Open and Closed Sets in Metric Spaces

The open/closed dichotomy underlies essentially all of modern analysis: continuity of a function $f: X \to Y$ between metric spaces is equivalent to the preimage of every open set being open, a purely topological reformulation of the $\varepsilon$-$\delta$ definition. Compactness (every open cover has a finite subcover) and its practical consequences — the Extreme Value Theorem, existence of minimizers in optimization — are stated entirely in the language of open sets. In complex analysis, domains of holomorphy and the identity theorem rely on open connected sets; the maximum modulus principle is phrased via open sets on which a function attains no interior maximum. In differential geometry, manifolds are built by gluing open sets homeomorphic to $\mathbb{R}^n$, making the open-set toolkit the literal scaffolding of the definition of a manifold. In numerical analysis and optimization, closed sets (such as closed convex feasible regions) guarantee that limits of feasible sequences remain feasible, which is exactly why convergence proofs for gradient descent and projection algorithms invoke closedness. In probability theory, the Borel $\sigma$-algebra generating measurable sets is built starting from the open sets of a metric space, so every measure-theoretic statement about "measurable events" ultimately traces back to this article's definitions. In cryptography and coding theory, metrics on finite spaces (e.g. Hamming distance) still support the same ball/open-set machinery, used to define error-correcting sphere-packing bounds. These threads all begin with the same open ball defined here, and all continue directly into Topological Spaces and Open Sets and Closure, Interior, and Boundary.

Ask the tutor: Can you show me why every finite subset of a metric space must be closed, using the limit-point characterization directly?

Think of an open ball as "wiggle room": if $y \in B(x,r)$, you can nudge $y$ slightly in any direction (in a general metric space, "direction" just means "to any nearby point") and stay inside the ball, provided the nudge is small enough. Openness of a set $U$ says every point of $U$ has some amount of wiggle room, though the amount is allowed to shrink as you approach the "edge" of $U$ — as long as it never reaches zero for points actually inside $U$. Closed sets are the mirror image: a set is closed exactly when it has no missing edge — whenever a sequence of points in $F$ homes in on some target point of $X$, that target is already in $F$. This is why intervals like $[0,1]$ (which contain their endpoints) are closed, while $(0,1)$ (missing both endpoints) is open. The picture below shows the real line with these two intervals, highlighting the crucial difference: at the endpoints of $[0,1]$, no ball of any positive radius fits inside $(0,1)$, but every ball around an endpoint of $[0,1]$ pokes outside $[0,1]$ into the complement, and that is precisely what makes $[0,1]$'s complement open.

:::svg[Open interval $(0,1)$ has wiggle room at every point; closed interval $[0,1]$ contains its endpoints]

::: The second picture worth carrying around is the ball itself in $\mathbb{R}^2$: not always literally round if you change the metric, but always defined the same way. In the taxicab metric $d_1((x_1,y_1),(x_2,y_2)) = |x_1-x_2|+|y_1-y_2|$, the "ball" $B(0,1)$ is a diamond, not a disk — a reminder that "open ball" is a metric-dependent notion, even though the resulting open sets of $\mathbb{R}^2$ turn out to be the same collection for many common metrics (a fact revisited when comparing equivalent metrics).

Example 1: A half-open rectangle in $(\mathbb{R}^2, d_2)$ with the Euclidean metric. Let

We claim $A$ is neither open nor closed, and we locate exactly which boundary pieces belong to $A$. Not open: consider the point $p = (0, 3) \in A$ (on the left edge, $x=0$). For any $r>0$, the ball $B(p,r)$ contains points with negative $x$-coordinate, e.g. $q = (-r/2, 3)$, since $d_2(p,q) = r/2 < r$. Such $q \notin A$ because its $x$-coordinate is negative. Hence no ball around $p$ fits in $A$, so $A$ is not open. Not closed: consider $p' = (1, 1)$, which lies on the bottom edge $y=1$, excluded from $A$ (since $A$ requires $y>1$ strictly), so $p' \notin A$. But $p'$ is a limit point of $A$: for any $r>0$, the point $q' = (1, 1+\min(r/2, 4))$ satisfies $d_2(p',q') = \min(r/2,4) < r$ (for $r \le 8$; for $r>8$ take any point of $A$ near $p'$, which still exists since $A$ meets every ball around $p'$) and lies in $A$ (its $y$-coordinate is in $(1,5]$ and $x=1 \in [0,2)$). So every ball around $p'$ meets $A$, making $p'$ a limit point of $A$ not contained in $A$; by the sequential/limit-point characterization, $A$ is not closed. Boundary bookkeeping: the edges $x=0$ and $y=5$ belong to $A$ (matching the $\le$ signs), while $x=2$ and $y=1$ do not (matching strict inequalities). This asymmetric mix of open and closed edges is exactly why $A$ falls into neither the "open" nor "closed" category — a single strict inequality among several is enough to destroy openness, and a single non-strict inequality is enough to destroy closedness at the complementary edge.

Example 2: A ball in the space of bounded sequences $\ell^\infty$. Let $X = \ell^\infty$, the space of bounded real sequences $x = (x_1,x_2,\dots)$ with metric

(This is a metric: symmetry and positivity are immediate, and the triangle inequality follows from $|x_k - z_k| \le |x_k-y_k|+|y_k-z_k| \le d(x,y)+d(y,z)$ for every $k$, hence also for the supremum over $k$.) Consider the set

the closed unit ball centered at $\mathbf{0} = (0,0,\dots)$. We verify directly (not merely by name) that $C$ is closed, by showing $X \setminus C$ is open. Let $z \in X\setminus C$, so $\sup_k |z_k| > 1$; write $M := \sup_k |z_k| = 1 + \varepsilon$ for some $\varepsilon > 0$ (finite since $z \in \ell^\infty$ is bounded). We claim $B(z, \varepsilon) \subseteq X \setminus C$. Let $w \in B(z,\varepsilon)$, i.e. $d(z,w) = \sup_k|z_k - w_k| < \varepsilon$. Choose an index $k_0$ with $|z_{k_0}|$ close to $M$, say $|z_{k_0}| > M - \varepsilon/2$ (possible since $M$ is a supremum). Then by the reverse triangle inequality,

This alone does not immediately exceed $1$; we must be more careful and choose $k_0$ with $|z_{k_0}| > M - \delta$ for a small $\delta$ depending on $\varepsilon$, then bound $d(z,w)$ appropriately. Redo cleanly: given $w \in B(z,\varepsilon)$, for every index $k$,

Taking the supremum over $k$ on both extreme sides (valid since the inequality holds for every fixed $k$, so it holds for the index achieving the sup on the left, and the right side's sup is $M - \varepsilon$... but we must take sup of $|z_k|-\varepsilon$, which equals $M - \varepsilon$ since $\varepsilon$ is constant):

This only gives $\sup_k|w_k| \ge 1$, not strictly greater than $1$, so we tighten the radius: use $B(z,\varepsilon/2)$ instead. Repeating the computation with $\varepsilon/2$ in place of $\varepsilon$ gives

so $w \notin C$, i.e. $w \in X \setminus C$. This holds for every $w \in B(z,\varepsilon/2)$, so $B(z,\varepsilon/2) \subseteq X\setminus C$. Since $z \in X\setminus C$ was arbitrary, $X\setminus C$ is open, so $C$ is closed — exactly as its name "closed unit ball" promises, now verified from the definitions rather than assumed.