Let $P \subseteq \Bbb R$ be a non-empty perfect set. Then $|P|=2^{\aleph_0}$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
For any non-empty bounded set $A \subseteq \Bbb R$ , we define the diameter of $A$ , $\operatorname{diam} A$ , as $\sup A – \inf A$ . Let $\mathfrak{P}$ be the set of all non-empty perfect subsets of $\Bbb R$ .
Lemmas:
There exists a functions $g_1:\mathfrak P \to \mathfrak P$ and $g_2:\mathfrak P \to \mathfrak P$ such that, for all $P \in \mathfrak P$, $g_1(P) \subseteq P$, $g_2(P) \subseteq P$, and $g_1(P) \cap g_2(P) = \emptyset$.
There exists a function $\mathcal F:\mathfrak{P} \times \Bbb N^+ \to \mathfrak{P}$ such that, for all $(P,n) \in \mathfrak{P} \times \Bbb N^+$, $\mathcal F(P,n) \subseteq P$ and $\operatorname{diam}\mathcal F(P,n) \le \dfrac{1}{n}$.
Any non-empty system of closed and bounded sets with the finite
intersection property has a nonempty intersection.
Let $\mathcal S = \{0,1\}^{\Bbb N}$. We construct a system of perfect subsets of $P$ recursively as follows:
$$\begin{align}
P_{\langle \rangle} &= P\\
P_{\langle s_0, \cdots, s_n, 0 \rangle} &= \mathcal F(g_1(P_{\langle s_0, \cdots, s_n \rangle}),n+1)\\
P_{\langle s_0, \cdots, s_n, 1 \rangle} &= \mathcal F(g_2(P_{\langle s_0, \cdots, s_n \rangle}),n+1)\\
\end{align}$$
For any $f \in \mathcal S$, let $P_f = \bigcap_{n \in \Bbb N} P_{f \restriction n}$. Then $P_f \neq \emptyset$ by Lemma 3. Moreover, $P_f$ contains a unique element: If $x,y \in P_f$, $|x-y| \le \operatorname{diam} P_f \le \operatorname{diam} P_{f \restriction n} \le \dfrac{1}{n}$ for all $n \in \Bbb N^+$ by Lemma 2, so $x=y$. Let $d_f$ be the unique element of $P_f$.
We define a function $\mathcal G:\mathcal S \to P$ by $$\mathcal G(f)=d_f$$
Clealy, $\mathcal G$ is injective and thus $2^{\aleph_0}=|\mathcal S| \le |P| \le |\Bbb R| = 2^{\aleph_0}$. Hence $|P|=2^{\aleph_0}$.
Update: Thank to Henno Brandsma's remarks, I added important details here.
- $\mathcal G$ is injective
Assume that $f,g \in \mathcal S$ such that $\mathcal G(f)=\mathcal G(g)$. Then $d_f=d_g=d$. Thus $d \in P_f = \bigcap_{n \in \Bbb N} P_{f \restriction n}$ and $d \in P_g = \bigcap_{n \in \Bbb N} P_{g \restriction n}$. It follows that $d \in P_{f \restriction n}$ and $d \in P_{g \restriction n}$ for all $n \in \Bbb N$. Hence $d \in P_{f \restriction n} \cap P_{g \restriction n}$ for all $n \in \Bbb N$.
Clearly, $f \restriction 0 = \emptyset = g \restriction 0$. Assume the contrary that $f \restriction n \neq g \restriction n$ for some $n \in \Bbb N$. Let $n_0+1$ be the least of such $n$'s. Thus $f \restriction n_0 = g \restriction n_0 = h$ and $f \restriction n_0+1 \neq g \restriction n_0+1$. We assume WLOG that $f \restriction n_0+1 = h \cup \{(n_0+1,0)\}$ and $g \restriction n_0+1 = h \cup \{(n_0+1,1)\}$. Thus $P_{f \restriction n_0+1}=\mathcal F(g_1(P_h),n_0+1)$ and $P_{g \restriction n_0+1} = \mathcal F(g_2(P_h),n_0+1)$. By the definition of $\mathcal F$, $P_{f \restriction n_0+1} \cap P_{g \restriction n_0+1} = \emptyset$. This is a contradiction. Hence $f \restriction n = g \restriction n$ for all $n \in \Bbb N$ and thus $f=g$.
- $P_f \neq \emptyset$ for all $f \in \mathcal S$
It follows from the construction of $\mathcal F$ that $P_{f \restriction n+1} \subseteq P_{f \restriction n}$ for all $n \in \Bbb N$. Thus $P_f = \bigcap_{n \in \Bbb N} P_{f \restriction n}$ $= P_{f \restriction 0} \cap \bigcap_{n \in \Bbb N^+} P_{f \restriction n} = P_{\langle \rangle} \cap \bigcap_{n \in \Bbb N^+} P_{f \restriction n} =P \cap \bigcap_{n \in \Bbb N^+} P_{f \restriction n}= \bigcap_{n \in \Bbb N^+} P_{f \restriction n}$. From the construction of $\mathcal F$, we get that $P_{f \restriction n}$ is bounded and perfect (thus closed) for all $n \in \Bbb N^+$.
Next we prove that $\{P_{f \restriction n} \mid n \in \Bbb N\}$ has the finite intersection property. Let $N$ be a nonempty finite subset of $\Bbb N^+$ and $n_0=\max N$. Then $P_{f \restriction n_0} \subseteq P_{f \restriction n}$ for all $n \in N$. Thus $\bigcap_{n \in N} P_{f \restriction n} = P_{f \restriction n_0} \neq \emptyset$.
Best Answer
Your proof is the standard Cantor tree argument (and shows in fact that the set $P_f$ is homeomorphic to $\{0,1\}^\mathbb{N}$, if you take bit more time to show $\mathcal{G}$'s continuity, which you don't need for the pure cardinality argument).
You might want to add an argument why $\mathcal{G}$ is injective (it is, but you must say more than "clearly". You apply lemma 3 without justifying it (it is justified, but prove by induction that the sets are nested along a path mapped by $f$, which implies nestedness).
The proof is fine beyond these nitpicks.