Open and Closed Sets in R^n

The vocabulary of open and closed sets is the load-bearing frame for essentially all of modern analysis. In multivariable calculus itself, the definition of continuity of $f:\mathbb{R}^n \to \mathbb{R}^m$ can be rephrased purely in terms of open sets ("preimages of open sets are open"), which is exactly the bridge to Limits and Continuity in Several Variables. Compact sets — closed and bounded subsets of $\mathbb{R}^n$ by the Heine–Borel theorem — are built directly from closedness, and drive the extreme value theorem and uniform continuity results developed in Compactness in R^n. Beyond calculus, this language recurs across mathematics and its applications:

• Differential geometry and manifolds: manifolds are built by gluing together pieces that look like open subsets of $\mathbb{R}^n$, and smoothness, charts, and tangent spaces are all phrased locally using open neighborhoods.
• Optimization and convex analysis: constraint sets in linear and convex programming are frequently closed (feasibility regions defined by $\le$ inequalities), and existence of optimal solutions typically relies on closedness plus boundedness (compactness).
• Numerical analysis and PDE theory: boundary conditions for partial differential equations are stated on the boundary $\partial \Omega$ of a domain $\Omega \subset \mathbb{R}^n$, and whether $\Omega$ is open (interior of the domain) versus its closure matters for well-posedness of Dirichlet and Neumann problems.
• Probability and measure theory: open and closed sets generate the Borel $\sigma$-algebra on $\mathbb{R}^n$, the foundation for defining probability measures and random vectors rigorously.
• Computer science and cryptography: in computational geometry, "is this point interior/boundary/exterior to a region" tests drive collision detection and mesh algorithms; in cryptographic lattice problems, closedness of certain sets (e.g. lattices themselves, which are closed and discrete) underlies security reductions for lattice-based cryptosystems.
• Dynamical systems: invariant sets, attractors, and basins of attraction are frequently characterized as closed (attractors) or open (basins), governing long-term qualitative behavior of iterated maps and flows.

:::quiz[Let $A = (0,1] \subset \mathbb{R}$. What is $\partial A$?]

• $\emptyset$
• $\{0\}$
• $\{1\}$
• $\{0,1\}$ ✓

Ask the tutor: Can you show me why the closure of a set equals the union of the set with all its limit points, using a concrete non-open, non-closed example in $\mathbb{R}^3$?

Picture $\mathbb{R}^2$. The open ball $B(x,r)$ is a disk with center $x$ and radius $r$, not including the circular rim. Openness of a set $U$ says: no matter which point of $U$ you pick, even one extremely close to where $U$ "ends," there is still some (possibly tiny) disk of safety around it that stays inside $U$. This rules out boundary points belonging to $U$ — if $x$ were on the edge, every disk around $x$ would poke outside $U$, no matter how small. Closedness is a different, complementary demand: it says $C$ swallows all the points it is "approaching." If you can find a sequence of points in $C$ marching toward some limit $x$, closedness insists $x$ itself is already in $C$. This is why $[0,1]$ is closed in $\mathbb{R}$ (a sequence approaching $0$ or $1$ from inside has its limit already included) while $[0,1)$ is not (points $1 - 1/k \to 1$, but $1 \notin [0,1)$). The three pieces — interior, boundary, exterior (the interior of the complement) — always partition $\mathbb{R}^n$ relative to a set $A$, regardless of whether $A$ itself is open, closed, or neither. The closed disk's boundary is its rim; the open disk's boundary is also that same rim, even though the rim is not part of the open disk. Boundary is a property of where a set sits in space, not of what the set happens to contain.

Example 1 — An open rectangle in $\mathbb{R}^2$ minus a boundary segment. Let

This set is neither open nor closed; we verify this carefully and compute $\operatorname{int}(A)$, $\overline{A}$, and $\partial A$. $A$ is not open. Consider the point $p = (1.5, -1) \in A$ (note $y=-1$ satisfies $-1 \le y$). For any $r>0$, the point $(1.5, -1 - r/2)$ lies in $B(p,r)$ since its distance to $p$ is $r/2 < r$, but its $y$-coordinate $-1-r/2 < -1$ violates $y \ge -1$, so this point is not in $A$. Hence no ball around $p$ fits inside $A$, so $p$ is not an interior point, so $A$ is not open. $A$ is not closed. The point $q = (1.5, 2)$ is a limit point of $A$: for any $r>0$, the point $(1.5,\, 2 - \min(r,1)/2)$ lies within $B(q,r)$ (distance $< r$) and lies in $A$ (its $y$-coordinate is in $[-1,2)$ for small enough choice, and its $x$-coordinate is $1.5\in(0,3)$), and it differs from $q$. So $q \in A'$. But $q = (1.5,2) \notin A$ since $A$ requires $y < 2$ strictly. So $A$ fails to contain a limit point, hence is not closed. Interior. We claim

Call this set $V$. First, $V$ is open: for $(x,y) \in V$, let $d = \min(x, 3-x, y-(-1), 2-y) > 0$ (all four quantities are strictly positive since the four inequalities are strict), and one checks $B((x,y), d) \subseteq V$ by the triangle inequality argument identical to part (1) of the theorem restricted coordinatewise (if $\|(x',y')-(x,y)\|<d$ then $|x'-x|<d$ and $|y'-y|<d$, forcing $x'\in(0,3)$ and $y'\in(-1,2)$). Since $V \subseteq A$ and $V$ is open, $V \subseteq \operatorname{int}(A)$ by part (7) of the theorem. Conversely, any point of $A \setminus V$ has $y = -1$ exactly (the only way to be in $A$ but not $V$, given the $x$-range is identical), and by the first bullet argument above, no such point is an interior point of $A$. So $\operatorname{int}(A) \subseteq V$. Hence $\operatorname{int}(A) = V$. Closure. We claim

the full closed rectangle. $R$ is closed (an intersection of four closed half-planes, each closed by an argument like the one for $\overline{A}$ in the theorem, or directly: complements are open half-planes). Since $A \subseteq R$ and $R$ is closed, $\overline{A} \subseteq R$ by minimality of closure (theorem, part 7). For the reverse inclusion, every point of $R$ is either in $A$ already, or is a limit point of $A$: e.g. $(0,y)$ for $y\in[-1,2]$ is approached by $(1/k, y')\in A$ with $y'$ chosen appropriately close to $y$ inside $[-1,2)$; similarly for the other three edges and the four corners, in each case one exhibits a sequence of points of $A$ (with $x$ and $y$ nudged strictly inside the open ranges) converging to the boundary point. Hence $R \subseteq \overline{A}$, giving $\overline{A} = R$. Boundary. $\partial A = \overline{A} \setminus \operatorname{int}(A) = R \setminus V$, which is exactly the perimeter of the rectangle $[0,3]\times[-1,2]$ — all four edges and corners, a one-dimensional curve in $\mathbb{R}^2$, regardless of which edges happened to be included in $A$ itself. This illustrates a key point: boundary depends on the closure and interior, not on set membership details along the edge.

Example 2 — A punctured ball with an isolated point, in $\mathbb{R}^3$. Let

where $c = (1,0,-1)$ and $p = (10, 10, 10)$ is a far-away isolated extra point (note $\|p - c\| = \sqrt{81+100+121} = \sqrt{302} \approx 17.38 \gg 2$, so $p$ genuinely sits outside the shell). Claim: $p$ is an isolated point of $B$, hence not a limit point of $B$, and $B$ is not closed. Let $\rho = \|p - c\| - 2 = \sqrt{302} - 2 > 0$. We claim $B(p,\rho) \cap B = \{p\}$. Indeed, suppose $z \in B(p,\rho)$ with $z \ne p$; if $z$ also belonged to the shell part of $B$, then $1 < \|z-c\|<2$. But by the triangle inequality,

contradicting $\|z-c\|<2$. So $z$ cannot lie in the shell, and $z \ne p$, so $z \notin B$. This shows $B(p,\rho)\cap B=\{p\}$, so $p$ is isolated: no ball around $p$ meets $B$ in a point other than $p$ itself, so $p \notin B'$ (it is not a limit point of $B$, despite belonging to $B$). Claim: $B$ is not closed. Consider $x_0$ with $\|x_0 - c\| = 2$, e.g. $x_0 = c + (2,0,0) = (3,0,-1)$. For any $r>0$, the point $c + (2 - \min(r,1)/2)\cdot(1,0,0)$ lies within the shell (its distance from $c$ is $2 - \min(r,1)/2 \in (1,2)$ for small enough radius) and within distance $\min(r,1)/2 < r$ of $x_0$, and is different from $x_0$. So $x_0 \in B'$. But $\|x_0-c\|=2$ fails the strict inequality $\|x-c\|<2$ needed to be in the shell, and clearly $x_0 \ne p$, so $x_0 \notin B$. A limit point outside the set means $B$ is not closed. Interior, closure, boundary. By an argument parallel to Example 1 (applied to the radial variable $\|x-c\|$ instead of coordinates), one finds:

Note $p \in \partial B$: $p \in \overline B$ (trivially, as $p\in B\subseteq\overline B$) but $p \notin \operatorname{int}(B)$ (just argued), so by definition $p \in \overline{B}\setminus \operatorname{int}(B) = \partial B$. This shows an isolated point of a set is always a boundary point of that set — a useful general fact following directly from the definitions.