Can you express $[0,1]$ as a countable disjoint union of closed sets, other than the trivial way of doing this?
Is $[0,1]$ a countable disjoint union of closed sets?
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3Whats the trivial way. – 2010-10-08
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5@Chandru1: I assume he's asking us for the sets to be proper and nonempty. – 2010-10-08
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0@Andy: it is impossible with open intervals. A countable union of open intervals is open. – 2010-10-08
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3No continuum can be written as a countable union of disjoint closed sets. – 2010-10-08
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0Non-trivially that is. – 2010-10-08
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1See also [this MO thread](http://mathoverflow.net/questions/48970) – 2011-12-29
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3@anonymous : the trivial way is using one set: $[0,1]$ itself. – 2013-11-20
7 Answers
The answer is no. In fact, as Steve D said, we have a theorem that holds for a wide class of spaces, which includes closed intervals, circles, balls and cubes. It was proved by Sierpiński in $1918$ $[1]$. You can find the proof in the book "General Topology" by Ryszard Engelking, but I'll post here since it's not easy to find it online. First a definition: a topological space is called a continuum if it is a compact connected Hausdorff space. The precise statement is the following:
Theorem (Sierpiński). If a continuum $X$ has a countable cover $\{X_i\}_{i=1}^{\infty}$ by pairwise disjoint closed subsets, then at most one of the sets $X_i$ is non-empty.
In order to prove this we'll need the following lemmas:
Lemma $1$. Let $X$ be a continuum. If $F$ is a non-trivial closed subset of $X$, then for every component $C$ of $F$ we have that $\text{Bd}(F) \cap C$ is non-empty.
Proof. Let $x_0$ be in $C$. Since $X$ is Hausdorff compact, quasicomponents coincide with components, so $C$ is the intersection of all open-closed sets in $F$ which contain $x_0$. Suppose that $C$ is disjoint from $\text{Bd}(F)$. Then, by compactness of $\text{Bd}(F)$, there is one open-closed set $A$ in $F$ containing $x_0$ and disjoint from $\text{Bd}(F)$. Take an open set $U$ such that $A = U \cap F$. Thus the equality $A \cap \text{Bd}(F) = \emptyset$ implies that $A = U \cap \text{Int}(F)$, so $A$ is open in $X$. But $A$ is also closed in $X$, and contains $x_0$, so $A=X$. But then $\text{Bd}(F) = \emptyset$, which is not possible since $F$ would be non-trivial open-closed in $X$. $\bullet$
Lemma $2$. If a continuum $X$ is covered by pairwise disjoint closed sets $X_1, X_2, \ldots$ of which at least two are non-empty, then for every $i$ there exists a continuum $C \subseteq X$ such that $ C \cap X_i = \emptyset$ and at least two sets in the sequence $C \cap X_1, C \cap X_2, \ldots$ are non-empty.
Proof. If $X_i$ is empty then we can take $C = X$; thus we can assume that $X_i$ is non-empty. Take $j \ne i$ such that $X_j \ne \emptyset$. Since $X$ is Hausdorff compact, there are disjoint open sets $U,V \subseteq X$ satisfying $X_i \subseteq U$ and $X_j \subseteq V$. Let $x$ be a point of $X_j$ and $C$ the component of $x$ in the subspace $\overline{V}$. Clearly, $C$ is a continuum, $ C \cap X_i = \emptyset$ and $ C \cap X_j \ne \emptyset$. By the previous lemma, $C \cap \text{Bd}( \overline{V}) \ne \emptyset$ and since $X_j \subseteq \text{Int}(\overline{V})$, there exist a $k \ne j$ such that $C \cap X_k \ne \emptyset$. $\bullet$
Now we can prove the theorem:
Proof. Assume that at least two of the sets $X_i$ are non-empty. From lemma $2$ it follows that there exists a decreasing sequence $C_1 \supseteq C_2 \ \supseteq \ldots$ of continua contained in $X$ such that $C_i \cap X_i = \emptyset$ and $C_i \ne \emptyset$ for $i=1,2, \ldots$ The first part implies that $\bigcap_{i=1}^{\infty} C_i = \emptyset$ and from the second part and compactness of $X$ it follows that $\bigcap_{i=1}^{\infty} C_i \ne \emptyset$. $\bullet$
The Hausdorff hypothesis is fundamental. For example, consider $X$ a countable infinite set with the cofinite topology. Then $X$ is compact, connected and a $T_1$-space. However, we can write $X$ as a disjoint union of countable singletons, which are closed.
$[1]$ Sierpiński, W: Un théorème sur les continus, Tôhoku Math. J. 13 (1918), 300–305.
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0Is Lemma 1 false if X=F? Do we really need to know X is Hausdorff in Lemma 1? Assume F is a proper subset of X. We need only know the general topology facts that the closure of a connected subspace is connected, and that the components of an open subspace are open. If C is a subspace of int(F), then C is a subspace of exactly one component U of int(F) (since C is connected). The closure of U is strictly larger than U (since X is connected). Thus U closure is a strictly larger connected subpace containing C, and we have a contradiction. – 2016-09-21
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1@Nuno, I can understand lemma 1 and lemma 2,but how to construct a decreasing $C_n$ having the desired property. Can you give a detailed construction, – 2017-08-06
I actually was just challeneged to do this problem today, and thought I would post my solution here for others to use in the future, since it seems like a typical kind of question. Nuno's answer requires more background and machinery, while a more straight-forward approach is perhaps desired.
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2Unless I've misunderstood something, you only prove that $[0,1]$ is not a non-trivial countable union of disjoint closed *intervals*. – 2014-09-22
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0Hey Taylor. Thanks for this comment. Do you happen to have a reference for someone who takes the last of your mentioned approaches? I can't seem to find one, and I'm curious about the details. Or do you have a hint as to how to construct the homeomorphism there? I can see how your set $S$ is homeomorphic to $C$, but I'm not seeing $(0,1)$ yet. – 2015-09-09
I thought a bit about this problem and wanted to clarify some misconceptions.
Suppose that $[0,1]=\cup_i F_i$. A popular approach (that appears fruitless) is to construct a compact set $[0,1]\supset K = \cup_i \partial F_i $, and then use the Baire Category Theorem to finish.
However, note that you can replace the set $[0,1]$ with any compact set $K'$ and leave the rest of the proof unchanged to "prove" the same result for any compact set. But this is absurd, take $K' = \{0,1\} = \{0\} \cup \{1\}$.
So where is the flaw? The sets $\partial F_i$ are nowhere dense, as elements of the space $([0,1],\tau)$. But they need not be nowhere dense as elements of the subspace topology $\left(K,\tau_{|K}\right)$.
@khaghans, I'm looking at you ಠ_ಠ
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0Are you asking or answering? – 2012-06-28
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0I am pointing out that one of the proofs posted here is incorrect, it would be nice to put some warning by that post. I don't want others to stumble on this, as I did at first. Of course, if I am making a mistake and the proof indeed holds, please explain the error in my ways. That's all. – 2012-07-20
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0For the $[0,1]$ case, you can show that none of the boundaries has nonempty interior relative to the union of the boundaries. It's not terribly hard to do that. Had khaghans done that, the proof would have been correct. – 2013-12-20
I have a different proof. Consider an arbitrary covering of $(0, 1)$ by pairwise disjoint, closed intervals $I_x$, each $I_x$ is in $(0, 1)$. Denote the covering set by $L$. One can show that $L$ is a linear continuum and hence uncountable.
Since the closed intervals are pairwise disjoint, there is a natural total order amongst them. Since $L$ covers $(0, 1)$, for any $I_x < I_y$ in $L$, there exists an $I_z$ in $L$ such that $I_x < I_z < I_y$.
It remains to show that $L$ has the least upper bound property. Suppose a subset of $L, S$, is bounded above by an $I_x$. There must exist a least upper bound of the right endpoints of the intervals in this subset. Call it $a$. Obviously $a$ is either to the left of $I_x$ or in $I_x$. Hence $a$ is in $(0, 1)$. If $a$ is the right endpoint of an interval in $S$, that interval is the least upper bound of $S$. Otherwise, there exists an $I_y$ in $L - S$ that covers $a$. Since $a$ is the least upper bound of the right ends of intervals in $S$, a must be the left endpoint of $I_y$. Hence $I_y$ is the least upper bound. Therefore $L$ has the least upper bound property.
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0Note I asked about closed sets, not closed intervals. – 2015-05-02
No. $[0,1]$ cannot be written as the union of countable disjoint closed intervals . Try to use the Baire Category theorem. You may also refer this post http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/
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0Note that I asked about closed sets, not closed intervals. – 2010-10-08
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0@Kevin: Aren't closed sets closed intervals? – 2010-10-08
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0@Chandru: it depends on the topology you have on R. @Kevin: you should clarify that...i.e. with the usual topology on R (the one induced by the euclidean metric) closed sets are only closed intervals. (Or a finite uninion of them) – 2010-10-08
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0@Andy: Absolutely! – 2010-10-08
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9@Andy: so you're saying the Cantor set is a finite union of closed intervals?? – 2010-10-08
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3@Andy,@Chandru: It should be fair to assume here that Kevin is talking about the usual topology on R. One way to characterize closed sets is as complements of countable unions of open intervals, but being closed basically just means containing all limit points. This is far from the same as being finite unions of closed intervals. To add an example to Pete's, consider {0,1,1/2,1/3,1/4,...}. – 2010-10-08
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0@Pete: No no no, my comment was only about the result that "a finite union of closed sets is closed"...that's what I meant to say anyway...I'm sorry if I didn't state it clearly – 2010-10-08
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1Or the set {1,2,...}. – 2010-10-08
The answer to the question as stated is no, as others have explained. However, if we relax the hypothesis from disjoint to non-overlapping, then the answer is yes.
Two intervals $I_1$ and $I_2$ are non-overlapping if $I_1^{\circ}\cap I_2^{\circ}=\emptyset$; that is, if their interiors are disjoint. If the intervals are closed and non-overlapping, then they intersect at most in their boundaries. For example, in $\mathbb{R}$, the intervals $\left[0,\frac{1}{2}\right]$ and $\left[\frac{1}{2},1\right]$ are non-overlapping, but clearly not disjoint as they share the point $\frac{1}{2}$.
A partial solution:
As Sargera began, if such a cover exists, it must be countably infinite. Let $[0,1]=\displaystyle\bigcup_{n=1}^{\infty}[a_n, b_n] = \displaystyle\bigcup_{n=1}^{\infty}I_n$ By the axiom of choice, construct $(x_n)$ by $x_n \in [a_n, b_n].$ Since $[0,1]$ is compact, there exists $(x_{n_k})$ such that $x_{n_k} \to x\in [0,1].$ Then there exists unique $s\in \mathbb{N}$ such that $x\in [a_s, b_s].$ If this interval is non-degenerate then corresponding to $\varepsilon = \frac{1}{2}\min\{x-a_s, b_s-x\}$ there exists $m \in \mathbb{N}$ such that $|x_{n_k}-x|< \varepsilon$ for $k \geq m.$ In particular $x_{n_{m}} \in (a_s, b_s)\subset [a_s, b_s] = I_s$ and since $x_{n_{m}} \in I_{n_{m}}$ by construction, we obtain a contradiction to the fact that the $I_n$ are disjoint.
If $I_s$ is degenerate, I haven't thought of a workaround yet. Maybe someone could suggest one.
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0Note the question is about closed *sets*, not closed intervals. – 2018-06-07
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0Oh good point, I misread that. I'll keep this answer until I find a more general approach. Any idea how to finish the proof in this set up when $I_s$ consists of a single point? – 2018-06-07