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!

  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Stefan Mesken Nov 13 '17 at 23:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.