# Help proving Cantor Intersection Theorem using Bolzano-Weierstrass Theorem

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.

• Just a first thought (I haven't digested the whole proof yet): Bolzano-Weierstrass doesn't guarantee an $x_j \in A$ which is a cluster point of $A$, but rather, gives some $x\in \mathbb{R}^p$ which is a cluster point of $A$. This also means your second paragraph is now irrelevant. Assuming you've already proven that compact sets are closed, that that closed sets contain their limit points, you can use the two facts to argue that $x\in C_i$ for all $i$. – Jason DeVito Feb 23 '14 at 19:09
• You do not need to consider proper subsets, $x_n\in C_n$ is enough, avoiding $C_{n+1}$ is not necessary. And why do you want to turn the constructed sequence into a set? Bolzano-Weierstraß is about sequences in compact sets. – Lutz Lehmann Feb 23 '14 at 19:10
• @Jason DeVito: I had completely misunderstood the Theorem. Just read through it again. I think I have an alternate argument. I'll post it as an answer. And yes I have already proven Heine-Borel Theorem. . – Ishfaaq Feb 23 '14 at 19:25
• @LutzL: The book I'm reading asserts the Theorem well before introducing sequences and is described purely in topological/set theoretic language. And hence I need $\{x_n\}$ to be infinite - a construction I failed to come up with if I chose points in $C_n$ arbtrarily – Ishfaaq Feb 23 '14 at 19:27
• How does it then come that you can work with sequences of nested subsets $C_n$? And construct the sequence elements $x_n$? Seems inconsistent. Perhaps you should document that restriction in your question, and also the formulation of Bolzano-Weierstraß that is used in the book. – Lutz Lehmann Feb 23 '14 at 19:32

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 $x \in C_n \; \forall n \in \Bbb N$

• I think it looks good now. – Jason DeVito Feb 23 '14 at 20:00
• @Jason DeVito: Thanks a lot for your time. How about $Q_1$ above? Any issues? – Ishfaaq Feb 23 '14 at 20:02
• I think Q1 is fine: By discarding and renumbering, you can definitely assume that the $C_i$ form a strictly decreasing chain of subsets. Note that it's possible that after discarding duplicates, only finitely many $C_i$ remain, so it's possible that you won't be able to use the Bolzano-Weierstrass theorem. On the other hand, when there are only finitely many $C_i$, the theorem is trivial anyway. One way to avoid all of this is just just take $x_i\in C_i$ without worrying if $C_{i+1}\subsetneq C_i$ or not. Letting $A = \{x_i\}$, either $A$ is infinite (and your argument works just fine)... – Jason DeVito Feb 23 '14 at 20:22
• or $A$ is finite. If $A$ is finite, then the $x_j$ of largest index in $A$ will be in all the $C_i$. – Jason DeVito Feb 23 '14 at 20:22
• I believe that any $x_j$ which is repeated infinitely will work. If $A$ is finite, some $x\in A$ is repeated infinitely many times via your choices. Now, given any $C_i$, because $x$ is repeated infinitely, there is a $j > i$ with $x = x_j$. Then $x = x_j \in C_j \subseteq C_i$, so $x\in C_i$ as well. – Jason DeVito Feb 24 '14 at 3:11