# Any closed subset in a separable metric space is the union of a perfect set and a set that is at most countable. Why does this matter?

I am writing regarding problem 2.28 in Rudin's "Principles of Mathematical Analysis". The statement the reader is asked to prove is as follows:

Every closed subset of a separable metric space is the union of a perfect set and a set that is at most countable.

This statement sounds like it should be important.

What came to mind to me was the definition of the functional limits and the ensuing definitions of continuity we encounter later in an undergraduate analysis course. The definition of continuity works "intuitively" with limit points in the domain, but less "intuitively" at isolated points. Since a perfect set is made up entirely of limit points (and is closed), it seems like this result should let us say something about "weird things" happening at only countably many places for functions defined on closed subsets of a separable metric space. I'm not really sure if this matters at all or is really comprehensible.

Thank you.

• It could be what you say. But also if the space is complete, then its perfect subsets have the same cardinal as $\mathbb{R}$, therefore this theorem proves that the continuum hypothesis is valid for closed subsets of $\mathbb{R}$. One can even use this to show it holds for borelian subsets Jan 12, 2018 at 7:30
• The various text-snippets you can read from a google-books search for Cantor-Bendixson should give you an idea of the importance of this result in mathematics. Jan 12, 2018 at 14:43

By the time you read the statement

$\tag 1 \text{Every closed subset of a separable metric space}$ $\qquad \qquad \text{is the union of a perfect set and a set that is at most countable.}$

you are far along the path of mathematically 'sculpturing' topological concepts. it is instructive to 'reverse engineer' this statement. I think you can safely say that if you do this, you will be forced to discover the foundations of point-set topology.

We all agree that the integers $\mathbb Z$ is a closed set of isolated points on the real number line $\mathbb R$. But this is just too easy to do, since you have the integers extending out to both $-\infty$ and $+\infty$. So can you find a countable (infinite) closed set of isolated points constrained to be inside the interval $[-1,+1]$?

Answering this question and the natural questions that arise as you proceed on your journey will be an awe inspiring mathematical challenge.

To answer the OP's question, reading and understanding (1) in a math book makes me think of mountaineers, and images like

The work, training and challenges of the accomplishment is safely behind them now.

The reader might be interested in exploring the perfect set property where they will find the following:

The Cantor–Bendixson theorem states that closed sets of a Polish space $X$ have the perfect set property in a particularly strong form; any closed set $C$ may be written uniquely as the disjoint union of a perfect set $P$ and a countable set $S$. Thus it follows that every closed subset of a Polish space has the perfect set property. In particular, every uncountable Polish space has the perfect set property, and can be written as the disjoint union of a perfect set and a countable open set.

Example: Consider $\text{2-dim Euclidean space, } \mathbb R^2$, which is a polish space. Define

$\quad C = \text{The Unit Circle } \cup \{(\pm \frac{1}{n},\, 0) \;|\; n \ge 1\} \cup \{(0,\, \pm \frac{1}{n}) \;|\; n \ge 1\} \cup \{(0,\,0)\}$

The set $C$ is closed in $\mathbb R^2$, and it can be written (uniquely) as the union of the unit circle, a perfect set, and the remaining countable points in $C$ with norm less than $1$. All these points not on the unit circle are isolated except for $(0,\,0)$, the origin.