Open/closed sets, limit points, closure, interior

In one sentence

Open and closed are not opposites, closure is not "the set plus its edge" in any naive sense, and almost every question in this topic is decided by producing one set in one metric where the option fails.

Why the exam asks it

Because this is where a candidate's mental picture is $\mathbb{R}$ and the question is about a metric space. Almost every false option in this topic is a true statement about $\mathbb{R}$ that stops being true the moment the metric changes — and the discrete metric, which takes ten seconds to apply, breaks a surprising number of them.

The idea

A metric on a set $X$ is $d : X \times X \to [0,\infty)$ with $d(x,y) = 0$ exactly when $x = y$, symmetry, and the triangle inequality. The open ball is $B(x,r) = \{y : d(x,y) < r\}$.

$U$ is open if every point of $U$ has a ball around it inside $U$. $F$ is closed if its complement is open.

Read that twice, because it is the source of the topic's first trap: closed is defined as complement-of-open, not as not-open. A set can be both — $\emptyset$ and $X$ always are — and a set can be neither, as $[0,1)$ in $\mathbb{R}$ is.

The unions and intersections rule

• Any union of open sets is open. Finite intersections of open sets are open.
• Any intersection of closed sets is closed. Finite unions of closed sets are closed.

The word "finite" is not a technicality.

an intersection of open sets that is not open. Dually, $\bigcup_{n=1}^{\infty}\left[\tfrac1n, 1\right] = (0,1]$, a union of closed sets that is not closed.

Limit points, closure, interior

$x$ is a limit point of $A$ if every ball around $x$ meets $A$ in a point other than $x$. The exclusion matters: it is what makes an isolated point of $A$ not a limit point of $A$, and it is why a finite set has no limit points at all.

• $\operatorname{int} A$: the union of all open sets inside $A$ — the largest one.
• $\overline{A}$: $A$ together with its limit points — equivalently the smallest closed set containing $A$.
• $\partial A = \overline{A} \setminus \operatorname{int} A$.

$A$ is open exactly when $A = \operatorname{int}A$, and closed exactly when $A = \overline{A}$.

$A$ is dense if $\overline{A} = X$. $\mathbb{Q}$ is dense in $\mathbb{R}$, and so is $\mathbb{R}\setminus\mathbb{Q}$ — two disjoint dense sets, which already kills several plausible statements about intersections of closures.

The discrete metric, your fastest counterexample

Here $B(x, 1/2) = \{x\}$, so every singleton is open, so every subset is open, and therefore every subset is closed as well. Any question of the form "in a metric space, is every open set …?" should be tested against this first.

It settles one classic immediately. In $\mathbb{R}$, the closure of an open ball is the closed ball of the same radius, and students carry that across as a fact about metric spaces. Take the discrete metric on a set with at least two points and $r = 1$:

The closure of the open ball is a single point; the closed ball is everything. In general $\overline{B(x,r)} \subseteq \{y : d(x,y) \le r\}$ and the inclusion can be strict.

Where intuition breaks

"A set is either open or closed." $[0,1)$ is neither, $\emptyset$ is both, and in the discrete metric everything is both. Tempting because the words are opposites in English and the definitions are not: closed means the complement is open, which is a statement about a different set.

"The closure of $B(x,r)$ is the closed ball of radius $r$." False in general, as above. True in $\mathbb{R}^n$ and in any normed space, which is exactly why it feels safe.

"An infinite intersection of open sets is open." $\bigcap(-1/n, 1/n) = \{0\}$. The finiteness in the axiom is doing real work: each finite intersection has a smallest radius to fall back on, and an infinite one need not.

"$\operatorname{int}(\overline{A}) = \operatorname{int}(A)$." Take $A = \mathbb{Q}$. Then $\operatorname{int}A = \emptyset$ while $\operatorname{int}(\overline{A}) = \operatorname{int}(\mathbb{R}) = \mathbb{R}$. About as far apart as two sets can be.

"$\overline{A \cap B} = \overline{A} \cap \overline{B}$." $A = \mathbb{Q}$, $B = \mathbb{R}\setminus\mathbb{Q}$: the left side is $\overline{\emptyset} = \emptyset$, the right side is $\mathbb{R}$. What is true is only $\overline{A\cap B} \subseteq \overline A \cap \overline B$.

"$\overline{\bigcup A_i} = \bigcup \overline{A_i}$." True for finitely many, false for infinitely many. Enumerate $\mathbb{Q} = \{q_1, q_2, \dots\}$: each $\overline{\{q_n\}} = \{q_n\}$, so the right side is $\mathbb{Q}$, while the left side is $\mathbb{R}$. The same "finite versus infinite" distinction, wearing different clothes.

"Every point of $\overline{A}$ is a limit point of $A$." Isolated points of $A$ are in $\overline A$ and are not limit points. $\overline A$ is $A$ union its limit points, and the union is not redundant.

"An infinite set has a limit point." Only with a hypothesis. $\mathbb{Z}$ in $\mathbb{R}$ has none; any infinite set in the discrete metric has none. Bolzano–Weierstrass needs boundedness and $\mathbb{R}^n$.

The exam's angle

1. Test the discrete metric first. It is the cheapest counterexample in the subject, and a large share of "in a metric space…" options die to it in one line.
2. Watch for finite against arbitrary. If an option says "any union" or "any intersection", check which side of the rule it is on. This is the single most common way these are made false.
3. Keep the four standard sets ready: $\mathbb{Q}$ (empty interior, closure $\mathbb{R}$), $\mathbb{Z}$ (closed, discrete, no limit points), $[0,1)$ (neither open nor closed), $\{1/n\}$ (limit point $0$ not in the set).
4. For closure identities, test with $\mathbb{Q}$ and its complement. Two disjoint dense sets break every identity that is really only an inclusion.
5. Distinguish "limit point of $A$" from "point of $\overline A$". Options trade on the difference, and the difference is exactly the isolated points.

The night before

• Open and closed are not opposites. Both, neither, and either are all possible.
• Arbitrary unions of open, finite intersections of open. Dually for closed.
• $\bigcap(-1/n,1/n) = \{0\}$; $\bigcup[1/n,1] = (0,1]$.
• $\overline{B(x,r)} \subseteq \{d \le r\}$, and the inclusion can be strict — discrete metric.
• Discrete metric: every set open, every set closed, no limit points, only finite sets compact.
• $\operatorname{int}\mathbb{Q} = \emptyset$, $\overline{\mathbb{Q}} = \mathbb{R}$. Both $\mathbb{Q}$ and its complement are dense.
• $\overline{A\cap B} \subseteq \overline A \cap \overline B$, one way only.
• Closure of a union: fine for finitely many, false for infinitely many.