IV.6 · Open & Closed Sets — Problem 6 #

← Back to IV. Open & Closed Sets, Problem 6

Problem. Let $T \subseteq S$ and write $T'$ for its set of limit points. (a) Show that $T'$ is closed. (b) Show that $p \in T'$ iff every ball $B_\epsilon(p)$ contains infinitely many points of $T$. (c) Find $T \subseteq \mathbb{R}$ with $T' \neq \varnothing$ but $T \cap T' = \varnothing$, and $T \subseteq \mathbb{R}$ with $T' = \mathbb{R}$.

Hint #

For (a), use the theorem that a set is closed iff it contains all its limit points: so you must show that a limit point of $T'$ is itself a limit point of $T$. Given $p \in (T')'$, find a point $q \in T'$ close to $p$, then a point of $T$ close to $q$, and arrange the radii so that this point lands inside your ball about $p$ without being $p$. For (b), if a ball contained only finitely many points of $T$, shrink it to exclude them all.

Solution #

(a) By the theorem characterizing closed sets, it suffices to show $(T')' \subseteq T'$. Let $p$ be a limit point of $T'$ and let $\epsilon > 0$. Then $B_\epsilon(p)$ contains some $q \in T'$ with $q \neq p$. Put $\delta = \min\{\rho(p,q),\ \epsilon - \rho(p,q)\} > 0$. The ball $B_\delta(q)$ lies inside $B_\epsilon(p)$ by the triangle inequality, and does not contain $p$ because $\rho(p,q) \geq \delta$. Since $q \in T'$, the ball $B_\delta(q)$ contains a point $t \in T$ with $t \neq q$; and $t \neq p$ as $p \notin B_\delta(q)$. So $B_\epsilon(p)$ contains a point of $T$ other than $p$. As $\epsilon$ was arbitrary, $p \in T'$.

(b) If every ball about $p$ contains infinitely many points of $T$, then each contains one different from $p$, so $p \in T'$. Conversely let $p \in T'$ and suppose some $B_\epsilon(p)$ contains only finitely many points of $T$; list the ones different from $p$ as $t_1, \ldots, t_k$ (possibly $k = 0$). Put $\delta = \min\{\epsilon, \rho(p,t_1), \ldots, \rho(p,t_k)\} > 0$. Then $B_\delta(p)$ contains no point of $T$ other than $p$, contradicting $p \in T'$.

(c) $T = \{\tfrac{1}{n} \mid n \in \mathbb{N}\}$ has $T' = \{0\}$ and $0 \notin T$. And $T = \mathbb{Q}$ has $T' = \mathbb{R}$, since every interval contains infinitely many rationals.

Long solution #

Strategy. Everything rests on the definition: $p \in T'$ iff every ball $B_\epsilon(p)$ contains a point of $T$ different from $p$. Part (a) is a “two-step approximation”: a limit point of limit points is approached by limit points, each of which is approached by points of $T$; the only care needed is to make sure the point of $T$ we find is not $p$ itself. Part (b) upgrades “a point other than $p$” to “infinitely many points” by a shrinking argument that uses the finiteness of a minimum. Part (c) is a matter of knowing two standard examples.

Part (a). We use the theorem from class: $A$ is closed iff $A' \subseteq A$. Applied to $A = T'$, we must show that every limit point of $T'$ is a limit point of $T$.

Let $p \in (T')'$ and fix $\epsilon > 0$. Since $p$ is a limit point of $T'$, there is a point $q \in T'$ with $q \neq p$ and $\rho(p,q) < \epsilon$. Let

$$ \delta = \min\{\, \rho(p,q),\ \epsilon - \rho(p,q) \,\}, $$

which is positive because $0 < \rho(p,q) < \epsilon$. Two things are true of the ball $B_\delta(q)$:

$B_\delta(q) \subseteq B_\epsilon(p)$: if $\rho(q,y) < \delta$ then $\rho(p,y) \leq \rho(p,q) + \rho(q,y) < \rho(p,q) + (\epsilon - \rho(p,q)) = \epsilon$.
$p \notin B_\delta(q)$: because $\rho(q,p) = \rho(p,q) \geq \delta$.

Since $q \in T'$, the ball $B_\delta(q)$ contains a point $t \in T$ with $t \neq q$. By (1), $t \in B_\epsilon(p)$; by (2), $t \neq p$. So $B_\epsilon(p)$ contains a point of $T$ different from $p$. As $\epsilon > 0$ was arbitrary, $p \in T'$. Hence $(T')' \subseteq T'$ and $T'$ is closed. $\blacksquare$

Alternative. One can instead show the complement of $T'$ is open. If $p \notin T'$, some $B_\epsilon(p)$ contains no point of $T$ other than $p$. Then no point $q \in B_\epsilon(p)$ with $q \neq p$ can be in $T'$ either, since a small ball about $q$ inside $B_\epsilon(p)$ and avoiding $p$ contains no point of $T$ at all. So $B_\epsilon(p) \cap T' \subseteq \{p\}$, and if $p \in T'$ we would have a contradiction, so in fact $B_\epsilon(p) \subseteq (T')^c$.

Part (b). ($\Leftarrow$) If every ball $B_\epsilon(p)$ contains infinitely many points of $T$, then in particular it contains at least one point other than $p$. So $p \in T'$.

($\Rightarrow$) Suppose $p \in T'$ but that some ball $B_\epsilon(p)$ contains only finitely many points of $T$. Remove $p$ from this list if it is there, and call the remaining points $t_1, \ldots, t_k$; each satisfies $\rho(p,t_i) > 0$. Define

$$ \delta = \min\{\, \epsilon,\ \rho(p,t_1),\ \ldots,\ \rho(p,t_k) \,\} > 0, $$

a minimum of finitely many positive numbers, hence positive. (If $k = 0$, take $\delta = \epsilon$.) Any point of $T$ in $B_\delta(p)$ lies in $B_\epsilon(p)$, so it is $p$ or one of the $t_i$; but $\rho(p,t_i) \geq \delta$ excludes every $t_i$. Thus $B_\delta(p)$ contains no point of $T$ other than $p$, contradicting $p \in T'$. $\blacksquare$

Consequences. A finite set has no limit points, because a ball cannot contain infinitely many of its points. So a finite set vacuously contains all its limit points and is closed, which recovers Problem IV.2. More generally, a set with no limit points at all, such as $\mathbb{Z} \subset \mathbb{R}$, is automatically closed.

Part (c). First example. Let $T = \{\tfrac{1}{n} \mid n \in \mathbb{N}\}$. The point $0$ is a limit point: given $\epsilon > 0$, choose $n > 1/\epsilon$; then $0 < \tfrac{1}{n} < \epsilon$, so $B_\epsilon(0)$ contains a point of $T$ other than $0$. No other real number is a limit point: a ball of small enough radius about $\tfrac{1}{n}$ contains no other point of $T$, a ball about a negative number or a number greater than $1$ can be chosen to miss $T$, and a number in $(0,1)$ that is not of the form $\tfrac{1}{n}$ lies strictly between two consecutive points $\tfrac{1}{n+1} < x < \tfrac{1}{n}$, so a small ball about it misses $T$. Thus $T' = \{0\}$, which is non-empty, while $0 \notin T$, so $T \cap T' = \varnothing$.

Second example. Let $T = \mathbb{Q}$. For any $p \in \mathbb{R}$ and $\epsilon > 0$, the interval $(p - \epsilon, p + \epsilon)$ contains a rational number different from $p$ (indeed infinitely many), by the density of $\mathbb{Q}$ in $\mathbb{R}$. So every real number is a limit point of $\mathbb{Q}$ and $T' = \mathbb{R}$. The irrationals work equally well.

Remark. Compare the first example with the definition $\overline{T} = T \cup T'$: here $\overline{T} = T \cup \{0\}$, the set which will reappear in Problem V.7 as a compact set, while $T$ itself is not closed.