Here's Theorem 2.43 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Let $P$ be a non-empty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.
Here's the definition of a perfect set:
Let $(X,d)$ be a metric space, and let $P \subset X$. Then $P$ is perfect if it is closed (i.e. it contains all of its limit points) and every point of $P$ is also a limit point of $P$.
Now here's the proof Rudin gives:
Since $P$ has limit points, $P$ must be infinite. Suppose $P$ is countable, and denote the points of $P$ by $x_1, x_2, x_3, \ldots$. We shall construct a sequence $\{V_n\}$ of neighborhoods as follows:
Let $V_1$ be any neighborhood of $x_1$. If $V_1$ consists of all $y \in \mathbb{R}^k$ such that $\vert y – x_1 \vert < r$, the closure $\overline{V_1}$ of $V_1$ is the set of all $y \in \mathbb{R}^k$ such that $\vert y – x_1 \vert \leq r$.
Supose $V_n$ has been constructed, so that $V_n \cap P$ is not empty. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_n$, (ii) $x_n \not\in \overline{V_{n+1}}$, (iii) $V_{n+1} \cap P$ is not empty. By (iii), $V_{n+1}$ satisfies our induction hypothesis, and the construction can proceed.
Put $K_n = \overline{V_n} \cap P$. Since $\overline{V_n}$ is closed and bounded, $\overline{V_n}$ is compact. Since $x_n \not\in K_{n+1}$, no point of $P$ lies in $\cap_1^\infty K_n$. Since $K_n \subset P$, this implies that $\cap_1^\infty K_n$ is empty. But each $K_n$ is non-empty, by (iii), and $K_n \supset K_{n+1}$, by (i); this contradicts the Corollary to Theorem 2.36.
Finally, here's Theorem 2.36:
If $\{K_\alpha\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_\alpha\}$ is nonempty, then $\cap K_\alpha$ is nonempty.
And, here's the Corollary to Theorem 2.36:
If $\{K_n\}$ is a sequence of nonempty compact sets such that $K_n \supset K_{n+1}$ ($n=1, 2, 3, \ldots$), then $\cap_1^\infty K_n$ is not empty.
Now my question is, what exactly is the induction hypothesis that Rudin refers to in the proof? For $V_1$, he only says that it is any neighborhood of the point $x_1$.
What is the induction hypothesis? Please also proceed from $V_1$ to $V_2$, and then from $V_2$ to $V_3$ for the sake of illustration.
After reading David's comment, I've modified my proof as follows:
Since $P$ is nonempty, it has a point $a$, which is also a limit point of $P$ since $P$ is perfect. But no finite set in a metric space can have a limit point. So $P$ must be infinite. Suppose that $P$ is countable, and let's denote the points of $P$ by $x_1, x_2, x_3, \ldots$.
We shall construct a sequence $V_n$ of neighborhoods such that, for each $n \in \mathbb{N}$, the intersection $V_n \cap P$ is non-empty and $x_n \not\in V_{n+1}$.
Let $V_1$ be any neighborhood of $x_1$. Then we can show that if $$V_1 = \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – x_1 \vert < \epsilon_1 \ \right\},$$ where $\epsilon_1$ is some positive real number, then the closure $\overline{V_1}$ of $V_1$ is given by
$$ \overline{V_1} = \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – x_1 \vert \leq \epsilon_1 \ \right\}.$$Since $x_1$ is a limit point of set $P$, the nieghborhood $V_1$ of $x_1$ contains a point, say, $y_1$ of $P$ other than the point $x_1$ itself.
Now $y_1 \in V_1$ and $V_1$ is an open set in the metric space $\mathbb{R}^k$. So there is a positive real number $\delta_1$ such that $N_{\delta_1} (y_1) \subset V_1$, where
$$N_{\delta_1} (y_1) \colon= \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – y_1 \vert < \delta_1 \ \right\}.$$
Let
$$\epsilon_2 \colon= \frac 1 2 \min \left( \delta_1, \vert x_1 – y_1 \vert \right).$$
Then $\epsilon_2 > 0$.
Let
$$V_2 \colon= \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – y_1 \vert < \epsilon_2 \ \right\}.$$
Then
$$\overline{V_2} = \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – y_1 \vert \leq \epsilon_2 \ \right\}.$$
Thus, $V_2$ is a neighborhood such that $ y_1 \in V_2 \cap P$ so that $V_2 \cap P$ is non-empty, $\overline{V_2} \subset V_1$, but $x_1 \not\in \overline{V_2}$.If $ y_1 \not= x_2$, then let
$$\epsilon_3 \colon= \frac 1 2 \min \left( \epsilon_2 , \vert p – x_2 \vert \right).$$
Then $\epsilon_3 > 0$.
Let
$$V_3 \colon= \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – y_1 \vert < \epsilon_3 \ \right\}.$$
Then
$$\overline{V_3} = \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – y_1 \vert \leq \epsilon_3 \ \right\}.$$
Thus, $y_1 \in V_3 \cap P$, and so $V_3 \cap P$ is non-empty; moreover, $\overline{V_3} \subset V_2$ and $x_2 \not\in \overline{V_3}$.On the other hand, if $y_1 = x_2$, then since $y_1$ is a limit point of set $P$, and $V_2$ is a neighborhood of $y_1$, this neighborhood $V_2$ contains a point $y_2$, say, of $P$ other than the point $y_1 = x_2$ itself.
Now as $y_2 \in V_2$ and $V_2$ is an open set in $\mathbb{R}^k$, so there is some positive real number $\delta_2 > 0$ such that
$$N_{\delta_2 } (y_2) \subset V_2.$$
So if we take
$$\epsilon_3 \colon= \frac 1 2 \min \left( \delta_2 – \vert y_2 – x_2 \vert , \ \vert y_2 – x_2 \vert \right),$$
then $\epsilon_3 > 0$.Let
$$V_3 \colon= \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – y_2 \vert < \epsilon_3 \ \right\}.$$
Then
$$\overline{V_3} \colon= \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – y_2 \vert \leq \epsilon_3 \ \right\}.$$
Thus, $y_2 \in V_3 \cap P$ and so $V_3 \cap P$ is non-empty; moreover, $\overline{V_3} \subset V_2$, and $x_2 \not\in V_3$.Thus, in either case we have obtained a neighborhood $V_3$ such that $V_3 \cap P$ is non-empty, $\overline{V_3} \subset V_2$, but $x_2 \not\in V_3$.
The step from $V_2$ to $V_3$ is redundant in the formal presentation of the proof, but this step perhaps more vividly illustrates how to proceed.
Now suppose that a neighborhood $V_n$ ($n= 3, 4, 5, \ldots$) has been constructed such that $x_{n-1} \not\in \overline{V_n}$, $\overline{V_n} \subset V_{n-1}$, and $V_n \cap P$ is non-empty.
Let
$$V_n \colon= \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – p \vert < \epsilon \ \right\},$$
where $\epsilon$ is some positive real number and $p$ is some point in $\mathbb{R}^k$.We now construct $V_{n+1}$ such that $V_{n+1} \cap P$ is non-empty, $\overline{V_{n+1}} \subset V_n$, and $x_n \not\in \overline{V_{n+1}}$.
Suppose that $q \in V_n \cap P$. As $q \in V_n$ and $V_n$ is open, there is some positive real number $\delta_n$ such that
$$N_{\delta_n} (q) \subset V_n,$$
where
$$N_{\delta_n} (q) \colon= \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – q \vert < \delta_n \ \right\}.$$If $q \not= x_n$, then let's take
$$\epsilon_{n+1} \colon= \frac 1 2 \min \left( \delta_n , \vert q – x_n \vert \right).$$
So $\epsilon_{n+1} > 0$, and let
$$V_{n+1} \colon= \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – q \vert < \epsilon_{n+1} \ \right\}.$$
Then
$$\overline{V_{n+1}} \colon= \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – q \vert \leq \epsilon_{n+1} \ \right\}.$$Thus, $V_{n+1}$ is a neighborhood such that $q \in V_{n+1} \cap P$ so that $V_{n+1} \cap P$ is non-empty, $\overline{V_{n+1}} \subset V_n$, and $x_n \not\in \overline{V_{n+1}}$.
On the contrary, if $q = x_n$, then since $q$ is a limit point of $P$, the neighborhood $N_{\delta_n} (q)$ contains a point $b$, say, of $P$ other than the point $q = x_n$ itself.
Now if we take
$$\epsilon_{n+1} \colon= \frac 1 2 \min \left( \delta_n – \vert b – x_n \vert, \ \vert b – x_n \vert \right),$$
then $\epsilon_{n+1} > 0$. Let
$$V_{n+1} \colon= \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – b \vert < \epsilon_{n+1} \ \right\}.$$
Then
$$\overline{V_{n+1}} = \left\{ \ y \in \mathbb{R}^k \ \colon \ \vert y – b \vert = \epsilon_{n+1} \ \right\}.$$
So, $b \in V_{n+1} \cap P$, which implies that $V_{n+1} \cap P$ is non-empty, $\overline{V_{n+1}} \subset V_n$, and $x_n \not\in \overline{V_{n+1}}$.Thus, from $V_n$, in either case, we have constructed a neighborhood $V_{n+1}$ such that $V_{n+1} \cap P$ is non-empty, $\overline{V_{n+1}} \subset V_n$, but $x_n \not\in \overline{V_{n+1}}$.
Thus, we have inductively obtained a sequence $\{V_n\}_{n \in \mathbb{N}}$ of neighborhoods such that, for each $n \in \mathbb{N}$, the intersection $V_n \cap P$ is non-empty,
$\overline{V_{n+1}} \subset V_n$, and $x_n \not\in \overline{V_{n+1}}$.Put $K_n \colon= \overline{V_n} \cap P$ for each $n \in \mathbb{N}$.
Let $n \in \mathbb{N}$ be arbitrary. Now $\overline{V_n}$, being a closed and bounded subset of $\mathbb{R}^k$, is compact. Moreover, as both $P$ and $\overline{V_n}$ are closed, so is $K_n$. Thus, $K_n$, being a closed subset of the compact set $\overline{V_n}$, is also compact. Since $V_n \cap P$ is non-empty, so $K_n$ is also non-empty. Moreover, as $$\overline{V_{n+1}} \subset V_n \subset \overline{V_n},$$
so we have $K_{n+1} \subset K_n$. Finally, as
$$P = \left\{ \ x_1, x_2, x_3, \ldots \ \right\}$$
and as $x_n \not\in \overline{V_{n+1}}$, so $x_n \not\in K_{n+1}$ and hence
$$x_n \not\in K_1 \cap K_2 \cap K_3 \cap \ldots,$$
which implies that this intersection is empty.Let $m_1, m_2, \ldots, m_r \in \mathbb{N}$, and let's take
$$m \colon= \max \left( m_1, \ldots, m_r \right).$$
Then
$$K_{m_1} \cap \ldots \cap K_{m_r} = K_m,$$
which is non-empty.Thus, $\{K_n\}_{n\in\mathbb{N}}$ is a sequence of non-empty compact sets in the metric space $\mathbb{R}^k$ such that the intersection of any finitely many of these sets is non-empty but the intersection $\cap_{n \in \mathbb{N}} K_n$ is empty.
But this cannot hold in any metric space, by Theorem 2.36 in Rudin. So our suppose that $P$ is a non-empty perfect set in $\mathbb{R}^k$ and $P$ is also countable is wrong. Hence Theorem 2.43 in Rudin holds.
Is the above proof correct?
If so, is the presentation clear enough (or any clearer than Rudin's presentation)?
Where does this proof need improvement?
Best Answer
I don't think this is actually a proof by induction, but rather a definition by induction of a sequence of open balls $V_n$ that have nonempty intersection with $P$.
The facts that are needed (once the sequence has been constructed) about the sets $V_n$ are read directly from the inductive step of the construction and do not need to be proved by induction.
However, the definition of $V_{n+1}$ would not make sense if it was not already known that $V_n \cap P \ne \varnothing$. Therefore it must be verified that $V_{n+1} \cap P \ne \varnothing$ in order for the construction to continue. (The verification is trivial because this results from (iii) in the choice of $V_{n+1}$.) This is what Rudin refers to as the "induction hypothesis."
On the other hand, points (i) and (ii) being true at step $n$ are not needed in order for the construction to continue at step $n+1$.