TST ANA2 Quiz 1

The Cantor Bendixson Theorem

Definition 291.1(Definition).

A point $p$ in a metric space $X$ is said to be a condensation point of a set $E\subseteq X$ if every open neighborhood of $p$ contains uncountably many points of $E$. Note that condensation points are limit points.

Let $X$ be a second countable metric space with countable basis $\mathcal{B}=\{B_n{\}}_{n=1}^{\infty }$ and $E\subseteq X$.

Exercise 291.2(Exercise).

Let $W$ be the union of those $B_n$ for which $E\cap B_n$ is at most countable and let $P=W^c$ ($P$ may be empty). Prove that $P$ is the set of all condensation points of $E$, with at most countably many points of $E$ not in $P$.

Proof.

Suppose$p$ is a condensation point of $E$. If $p\in B_n$ for some $n$, then $B_n\cap E$ is uncountable by definition. Thus, $p\notin W$, so $p\in W^c=P$. Conversely, if $p\in P$, then every $B_n\in \mathcal{B}$ containing $p$ contained uncountably many points of $E$. It follows that any open set containing $p$ contains uncountably many points of $E$, and $p$ is a condensation point of $E$. It is clear that $W$ can only contain countably many points of $E$, which are the ones not in $P$.□

Intersecting $P$ with $E$ and taking $X={\mathbb{R}}^n$, one obtains as a corollary that an uncountable subset of ${\mathbb{R}}^n$ contains uncountably many condensation points.

Exercise 291.3(Exercise).

Show that $P$ is closed in $X$ and therefore perfect (i.e. $P$ is closed and every point of $P$ is a limit point of $P$). As a result, if $E$ is closed, show that $P\subseteq E$, hence every closed subset of a second countable metric space is expressible as a disjoint union of a perfect set and a set which is at most countable.

Proof.

If $p$ is a limit point of $P$, every neighborhood of $p$ contains a point of $P$, and hence has uncountably many points of $E$, making $p$ a condensation point of $E$. Thus, $P$ is closed.

Next, if $p\in P$, every neighborhood of $p$ contains uncountably many points of $E$. Since only countably many points of $E$ are not in $P$, this implies every neighborhood of $p$ contains points of $P$ - $p$ is a limit point of $P$. Thus, $P$ is perfect.

If $E$ is closed, it is obvious that $P\subseteq E$.□

Exercise 291.4(Exercise).

Assume that $X$ is complete and let $E$ be a closed subset of $X$. Let $C$ and $Q$ be disjoint sets that are at most countable and perfect respectively, such that $E=C\cup Q$. Prove that every point of $Q$ is a condensation point of $E$.

Proof.

FTSOC, assume that some$x\in Q$ is not a condensation point of $E$. Then there exists an open neighborhood $U$ of $x$ containing only countably many points $\{x_n{\}}_{n=1}^{\infty }$ of $E$. WLOG, assume $\{x_n{\}}_{n=1}^{\infty }\subseteq Q$ ($Q$ is perfect!). WLOG, let $x_1=x$. Take a small enough open ball $U_2$ around $x_2$ whose closure lies entirely inside $U$ and does not contain $x$, and pick the smallest index $n$ such that $x_n$ lies in $U_2$ (since $x_2\in Q$, it is also a limit point of $Q$). Repeat this procedure ensuring that the radii of the balls go to $0$. By the Cantor intersection theorem, the subsequence of $\{x_n{\}}_{n=1}^{\infty }$ thus obtained must converge to some point in $E$. But, we’ve knocked off every point in $\{x_n{\}}_{n=1}^{\infty }$ when constructing the contracting sequence ($x_i\notin U_{i+1}$), so this is impossible!□

Exercise 291.5(Cantor Bendixson Theorem).

Finally, show that every condensation point of $E$ is contained in $Q$, hence $Q$ is the set of condensation points of $E$. This shows that a closed subset of a complete, second countable metric space is uniquely expressible as a disjoint union of a perfect set and a set that is at most countable.

Proof.

If$p$ is a condensation point of $E$, every neighborhood of $p$ contains uncountably many points of $E$, hence must contain points of $Q$. Thus, $p$ is a limit point of $Q$, and $p\in Q$.□

[!Exercise]
Find a counterexample to unique expressibility if $X$ is not assumed to be complete.