IV.2 · Open & Closed Sets — Problem 2 #

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Problem. Let $(S,\rho)$ be a metric space and $p \in S$. Prove the singleton $\{p\}$ is closed, and deduce that every finite set $\{p_1,\dots,p_n\}$ is closed.

Hint #

Show the complement $S \setminus \{p\}$ is open: given any $q \neq p$, the distance $r = \rho(p,q)$ is strictly positive, so the ball $B_r(q)$ misses $p$. For the finite set, write it as a finite union of singletons and quote Problem 1(c).

Solution #

If you recall from class that a singleton set in a metric space is always compact, and that compact sets are always closed, then the problem follows. Alternately, consider the complement of $\{p\}$ in $S$. For any point $q$ of this complement, the ball $B_r(q)$ of radius $r = \rho(p,q)$ does not contain $p$ and is therefore itself contained entirely in the complement of $\{p\}$. But then by definition, the complement of $\{p\}$ must be open, forcing $\{p\}$ to be closed.

For the second statement, we can use the third part of the previous problem: $\{p_1,\dots,p_n\} = \bigcup_{i=1}^n \{p_i\}$ is a finite union of closed sets, hence closed.

Long solution #

Strategy. To show $\{p\}$ is closed we must show its complement $S \setminus \{p\}$ is open, and openness is a pointwise job: we pick an arbitrary point $q$ of the complement and must produce a radius $r > 0$ whose ball $B_r(q)$ avoids $p$ entirely. What radius could work? The only obstruction is the single point $p$, which sits at distance $\rho(p,q)$ from $q$. Any ball around $q$ of radius at most that distance cannot reach $p$ — so we simply take $r = \rho(p,q)$, which is positive precisely because $q \neq p$. The finite-set statement then costs nothing: a finite set is a finite union of singletons, and Problem 1(c) says finite unions of closed sets are closed.

A point q with the open ball of radius rho(p,q) around it; p lies on the boundary circle, outside the open ball

The choice of radius: taking $r = \rho(p,q)$ puts $p$ exactly on the boundary circle of $B_r(q)$. Since the open ball consists of points at distance strictly less than $r$ from $q$, the point $p$ is excluded, and the whole ball stays inside $S \setminus \{p\}$.

Definitions. For $q \in S$ and $r > 0$, the open ball is $B_r(q) = \{\, x \in S : \rho(q,x) < r \,\}$. A set $U \subseteq S$ is open iff for every $q \in U$ there is some $r > 0$ with $B_r(q) \subseteq U$. A set is closed iff its complement is open. We will also use two metric axioms: positivity ($\rho(a,b) > 0$ whenever $a \neq b$) and symmetry ($\rho(a,b) = \rho(b,a)$).

Part 1: the singleton $\{p\}$ is closed.

By definition, $\{p\}$ is closed iff its complement

$$ U := S \setminus \{p\} = \{\, x \in S : x \neq p \,\} $$

is open. So this is what we prove. Unwinding the definition of “open,” we must show: for every $q \in U$ there exists $r > 0$ such that $B_r(q) \subseteq U$.

Let $q \in U$ be arbitrary. Since $q$ lies in the complement of $\{p\}$, we know $q \neq p$. By the positivity axiom of the metric, distinct points are at strictly positive distance, so the number

$$ r := \rho(p,q) $$

satisfies $r > 0$; it is therefore a legitimate radius. (This is the one place where $q \neq p$ is used, and it is essential: if $r$ were $0$, the “ball” $B_0(q)$ would be empty and useless.)

Claim: $B_r(q) \subseteq U$, i.e. the ball contains no point equal to $p$.

Proof of claim. Let $x \in B_r(q)$ be arbitrary. By the definition of the open ball,

$$ \rho(q,x) < r = \rho(p,q). \tag{*} $$

Suppose, for contradiction, that $x = p$. Substituting $x = p$ into the left side of $(*)$ and using symmetry of the metric,

$$ \rho(q,x) = \rho(q,p) = \rho(p,q) = r, $$

so $(*)$ would read $r < r$ — impossible. Hence $x \neq p$, which means $x \in S \setminus \{p\} = U$. Since $x \in B_r(q)$ was arbitrary, $B_r(q) \subseteq U$, proving the claim.

We have shown that every point $q \in U$ admits a ball $B_r(q) \subseteq U$ (with $r$ depending on $q$, which is allowed — openness only asks for some radius at each point). Therefore $U$ is open, and its complement $\{p\}$ is closed.

Part 2: every finite set is closed.

Let $F = \{p_1, p_2, \dots, p_n\}$ be any finite subset of $S$. A finite set is exactly the union of its one-point subsets:

$$ F = \{p_1\} \cup \{p_2\} \cup \cdots \cup \{p_n\} = \bigcup_{i=1}^{n} \{p_i\}. $$

(Both sides have the same elements: $x \in F$ iff $x = p_i$ for some $i$ iff $x \in \{p_i\}$ for some $i$.) By Part 1, each singleton $\{p_i\}$ is closed. The index set $\{1, 2, \dots, n\}$ is finite, so $F$ is a finite union of closed sets, and Problem 1(c) tells us exactly that such a union is closed. Hence $F$ is closed.

The degenerate case $n = 0$ (the empty set) is also covered: $\varnothing$ is closed by Problem 1(b).

Remark. Finiteness matters in Part 2 for the same reason as in Problem 1(c): an infinite set of points need not be closed. For instance, in $\mathbb{R}$ the set $\{1, \frac{1}{2}, \frac{1}{3}, \dots\} = \{1/n : n \in \mathbb{Z}^{+}\}$ is a union of closed singletons, but it is not closed, because $0$ is a limit point of the set that does not belong to it. Note also that this proof never used the triangle inequality — positivity and symmetry alone suffice. $\blacksquare$