Came across the following exercise in Bartle's Elements of Real Analysis and am quite unsure about my solution. Would greatly appreciate it if someone could take a look at it. The Bolzano – Weierstrass as mentioned in the text is "Every bounded infinite subset of $\Bbb R^p$ has a cluster point".
Prove the Cantor Intersection Theorem by selecting a point $x_n$
from $C_n$ and then applying the Bolzano – Weierstrass Theorem to the
set $\{x_n \ | \ n \in \Bbb N \}$.
My Attempt:
Let $C_1 \supset C_2 \supset \ … \ \supset C_n \supset \ … $ be a sequence of non-void compact sets in $\Bbb R^p$. Without loss of generality we may assume that each set in the sequence is a "proper" subset of its predecessor. Then we can choose $x_n \in C_n \setminus C_{n + 1}$for every $n \in \Bbb N$. Then the set $ A = \{x_n \ | \ n \in \Bbb N \}$ will have infinitely many elements. $A$ is bounded since $A \subseteq C_1 $. Therefore by the Bolzano – Weierstrass Theorem $\exists x_j \in A$ such that every neighbourhood of $x_j$ contains at least one element in $A$ distinct from $x_j$.
Suppose $x_j \not \in C_n$ for certain $n \in \Bbb N$. Let $m = \text{Min} \{n \in \Bbb N \ | \ x_j \not \in C_n\}$. $m$ exists by the Well-Ordering Principle. Then $x_j \in (C_m)^C$. $ \ j \le m$ would lead to a contradiction since it would imply $x_j \in C_j \subset C_m.$ And $j \lt m – 1 \implies x_j \not \in C_{m – 1}$ again leading to a contradiction. $\therefore j = m – 1 \; (m \neq 1)$.
$(C_m)^C$ is a neighbourhood of the cluster point $x_j$ and hence $\exists x_i \in A$ such that $x_i \in (C_m)^C$. Clearly $i \lt m – 1$. Therefore the set $B = \{ x_i \ | \ x_i \in A \cap (C_m)^C \}$ is finite. Let $r = \text{Min} \{ r_i \ | \ r_i = \left| \left|{x_j – x_i}\right| \right|, x_i \in B \}$. The set $\{ y \in \Bbb R^p \ | \ \left| \left|{x_j – y}\right| \right| \lt r \} \cap (C_m)^C$ is a neighbourhood of $x_j$ (since it is the intersection of two open sets) and contains no points in $A$ distinct from $x_j$ leading to a contradiction.
$\implies x_j \in C_n \; \forall n \in \Bbb N$
Q.E.D.
$Q_1:$ Is my restriction to consider only proper subsets fair enough? My argument is given any sequence of nested compact sets we remove the additional equal ones in the sequence and find a common point. Then it will be a common point to the original sequence too. But I need the set $\{x_n\}$ to be infinite. Will that be affected?
$Q_2:$ Is my proof correct in general? Are the arguments good enough?
Please comment. Any help is appreciated. Thanks in advance.
Best Answer
Due to the glaring error pointed out by @Jason DeVito I'm gonna post here a corrected answer to the above exercise. Still needs some verification though I think and would love some opinion on it.
... Continuing after paragraph 1 above in my solution.
Let $x$ be the cluster point of $A$ guaranteed by the Bolzano - Weierstrass Theorem.
Suppose $x$ is an exterior point for a set $C_m$ in the sequence. Then there is a neighbourhood of $x$ entirely contained in $(C_m)^C$. This neighbourhood must contain some $x_i \in A$ such that $x_i \neq x$. Now, $i \ge m$ would lead to a contradiction since $x_i \in C_i \subset C_m$. Therefore, $i \lt m$. Let $r = \text{Min} \{ r_i \ | \ r_i = \left| \left|{x - x_i}\right| \right|, x_i \in A \cap (C_m)^C \}$. Then the set $\{ y \in \Bbb R^p \ | \ \left| \left|{x_j - y}\right| \right| \lt r \} \cap (C_m)^C$ is a neighbourhood of $x$ and contains no elements in $A$ contradicting the fact that $x$ is a cluster point of $A$.
Therefore, $x$ is either an interior or boundary point of $C_m$. Since $C_m$ is closed it contains all of its boundary points and hence $x \in C_m$.
Since $C_m$ was arbitrary $x \in C_n \; \forall n \in \Bbb N $