# Counterexamples about set of condensation points in metric spaces

Let $S^*$ be the set of condensation points of $S$, and $S'$ be the set of cluster points of $S$ . Give an example of $S$ such that $S^{**}$ is not equal to $S^*$ and an example of $S$ such that $S^*$ is neither contained in nor contains $(S')^*$.

P.S. The problem comes with a hint which suggests considering space of functions $\left[a,b\right]$ to $\left[0,1\right]$ with sup metric and the set of "delta-functions with rational values", of which I have no idea.

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The first question has many easy answers: all you need is a space that contains a set with only one condensation point. For instance, let $S$ be any uncountable set, let $p$ be any point not in $S$, and let $X = S \cup \{p\}$. Make each point of $S$ an isolated point of $X$, and say that a set $A$ containing $p$ is a nbhd of $p$ iff $S \setminus A$ is finite. (In other words, $S$ is an uncountable discrete space, and $X$ is its one-point compactification.) Clearly $S^* = \{p\}$, and $(S^*)^* = \{p\}^* = \varnothing \ne S^*$.

I’ll modify the example a bit to answer the second question. First I’ll give $p$ more nbhds: now let $\{X \setminus A: A \subseteq S\text{ is countable}\}$ be a local base at $p$. ($X$ is now the one-point ‘Lindelöfization’ of $S$.) It’s still true that $p$ is the unique condensation point of $S$ in $X$. Since $S$ is uncountable, it can be partitioned into uncountably many countably infinite subsets, so let $S = \bigcup\{S_i:i \in I\}$, where $I$ is some uncountable index set, and each $S_i$ is countably infinite. For each $i \in I$ let $q_i$ be a distinct new point not in $X$, and let $Y = X \cup \{q_i:i \in I\}$. Basic nbhds of the point $q_i$ are the sets of the form $\{q_i\} \cup (S_i \setminus F)$ such that $F$ is a finite subset of $S_i$. (In other words, $\{q_i\} \cup S_i$ is the one-point compactification of the the countably infinite discrete space $S_i$.)

To sum up, $Y = S \cup \{p\} \cup \{q_i:i \in I\}$, and a base for $Y$ is given by \begin{align*} \mathscr{B} = \left\{\{s\}:s \in S \right\} &\cup \left\{\{p\} \cup (S \setminus F):F \subseteq S\text{ is countable} \right\}\\ &\cup \left\{\{q_i\}\cup (S_i \setminus F):i \in I \land F \subseteq S_i\text{ is finite} \right\}. \end{align*} (It’s easy to check that this is a base for a topology.)

This almost works. Clearly $S' = Y \setminus S$: the point $p$ and the points $q_i$ ($i \in I$) are the cluster points of $S$. But each $q_i$ has a countable nbhd and therefore cannot be a condensation point of $S$, so $S^* = \{p\}$. Unfortunately, $S'$ has no condensation points, so $(S')^*$ is a subset of $S^*$, albeit a proper one.

The natural idea for fixing this is to give $\{q_i:i \in I\}$ a new condensation point that isn’t a condensation point of $S$. Unfortunately, any open set that contains uncountably many of the $q_i$ automatically contains uncountably many point of $S$, so this idea doesn’t work as it stands. A slightly more complicated version of it can be made to work, however.

Fix $i_0 \in I$. It’s well known that we can find in $S_{i_0}$ an uncountable family $\mathscr{A}$ of almost disjoint sets, i.e., of sets whose pairwise intersections are finite.

(For example, let $\varphi:\mathbb{Q} \to S_{i_0}$ be a bijection, for each $r \in \mathbb{R}$ let $\langle q_r(n):n \in \omega\rangle$ be a monotone increasing sequence of rationals converging to $r$ and $A_r = \{\varphi(x_r(n)):n\in\omega\}$, and take $\mathscr{A} = \{A_r:r \in \mathbb{R}\}$.)

For each $A \in \mathscr{A}$ let $q_A$ be a new point, and let $Z = Y \cup \{q_A:A \in \mathscr{A}\}$. Points of $Y \setminus \{q_{i_0}\}$ retain their basic open nbhds. For each $A \in \mathscr{A}$ the basic open nbhds of $q_A$ are the sets $\{q_A\} \cup (A \setminus F)$ for $F$ a finite subset of $A$; note that each of these sets is contained in the countable set $\{q_A\} \cup S_{i_0}$, so $q_A$ is not a condensation point of $S$. Finally, we change the topology at $q_{i_0}$ by taking its basic open nbhds to be the sets of the form $$\{q_{i_0}\} \cup \bigcup\{B_A:A \in \mathscr{A} \setminus \mathscr{F}\},$$ where $\mathscr{F}$ is a finite subset of $\mathscr{A}$ and $B_A$ is an open nbhd of $q_A$ for each $A \in \mathscr{A} \setminus \mathscr{F}$. Each of these nhbds also intersects $S$ in a subset of $S_{i_0}$, so $q_{i_0}$ is also not a condensation point of $S$; $q_{i_0}$ is a condensation point of $\{q_A:A \in \mathscr{A}\}$, however.

It should now be easy to see that $S^*$ is still just $\{p\}$, $S' = Z \setminus S$, and $(S')^* = \{q_{i_0}\}$, so that neither of $S^*$ and $(S')^*$ is a subset of the other.

(I found it easiest to ignore the hint.)

Added: My original answer was intended to illustrate one possible approach to the problem, but it occurs to me that an explicit description of a space like $Z$ may be helpful. Let $\mathbb{R}^+$ be the set of positive reals and $\mathbb{R}_0^+$ the set of non-negative reals, and define $\mathbb{Q}^+$ and $\mathbb{Q}_0^+$ similarly.

Let $Z_0 = (\mathbb{Q}_0^+\times\mathbb{R}_0^+) \cup (\mathbb{R}_0^+\times\{0\})$. Let $S = \mathbb{Q}^+\times\mathbb{R}_0^+$; points of $S$ are isolated in $Z_0$.

Let $P = \mathbb{P}^+ \times \{0\}$, where $\mathbb{P}^+$ is the set of positive irrational numbers. For each $p \in \mathbb{P}^+$ let $\langle q_p(n):n \in \omega \rangle$ be a monotone increasing sequence of positive rational numbers converging to $p$ in the usual topology on $\mathbb{R}$. If $p \in \mathbb{P}^+$, the sets $B(p,n) \triangleq \{(p,0)\} \cup \{(q_p(k),0):k\ge n\}$ form a local base at $(p,0)$.

For each $r \in \mathbb{R}^+$ and each finite subset $F$ of $\mathbb{Q}_0^+$ let $B(r,F) = \{(0,r)\} \cup \left((\mathbb{Q}_0^+ \setminus F)\times \{r\}\right)$; the sets $B(r,F)$ form a local base at $(r,0)$.

For each countable $F \subseteq \mathbb{P}^+$ and each function $\varphi:\mathbb{P}^+ \to \omega$ let $$B(F,\varphi) = \{(0,0)\} \cup \bigcup\limits_{p\in\mathbb{P}^+\setminus F}B(p,\varphi(p));$$ the sets $B(F,\varphi)$ form a local base at $(0,0)$.

Finally, let $Z = Z_0 \cup \{p_\infty\}$, where $p_\infty$ is a new point, and for each countable subset $C$ of $S$ let $B(C) = \{p_\infty\} \cup (S \setminus C\}$; the sets $B(C)$ form a local base at $p_\infty$.

Then $S' = Z \setminus S$, $S^* = \{p_\infty\}$, and $(S')^* = \{(0,0)\}$.

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I think that the asker looks for metrisable counterexamples. At first I did not notice this either, but then I happened to glance at the title where he mentions metric spaces. –  LostInMath Aug 25 '11 at 9:28

I’ll let my first answer stand, since the ideas may be useful to someone, but here’s a metric space example based on the hint. Let $X$ be the set of functions from $[a,b]$ to $[0,1]$ with the sup metric topology. For $f \in X$ and $\epsilon > 0$ let $N(f,\epsilon)$ be the open ball of radius $\epsilon$ about $f$.

Let $z \in X$ be the function that is identically $0$. For each $c \in [a,b]$ and $r \in (0,1]$ let $f_{c,r} \in X$ be defined by $$f(x) = \begin{cases}r,&\text{if }x=c\\0,&\text{otherwise}. \end{cases}$$ Let $D = \{f_{c,r}:c \in [a,b] \land r \in (0,1]\}$, and let $S = \{f_{c,q} \in D:q \in \mathbb{Q}\}$. ($D$ is the set of delta-functions mentioned in the hint; $S$ contains those with rational values.)

Fix $g\in X$. If there are distinct $c,d \in [a,b]$ such that $g(c) \ne 0 \ne g(d)$, let $\epsilon = \min\{g(c),g(d)\}$; clearly $N(g,\epsilon) \cap S = \varnothing$, so $g$ is not a cluster point of $S$. If $g = f_{c,r} \in D$, then $N(g,r) \cap S$ is contained in the countable set $\{f_{c,q}:q \in (0,1] \cap \mathbb{Q}\}$, so $g$ is not a condensation point of $S$; it is, however, the limit of a sequence in $\{f_{c,q}:q \in (0,1] \cap \mathbb{Q}\}$, since $r$ is the limit of a sequence in $(0,1] \cap \mathbb{Q}$, so $g$ is a cluster point of $S$. The only remaining possibility is that $g=z$. Given $\epsilon > 0$, let $q$ be a positive rational less than $\min\{\epsilon,1\}$; then $N(z,\epsilon) \cap S \supseteq \{f_{c,q}:c \in [a,b]\}$, which is of course uncountable, so $z$ is a condensation point of $S$.

We’ve now shown that $S^* = \{z\}$ and $S' = D \cup \{z\}$. Clearly $(S^*)^* = \varnothing \ne S^*$; this takes care of the first question. Unfortunately, it’s easy to see that $(S')^* = S'$, which is a superset of $S^*$, albeit a proper one. To fix this, let $S_0 = \{f_{c,q} \in S:q > 1/2\} \cup \{f_{c,2^{-n}}:n \in \omega\}$. Using the arguments of the last paragraph it’s easy to check that $S_0^*= \{z\}$ and $S_0' = \{z\} \cup \{f_{c,r} \in D:r \ge 1/2\}$; and since $z$ is isolated in $S_0'$, $(S_0')^* = \{f_{c,r} \in D:r \ge 1/2\}$, which neither contains nor is contained in $S_0^*$.

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