# Construct a sequence whose limit points are exactly a given closed set $F$

Given a closed set $F$ of $\mathbb{R}$, show there exists a sequence in $\mathbb{R}^\mathbb{N}$ whose set of limit points is exactly $F$.

Under what conditions does this hold if we replace $\mathbb{R}$ by a metric space $E$?

After thinking for some time about the first question my professor showed me a means of constructing the sequence, but it was completely unmotivated. So some motivation would be nice when describing your construction!

• You wrote 'set of limit points' but I'm certain you meant to say 'set of accumulation points' since a sequence has at most one limit point. – Stefan Mesken Nov 13 '17 at 23:40

Let $(X; d)$ be a separable metric space and let $C$ be closed in $(X;d)$. Fix a countable basis $B$ for the topology of $(X;d)$ and fix a countable dense set $D$ such that for all $O \in B$ $$O \cap C \neq \emptyset \implies D \cap C \neq \emptyset.$$ (This is possible since we may just add those countably many points to a given dense set.) Now let $$D' := D \cap C = \{d_n \mid n \in \mathbb N\}.$$ and fix a function $$x \colon \mathbb N \to D'$$ such that $x^{-1}[\{d\}]$ is infinite for all $d \in D'$. (This can easily be arranged, e.g. by setting $f(p_n^k) = d_n$ for all $k$, where $p_n$ is the $n$th prime number.)

Let us write $$(x_n \mid n \in \mathbb N) := (x(n) \mid n \in \mathbb N).$$

Since $\{x_n \mid n \in \mathbb N \} \subseteq C$ and $C$ is closed, every accumulation point of $(x_n \mid n \in \mathbb N)$ is in $C$.

Conversely, if $x \in C$, fix a strictly increasing function $$f \colon \mathbb N \to \mathbb N$$ such that for all $n \in \mathbb N$ $$x_{f(n)} \in B(x, 2^{-n}) := \{ y \in X \mid d(x,y) < 2^{-n} \}.$$ This is possible because there is some $O \in B$ with $x \in O \subseteq B(x, 2^{-n})$ and for this $O$ there is, by our choice of $D$ some $x_{m} \in D \cap C$. Let $f(n)$ be large enough such that $f$ stays strictly increasing and $x_{f(n)} = x_m$.

Now the subsequence $$(x_{f(n)} \mid n \in n \in \mathbb N)$$ witnesses that $x$ is an accumulation point of $(x_n \mid n \in \mathbb N)$.

Note that this is optimal: If $(X;d)$ is a metric space and $(x_n \mid n \in \mathbb N)$ is a sequence such that $X$ is equal to the set of accumulation points of $X$, then $$\{ x_n \mid n \in \mathbb N \}$$ is a countable dense subset of $(X;d)$ and hence $(X;d)$ is separable.

Proof. Let $O$ be a nonempty open subset of $(X;d)$ and fix $x \in O$. Since $x$ is an accumulation point of $(x_n \mid n \in \mathbb N)$ there is some $n \in \mathbb N$ such that $x_n \in O$. Q.E.D.

• As far as the motivation goes... Well, I'm afraid I don't really have one. I started generalizing this little fact to perfect Polish spaces (since they are very similar to $\mathbb R$ from a topological point of view) and then observed that all I needed was a countable dense set. – Stefan Mesken Nov 13 '17 at 23:27