IV.3 · Open & Closed Sets — Problem 3 #

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Problem. Let $\overline{T}$ denote the closure of a set $T$ in a metric space $(S,\rho)$. Show that $\overline{T}$ is the smallest closed set containing $T$: it is closed, it contains $T$, and it is contained in every closed set that contains $T$.

Hint #

“Smallest closed set containing $T$” really means two things: (i) $\overline{T}$ is itself closed and contains $T$, and (ii) if $U$ is any closed set with $T \subseteq U$, then $\overline{T} \subseteq U$. For (ii), use $\overline{T} = T \cup T'$ (the set together with its limit points) and the fact that a closed set contains all of its own limit points.

Solution #

Assume that $U$ is a closed set which contains $T$ and is itself contained in $\overline{T}$, that is, $T \subseteq U \subseteq \overline{T}$. Assume that there is a point $p$ in $\overline{T}$ which is not in $U$. Because $\overline{T} = T \cup T'$, the point $p$ must lie in $T$ or $T'$. The first possibility cannot arise, since if $p$ is in $T$, then it must lie in $U$ (because $T \subseteq U$).

If $p$ lies in $T'$, then it is a limit point of $T$, and therefore a limit point of $U$ (since $T \subseteq U$). But we are assuming that $U$ is closed, and therefore it must contain all of its limit points. Therefore $p$ must lie in $U$, a contradiction. Therefore $U$ must equal $\overline{T}$, as desired.

Long solution #

Strategy. The phrase “smallest closed set containing $T$” packs three separate claims into five words, and the proof is simply to verify them one at a time: $\overline{T}$ contains $T$; $\overline{T}$ is itself closed; and $\overline{T}$ fits inside every closed set that contains $T$. The first is immediate from the definition $\overline{T} = T \cup T'$. The second requires showing that a limit point of $\overline{T}$ is already a limit point of $T$ itself — the one genuinely technical step, where we must nest a small ball inside a bigger one using the triangle inequality. The third is the picture below: a limit point of $T$ is automatically a limit point of any superset $U \supseteq T$, and a closed $U$ must contain its limit points, so $U$ has no choice but to contain all of $\overline{T}$.

The set T with dashed boundary, a limit point p of T with small balls all meeting T, inside a larger closed set U

Minimality in a picture: $T$ (orange, dashed boundary — its boundary points may be missing from $T$) sits inside a closed set $U$ (gray). A limit point $p$ of $T$ has points of $T$ in every ball around it, however small; those points also lie in $U$, so $p$ is a limit point of $U$, and closedness forces $p \in U$.

Definitions. A point $p$ is a limit point of $T$ if every ball $B_r(p)$ ($r > 0$) contains at least one point of $T$ other than $p$. Write $T'$ for the set of all limit points of $T$. The closure is $\overline{T} = T \cup T'$. A set is closed iff it contains all of its limit points (equivalently, iff its complement is open).

We verify three properties:

$T \subseteq \overline{T}$;
$\overline{T}$ is closed;
if $U$ is closed and $T \subseteq U$, then $\overline{T} \subseteq U$.

Property 3 is exactly the minimality statement: $\overline{T}$ sits inside every closed set containing $T$, so among all closed sets containing $T$ it is the smallest. (Such a smallest set is automatically unique: if $C$ were another closed set containing $T$ and contained in every such set, then $\overline{T} \subseteq C$ by minimality of $C$ applied to $\overline{T}$, and $C \subseteq \overline{T}$ by property 3 applied to $C$; hence $C = \overline{T}$.)

1. $T \subseteq \overline{T}$.

By definition $\overline{T} = T \cup T'$, and a union contains each of the sets being unioned, so $T \subseteq T \cup T' = \overline{T}$.

2. $\overline{T}$ is closed.

By the characterization in the box, it suffices to show that $\overline{T}$ contains all of its limit points: $(\overline{T})' \subseteq \overline{T}$. So let $p$ be a limit point of $\overline{T}$. Our plan is to show that $p$ is in fact a limit point of the original set $T$; then $p \in T' \subseteq \overline{T}$ and we are done.

To show $p \in T'$ we must check the definition: given an arbitrary radius $r > 0$, we must find a point of $T$, different from $p$, inside $B_r(p)$. Fix such an $r$.

Since $p$ is a limit point of $\overline{T}$, the ball $B_r(p)$ contains some point $x \in \overline{T}$ with $x \neq p$. We now want a small “buffer ball” around $x$ that stays inside $B_r(p)$ but excludes $p$. Here is the scratch work for choosing its radius. We need two things:

$B_s(x) \subseteq B_r(p)$: if $\rho(x,y) < s$, the triangle inequality gives $\rho(p,y) \leq \rho(p,x) + \rho(x,y) < \rho(p,x) + s$, so it suffices to have $\rho(p,x) + s \leq r$, i.e. $s \leq r - \rho(p,x)$. (This quantity is positive because $x \in B_r(p)$ means $\rho(p,x) < r$.)
$p \notin B_s(x)$: we need $\rho(x,p) \geq s$, i.e. $s \leq \rho(p,x)$. (This quantity is positive because $x \neq p$.)

Both demands are met by choosing

$$ s := \min\big\{\, r - \rho(p,x),\ \rho(p,x) \,\big\} > 0 . $$

Now use $x \in \overline{T} = T \cup T'$ and argue by cases:

If $x \in T$: then $x$ itself is a point of $T$ lying in $B_r(p)$, and $x \neq p$. Done.
If $x \in T'$: then $x$ is a limit point of $T$, so applying the definition of limit point with the radius $s$, the ball $B_s(x)$ contains a point $t \in T$ with $t \neq x$. By our first demand $t \in B_s(x) \subseteq B_r(p)$, and by our second demand $p \notin B_s(x)$ while $t \in B_s(x)$, so $t \neq p$. Thus $t$ is a point of $T$, different from $p$, inside $B_r(p)$. Done.

In both cases $B_r(p)$ contains a point of $T \setminus \{p\}$. Since $r > 0$ was arbitrary, $p$ is a limit point of $T$, i.e. $p \in T' \subseteq \overline{T}$. Hence $\overline{T}$ contains all of its limit points and is closed.

3. Minimality: if $U$ is closed and $T \subseteq U$, then $\overline{T} \subseteq U$.

Let $p \in \overline{T} = T \cup T'$ be arbitrary; we show $p \in U$ by cases.

If $p \in T$: then $p \in U$ immediately, because $T \subseteq U$.
If $p \in T'$: then $p$ is a limit point of $T$, so for every $r > 0$ the ball $B_r(p)$ contains a point $t \in T$ with $t \neq p$. Since $T \subseteq U$, this same $t$ belongs to $U$. So every ball around $p$ contains a point of $U$ other than $p$ — which is precisely the statement that $p$ is a limit point of $U$. (This is the general principle visible in the figure: enlarging a set can only create more limit points, never destroy them.) But $U$ is closed, so it contains all of its limit points; hence $p \in U$.

In either case $p \in U$. Since $p \in \overline{T}$ was arbitrary, $\overline{T} \subseteq U$.

Combining 1–3: $\overline{T}$ is a closed set containing $T$, and it is contained in every closed set containing $T$. That is, $\overline{T}$ is the smallest closed set containing $T$.

Remark. Property 3 gives a slick alternative description of the closure:

$$ \overline{T} = \bigcap \{\, U : U \text{ closed and } T \subseteq U \,\}, $$

the intersection of all closed sets containing $T$. Indeed, the intersection is closed by Problem 1(d), contains $T$, and by property 3 contains no more than $\overline{T}$ — while $\overline{T}$ is itself one of the sets being intersected. Some textbooks take this intersection as the definition of closure; this problem shows the two definitions agree. $\blacksquare$