# Prove that $[0,1]$ is not a disjoint union of a countable family of closed sets [closed]

This proof is probably too long, please don't downvote this. I'm just leaving it here in case somebody wants to actually verify its validity for me. I didn't use a hint in the book and decided to prove it my own way.

Outline of the proof:

• After trying to cover $$[0,1]$$ with 2 disjoint closed sets A and B, show that there will remain some open interval $$(c_{0}, c_{1}) \in [0,1]$$ that has to be covered, where $$c_{0} \in A$$ or $$B$$, $$c_{1} \in A$$ or $$B$$.
• Use some closed set to try cover this interval and obtain 2 intervals $$(c_{00}, c_{01}), (c_{10}, c_{11})$$ yet to cover, where all endpoints belong to closed sets (endpoints are not necessarily infs/sups of those closed sets, just points belonging to them).
• Repeat this process infinitely, proving that trying to cover each of $$n$$ intervals will create at least $$2n$$ yet uncovered open intervals (taking into account that total length of all open intervals has to approach 0).
• Prove that there will be an uncountable number of points left not belonging to any of the above closed sets after the above process, and name this set $$N$$. Name set of all endpoints of closed sets (i.e. $$c_{0}, c_{10}$$...) $$E$$.
• To make proving that $$N$$ is not a union of a countable number of closed sets easier, define a metric space $$S$$ with Euclidian metric which consists of a) set $$N^\prime$$ which contains same points as $$N \in \mathbb{R}$$, b) set $$E^\prime$$ which contains same points as $$E \in \mathbb{R}$$. Proceed to prove this space is complete, then assume $$N^\prime$$ is a union of a countable number of closed sets, and then use Baire Category Theorem to derive a contradiction, henceforth negating the assumption.
• Prove that $$N^\prime$$ not being a union of a countable number of closed sets in $$S$$ implies that $$N$$ is not a union of a countable number of closed sets in $$\mathbb{R}$$, which would imply that $$[0,1]$$ is not a union of a countable number of disjoint closed sets.

Proof:

1. Every nonempty bounded subset of $$\mathbb{R}$$ has inf/sup, and that inf/sup belongs to the subset if it is closed (if it's isolated from the set then it is a min/max of the set and is an isolated point of the set, if it's not isolated from the set then it's a limit point thus belongs to it).
2. Take a closed set $$A \subset \mathbb{R} \mid A \cap [0,1] \neq \emptyset$$. Take a random point $$x \in ([0,1] \setminus A), x \notin \{0,1\} \implies \exists \epsilon > 0 \mid B_{\epsilon}(x) \in ([0,1] \setminus A)$$, since if not, A will not be a closed set. Take $$(c_{0}, c_{1}) \subset [0,1] \mid c_{0} = \max(\sup(A \cap [0,x)),0), c_{1} = \min(\inf(A \cap (x,1]), 1)$$. If $$c_{0} \notin A$$ or $$c_{1} \notin A$$, then $$c_{0} = 0$$ or $$c_{1} = 1$$, so attempting to cover $$[0, c_{1})$$ or $$(c_{0}, 1]$$ with another closed set B will yield an open interval of form $$(c_{0}, c_{1})$$. Next:
3. Take arbitrary closed set $$C_{0} \cap (c_{0},c_{1}) \neq \emptyset$$. $$c_{0},c_{1}$$ are not limit points of $$C_{0}$$ as $$c_{0},c_{1} \notin C_{0}$$ since they belong to $$B$$ and $$A$$, also $$C_{0} \cap (c_{0},c_{1})$$ is bounded. Thus $$\exists c_{01}, c_{10} \in C_{0} | c_{0} < \inf(C_{0} \cap (c_{0},c_{1}))=c_{01} \in C_{0} \leq \sup(C_{0} \cap (c_{0},c_{1}))=c_{10} \in C_{1} < c_{1}$$. Denote $$c_{0}$$ as $$c_{00}$$ and $$c_{1}$$ as $$c_{11}$$, so an open set $$(c_{00}, c_{01}) \cup (c_{10}, c_{11})$$ is still not covered, and proceed further in this fashion.
4. In this proof only, we will denote $$\overline{d_{\overline{m1}}...d_{\overline{mn}}}$$ as $$d_{m1}...d_{mn}$$, a sequence of 0s and 1s, where m (first sub-subscript) denotes ordinal number of open interval in a union $$O^\prime_{n}$$ and second sub-subscript ranges from 1 to n where n is a step at which we are trying to cover all open sets. Let $$O_{d_{m1}...d_{m(n-1)}} = (c_{d_{m1}...d_{m(n-1)}0}, c_{d_{m1}...d_{m(n-1)}1})$$ and $$O^\prime_{n-1} = \bigcup\limits_{m=1; \{d_{m1}...d_{m(n-1)}:1 \leq k \leq n, d_{mk} \in \{0,1\}\}}^{2^{n-1}} O_{d_{m1}...d_{m(n-1)}}$$. Since at $$n$$-th step we have to cover $$2^{n-1}$$ open intervals, we can take a union of disjoint closed sets $$C^\prime_{n-1} = \bigcup\limits_{m=1; \{d_{m1}...d_{m(n-1)}:1 \leq k \leq n-1, d_{mk} \in \{0,1\}\}}^{2^{n-1}} C_{d_{m1}...d_{m(n-1)}}$$ and take its intersection with open set $$O^\prime_{n-1}$$ so that $$C_{d_{m1}...d_{m(n-1)}} \cap O_{d_{m1}...d_{m(n-1)}} \neq \emptyset$$. This implies $$\forall m, 1 \leq m \leq 2^{n-1}, \hspace{0.1cm} \exists c_{d_{m1}...d_{m(n-1)}01} = \inf(O_{d_{m1}...d_{m(n-1)}} \cap C^\prime_{n-1})$$ and $$\exists c_{d_{m1}...d_{m(n-1)}10} = \sup(O_{d_{m1}...d_{m(n-1)}} \cap C^\prime_{n-1}) \mid (c_{d_{m1}...d_{m(n-1)}00},c_{d_{m1}...d_{m(n-1)}01}) \cup (c_{d_{m1}...d_{m(n-1)}10},c_{d_{m1}...d_{m(n-1)}11}) \subset (O_{d_{m1}...d_{m(n-1)}} \setminus C^\prime_{n-1})$$. $$(c_{d_{m1}...d_{m(n-1)}00},c_{d_{m1}...d_{m(n-1)}01}) = (c_{d_{k1}...d_{k(n-1)}d_{kn}0},c_{d_{k1}...d_{k(n-1)}d_{kn}1}) = O_{d_{k1}...d_{k(n-1)}d_{kn}} \subset O^\prime_n$$. Denote $$O = \bigcup\limits_{m=0}^{\infty} O^\prime_{m}$$, $$C = \bigcup\limits_{n=0}^{\infty} C^\prime_{n}$$, $$E = \{c_{d_{m1}...d_{mn}d_{mn}d_{mn}...}: k \in \mathbb{N}, d_{mk} \in \{0,1\}\}$$. Also please note that for any endpoint in $$E$$ with finite subscript, such as $$c_{d_{m1}d_{m2}...d_{mn}}$$, $$c_{d_{m1}d_{m2}...d_{mn}} = c_{d_{m1}d_{m2}...d_{mn}d_{mn}d_{mn}...}$$, i.e. its finite subscript is equivalent to its infinite subscript which has last digit of its finite subscript repeating infinitely.
5. Since closed sets must cover entire $$[0,1], \lim_{n \to +\infty} \sum\limits_{m=1;\{d_{m1}...d_{m(n-1)}: d_{mk} \in \{0,1\}, 1 \leq k \leq n-1\}}^{2^{n-1}} |c_{d_{m1}...d_{m(n-1)}1}-c_{d_{m1}...d_{m(n-1)}0}| = 0$$. The number of closed sets will be countable (can get an injection to $$\mathbb{Q} \subset [0,1]$$: $$f(m,n)=(2m-1)/2^n$$ for $$m$$-th closed set inserted at $$n$$-th step, $$2m-1<2^n$$).
6. This infinite union of disjoint closed sets will not cover the entire $$[0,1]$$. Take an intersection of nested open intervals $$\bigcap\limits_{n=1; d_{q_{k}k} \in \{0,1\}, 1 \leq k \leq n}^{\infty} O_{d_{q_{1}1}...d_{q_{n}n}} \mid O_{d_{q_{1}1}...d_{q_{n-1}(n-1)}d_{q_{n}n}} \subset O_{d_{q_{1}1}...d_{q_{n-1}(n-1)}}$$ and $$\forall n_{1} \in \mathbb{N} \hspace{0.1cm} \exists n_{2} \in \mathbb{N}, n_{2} > n_{1} \mid d_{q_{n_{2}}n_{2}}=1-d_{q_{n_{1}}n_{1}}$$ and take a sequence of points $$(o_{d_{q_{1}1}...d_{q_{n}n}}), o_{d_{q_{1}1}...d_{q_{n}n}} \in O_{d_{q_{1}1}...d_{q_{n-1}n-1}}$$. $$(o_{d_{q_{1}1}...d_{q_{n}n}}) \rightarrow L \notin C$$. Proof: assume $$L \in (C \setminus E) \implies \exists e_{1}, e_{2} \in E \mid L \in (e_{1}, e_{2})$$, $$[e_{1}, e_{2}] \subset C \implies \exists \epsilon > 0 \mid B_{\epsilon}(L) \cap O = \emptyset \implies L \notin (C \setminus E)$$. Assume $$L=c_{d_{m_{1}1}...d_{m_{n}n}d_{m_{n+1}(n+1)}...} \in E \implies \exists N \in \mathbb{N} \mid \forall n,k \geq N, d_{m_{k}k}=d_{m_{n}n}$$. Take a sequence $$(o_{d_{m_{1}1}...d_{m_{n}n}})$$, $$o_{d_{m_{1}1}...d_{m_{n}n}} \in O_{d_{m_{1}1}...d_{m_{n-1}n-1}}$$, $$O_{m}=(O \supset O_{d_{m_{1}1}} \supset O_{d_{m_{1}1}d_{m_{2}2}} \supset ... \supset O_{d_{m_{1}1}...d_{m_{N-1}(N-1)}d_{m_{N}N}d_{m_{N+1}N}}...)$$. $$\exists N^\prime \geq N \mid d_{q_{N^\prime}N^\prime} = 1-d_{m_{N^\prime}N^\prime} \implies \exists \epsilon > 0 \mid \forall N^{\prime\prime} > N^\prime$$, $$|o_{d_{q_{1}1}...d_{q_{N^{\prime\prime}}N^{\prime\prime}}}-o_{d_{m_{1}1}...d_{m_{N^{\prime\prime}}N^{\prime\prime}}}| > \epsilon$$ since $$O_{d_{q_{1}1}...d_{q_{N^{\prime\prime}-1}N^{\prime\prime}-1}} \cap O_{d_{m_{1}1}...d_{m_{N^{\prime\prime}-1}N^{\prime\prime}-1}} = \emptyset$$. For the above $$\epsilon/2$$ there exist $$N_{q} \in \mathbb{N} \mid \forall n \geq N_{q}$$, $$|o_{q_{1}1q_{2}2...q_{n}n}-L| < \epsilon/2$$, $$N_{m} \in \mathbb{N} \mid \forall n \geq N_{m}$$, $$|o_{m_{1}1m_{2}2...m_{n}n}-L| < \epsilon/2$$. This implies $$\exists M = max(N^{\prime\prime}, N_{q}, N_{m}) \mid \forall n \geq M$$, $$\epsilon < |o_{m_{1}1...m_{M}M}-o_{q_{1}1...q_{M}M}| = |o_{m_{1}1...m_{M}M}-L+L-o_{q_{1}1...q_{M}M}| \leq |o_{m_{1}1...m_{M}M}-L|+|L-o_{q_{1}1...q_{M}M}| < \epsilon/2+\epsilon/2 = \epsilon$$ which is a contradiction $$\implies L \notin (E \cup (C \setminus E)) \implies L \notin C$$. Also, we can similarly prove that $$L = \bigcap\limits_{n=1; d_{q_{k}k} \in \{0,1\}, 1 \leq k \leq n}^{\infty} O_{d_{q_{1}1}...d_{q_{n}n}} \mid O_{d_{q_{1}1}...d_{q_{n-1}(n-1)}d_{q_{n}n}} \subset O_{d_{q_{1}1}...d_{q_{n-1}(n-1)}}$$ by assuming it doesn't belong to this intersection which will imply that there exists some $$\epsilon > 0$$ and some $$N \in \mathbb{N} \mid \forall n \geq N, |o_{q_{1}1...q_{n}n}-L| > \epsilon$$.
7. Denote $$N = \bigcup\limits_{\{d_{m1}d_{m2}...: d_{mk} \in \{0,1\}\}} (\bigcap\limits_{n=1; d_{mk} \in \{0,1\}, 1 \leq k \leq n}^{\infty} O_{d_{m1}...d_{mn}} \mid \forall t \in \mathbb{N} \hspace{0.1cm} \exists q \in \mathbb{N}, q > t \mid d_{mq}=1-d_{mt}) = \{o_{d_{m1}d_{m2}...}: \forall t \in \mathbb{N} \hspace{0.1cm} \exists q \in \mathbb{N}, q > t \mid d_{mq}=1-d_{mt}\}$$ (by the above proof of intersection of nested open intervals being non-empty and converging to a point L in the intersection). This union amounts to a union of uncountably infinite number of points with each point having as subscript an infinite sequence consisting of 2 subsequences of 0's and 1's. Since $$\forall L \in N, L \notin E \implies N \cap E = \emptyset$$. We have to prove $$N$$ is not a countable union of disjoint closed sets. Proof:
• Define a metric space $$S = (N \cup E, d)$$, $$d$$ is the Euclidian metric, $$N \cap E = \emptyset$$. It is complete which we prove next. $$\forall n \in \mathbb{N}, O_{d_{m1}d_{m2}...d_{mn}} = (c_{d_{m1}d_{m2}...d_{mn}0}, c_{d_{m1}d_{m2}...d_{mn}1}) \implies N \cup E = (\bigcup\limits_{\{d_{m1}d_{m2}...: d_{mk} \in \{0,1\}\}} (\bigcap\limits_{n=1; d_{mk} \in \{0,1\}, 1 \leq k \leq n}^{\infty} (c_{d_{m1}d_{m2}...d_{mn}0}, c_{d_{m1}d_{m2}...d_{mn}1}) \mid \forall t \in \mathbb{N} \hspace{0.1cm} \exists q \in \mathbb{N}, q > t \mid d_{mq}=1-d_{mt})) \cup \{c_{d_{m1}d_{m2}d_{m3}...d_{mn}d_{mn}d_{mn}...}: k \in \mathbb{N}, d_{mk} \in \{0,1\}\} = (\bigcup\limits_{\{d_{m1}d_{m2}...: d_{mk} \in \{0,1\}\}} (\bigcap\limits_{n=1; d_{mk} \in \{0,1\}, 1 \leq k \leq n}^{\infty} [c_{d_{m1}d_{m2}...d_{mn}0}, c_{d_{m1}d_{m2}...d_{mn}1}] \mid \forall t \in \mathbb{N} \hspace{0.1cm} \exists q \in \mathbb{N}, q > t \mid d_{mq}=1-d_{mt})) = (\bigcap\limits_{n=1}^{\infty} (\bigcup\limits_{d_{mk} \in \{0,1\}, 1 \leq k \leq n} [c_{d_{m1}d_{m2}...d_{mn}0}, c_{d_{m1}d_{m2}...d_{mn}1}])$$. It is closed since finite union of closed sets is closed, and intersection of an arbitrary number of closed sets is closed. Since $$S$$ is closed it is complete under the same Euclidian metric.
• $$N \in \mathbb{R}$$ is defined as $$N = \bigcap\limits_{n=1;\{d_{mk}: 1 \leq k \leq n, d_{mk} \in \{0,1\}\}}^{\infty} O_{d_{m1}d_{m2}...d_{mn}} = \bigcap\limits_{m=1}^{\infty} O_{m}^{\prime}$$. Take some $$m \in \mathbb{N}$$, $$O_{m}^{\prime}$$ open in $$\mathbb{R} \implies \forall x \in ((N \cup E) \cap O_{m}^{\prime}) \hspace{0.05cm} \exists \epsilon > 0 \mid B_{\epsilon}(x) \subset O_{m}^{\prime} \implies \forall y \in B_{\epsilon}(x), y \in (N \cup E)$$ or $$y \in (O_{m}^{\prime} \setminus (N \cup E)) \implies$$ because $$S$$ consists of points in $$N \cup E$$, $$N_{m}^{\prime} = (O_{m}^{\prime} \cap (N \cup E))$$ is open in $$S$$. If $$N \in \mathbb{R}, N=\bigcap\limits_{m=1}^{\infty} O_{m}^{\prime}=\bigcap\limits_{m=1}^{\infty} (O_{m}^{\prime} \cap (N \cup E))=\bigcap\limits_{m=1}^{\infty} N_{m}^{\prime}= N^{\prime}$$, so $$N^\prime$$ (denoted as $$N^{\prime}$$ to indicate it refers to $$S$$) can be expressed as intersection of a countable number of open sets in $$S$$. Denote $$E^{\prime}$$ to indicate it refers to $$S$$. Assume $$N^{\prime} = \bigcup\limits_{m=1}^{\infty} T_{m}, T_{m}$$ closed $$\implies (N^{\prime})^\complement = E^{\prime} = (\bigcup\limits_{m=1}^{\infty} T_{m})^\complement = \bigcap\limits_{m=1}^{\infty} E_{m}^{\prime}, \forall m \in \mathbb{N}, E_{m}^{\prime}$$ open. Both $$N$$ and $$E$$ are dense in $$S$$ since $$\forall n \in N$$ and $$\forall \epsilon > 0, \exists e \in E \mid e \in B_{\epsilon}(n) \implies$$ since space $$S$$ is complete, $$\overline{E} = (N \cup E) \implies \forall m \in \mathbb{N}, E_{m}^{\prime}$$ is dense in $$S$$. Similarly, $$\forall e \in E$$ and $$\forall \epsilon > 0, \exists n \in N \mid n \in B_{\epsilon}(e) \implies$$ since space $$S$$ is complete, $$\overline{N} = (E \cup N) \implies \forall m \in \mathbb{N}, N_{m}^{\prime}$$ is dense in $$S$$. Therefore we have $$E^{\prime} \cap N^{\prime} = (\bigcap\limits_{m=1}^{\infty} E_{m}^{\prime}) \cap (\bigcap\limits_{m=1}^{\infty} N_{m}^{\prime}) = \emptyset$$, but by Baire Category Theorem, countable intersection of dense open sets is a proper subset of a complete metric space. So $$N^{\prime}$$ is not a union of a countable number of closed sets in $$S$$. Therefore $$N$$ is not a union of a countable number of closed sets in $$\mathbb{R}$$. Proof: assume $$N = \bigcup\limits_{m=1}^{\infty} N_{m}, \forall m \in \mathbb{N} N_{m}$$ closed. Then every $$N_{m} \subset N \implies N_{m} \subset N^{\prime}$$ in $$S$$. Assume $$N_{m}^{\prime}$$ is not closed in $$S$$, then $$\exists (n_{k}) \rightarrow L \notin N_{m}^{\prime} \implies$$ since $$N_{m}$$ and $$N_{m}^{\prime}$$ consist of the same elements, $$(n_{k}) \rightarrow L \notin N_{m} \implies N_{m}$$ is not closed which is a contradiction. Hence $$N_{m}^{\prime}$$ is closed in $$S$$ and $$N^{\prime} = \bigcup\limits_{m=1}^{\infty} N_{m}^{\prime}$$ is a union of a countable number of closed sets in $$S$$ which contradicts our proof above. Therefore the assumption is wrong and $$N$$ cannot be expressed as a union of a countable number of closed sets in $$\mathbb{R}$$. Therefore $$N \cup C \cup B \cup A \neq [0,1]$$ and $$[0,1]$$ is not a countable union of disjoint closed sets.
• Uhh.... "here is my proof is it correct" kind of questions are generally not allowed and as you have noted your proof is very very long. However, if there is a specific part in the proof that you are unsure of we would be glad to help :). Sep 16 at 20:31
• Your post is quite complicated. In general one can use Sierpinski's theorem on continua Sep 16 at 20:33
• Spaces which can't be non-trivially partitioned into countable amount of disjoint closed sets are called $\sigma$-connected by some authors (e.g. Nadler). Sierpiński's theorem on continua says that all continua are $\sigma$-connected. One can summarize the statement you're proving by "$[0, 1]$ is $\sigma$-connected". Sep 16 at 20:36
• To get $\{$ and $\}$ in MathJax type \{ and \} (you are typing \\{ and \\}) Sep 17 at 20:14
• For a shorter proof, see the answer to this question: math.stackexchange.com/questions/3800398/…
– bof
Sep 19 at 3:11