# Every non-empty perfect set in $\mathbb{R}^{k}$ is uncountable

In the Principles of Mathematical Analysis, 3rd Ed. by Rudin, he mentions in a Theorem 2.43 on page 41 that every non-empty perfect set $$P$$ in $$\mathbb{R}^{k}$$ is uncountable.

In the proof, he assumes contrariwise that the points of $$P$$ are denumerable and labelled $$\overline{x_{1}}, \overline{x_{2}}, \overline{x_{3}}, ...$$ . He then constructs a sequence of neighborhoods $$\{V_{n}\}$$ by induction as follows:

Let $$V_{1}$$ be the neighborhood of $$\overline{x_{1}}$$, defined as $$V_{1} = \{\overline{y} \in \mathbb{R}^{k} : |\overline{x_{1}} - \overline{y}| < r\}$$, and the closure of $$V_{1}$$ denoted $$\overline{V_{1}}$$ defined $$\overline{V_{1}} = \{\overline{y} \in \mathbb{R}^{k} : |\overline{x_{1}} - \overline{y}| \leq r\}$$.

Now, suppose $$V_{n}$$ has been constructed, with $$V_{n} \cap P \neq \emptyset$$. Then, there exists $$V_{n+1}$$ such that:

1. $$\overline{V_{n+1}} \subset V_{n}$$ (the closure of $$V_{n+1}$$ is a proper subset of $$V_{n}$$)
2. $$\overline{x_{n}} \notin \overline{V_{n+1}}$$
3. $$\overline{V_{n+1}} \cap P \neq \emptyset$$.

The third condition ensures our induction hypothesis holds and allows for the construction to continue.

Now, let $$K_{n} = \overline{V_{n}} \cap P$$. Since $$\overline{V_{n}}$$ is closed and bounded, it must be compact as well (proved earlier). Since $$\overline{x_{n}} \notin K_{n+1}$$ by construction, no point of $$P$$ is in $$\cap_{1}^{\infty}K_{n}$$. Since $$K_{n} \subset P$$, this implies that $$\cap_{1}^{\infty}K_{n} = \emptyset$$. This fact in turn contradicts a corollary of an earlier theorem that such an intersection must be non-empty.

Where my understanding falls short is in the statement that is in bold. The contradiction relies on it. Now, since since $$P$$ is perfect, $$P$$ is closed and every point of $$P$$ must be a limit point of $$P$$. Further, since $$P$$ is non-empty, it must have infinitely many points (or else, it cannot have any limit points). Now, when I take a point $$x_{n+1}$$ inside the neighborhood $$V_{n}$$ of $$x_{n}$$, and construct a neighborhood $$V_{n+1}$$ in a manner that doesn't include $$x_{n}$$, $$V_{n+1}$$ still contains infinitely many points; it just doesn't contain $$x_{n}$$. So, when I continue this induction, doesn't each subsequent neighborhood $$V_{m}$$ have points in it? And, when I take an infinite intersection as suggested, it needn't contain the specific labelled points at the centre of the neighborhood; but it still does contain infinitely many points.

Where am I going wrong?

For some visual context, I am visualizing my argument as a set of circles with each successive circle (denoting $$V_{n}$$) drawn within the previous one but drawn so that it doesn't contain the centre of the previous one.

You are making things too complicated. $$\overline {x_n} \notin K_{n+1}=\overline {V_{n+1}} \cap P$$ by property 2). If $$\overline {x_i}$$ is a point of $$P$$ which belongs to every $$K_n$$ then $$\overline {x_i} \in K_{i+1} \subset \overline {V_{i+1}}$$ again contradicting 2).
• I see your point, sir. Could you explain then where my circles analogy fails? It is true that if I consider some point $x_{i}$, then, $K_{i+1}$ must not contain it by construction. However, $K_{i+1}$ will still have an infinite number of points in common with its superset, right? Nov 27, 2020 at 9:29