{cases}$ Then $B_1(x_0) = \{x_0\}, \quad\text{ so that }\quad \overline{B_1(x_0)} = \overline{\{x_0\}} = \{x_0\}.$ But $\tilde B_1(x_0) = X.$

℘ (Open) balls are open

Let $(X,d)$ be a metric space, $x_0$ a point in $X$, and $r > 0$. Then $B_r(x_0)$ is open in $X$ with respect to the metric $d$.

Proof

Proof

Pick $x \in B_r(x_0)$. Then \begin{align} d(x,x_0) < r &\quad\Longrightarrow\quad \exists\: \varepsilon > 0; \quad d(x,x_0) < r - \varepsilon\\ &\quad\Longrightarrow \quad d(y,x) < \varepsilon \quad\text{ implies }\quad d(y,x_0) \leq d(y,x) + d(x,x_0) < \varepsilon + (r - \varepsilon) = r. \end{align} This means: $y \in B_r(x_0)$ if $y \in B_\varepsilon(x)$, i.e. $B_\varepsilon(x) \subset B_r(x_0)$.

1)
An alternative to this approach is to take closed sets as complements of open sets. These two definitions, however, are completely equivalent. In particular, a set is open exactly when it does not contain its boundary.
2)
Equivalent norms induce the same topology on a space (i.e., the same open and closed sets). Since all norms on $\R^n$ are equivalent, it is unimportant which norm we choose.