Is $[0,1]$ a countable disjoint union of closed sets?

Can you express $[0,1]$ as a countable disjoint union of closed sets, other than the trivial way of doing this?

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.

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.

Proof. The first observation is that it doesn't matter whether this is proved for $[0,1),$ $(0,1)$, or $[0,1]$ (the latter case adding the restriction of non-trivial intervals). This is because if one assumes $[0,1)$ is the union of disjoint closed intervals, then one can simply remove the end-intervals to obtain a covering of an interval of the form $(a,b)$ $0<a<b<1$ by closed disjoint intervals. So assume that $$(0,1)=\bigcup\limits_{i=1}^{\infty}I_{i}.$$ First notice that $\{I_{i}\}$ cannot be finite, for the finite union of closed sets is closed; moreover, any two closed bounded sets have positive distance from each other, so a nesting argument shows that there must be at least countably many of them (this will be exploited later).

From this point, there are several ways to approach the problem. For example, define sets $$S_{j}=(0,1)-\bigcup\limits_{i=1}^{j}I_{i}.$$ Then we obtain a collection of open nested subsets $\{S_{j}\}_{j=1}^{\infty}\subset(0,1)$. If we can show $S=\bigcap_{j=1}^{\infty}S_{j}\neq\emptyset$, then we will have also shown that there must be at least one point in $(0,1)$ not covered by $\{I_{i}\}_{i=1}^{\infty}$, thus obtaining a contradiction. But showing directly whether a somewhat arbitrary countable intersection of open sets is empty or nonempty is difficult. So let's modify the $S_{j}$ to be $$S_{j}=(0,1)-\bigcup\limits_{i=1}^{j}I_{j}^{\circ}.$$ Then we obtain a collection of nested compact sets $\{S_{j}\}_{j=1}^{\infty}\subset[0,1)$. Cantor's intersection theorem then implies that $$S=\bigcap_{j=1}^{\infty}S_{j}\neq\emptyset.$$ Indeed, under our supposition, the intersection contains the end points of each $I_{i}$, or in other words, $$S=(0,1)-\bigcup\limits_{j=1}^{\infty}(a_{i},b_{i}=\{a_{i},b_{i}\}_{i=1}^{\infty}$$ where the $a_{i},b_{i}$ are the right and left endpoints, respectively, of each $I_{i}$. If we can then show $S$ is uncountable (for example, if $S$ was perfect), then we would obtain our desired contradiction, for the set of endpoints is clearly countable (this approach seems more tractable than trying to show there are some points in $S$ which are not endpoints of any $I_{i}$). But this is trivial. If $x\in S$, then $x=a_{i}$ or $x=b_{i}$ for some $i\in\mathbb{N}$. If $x=a_{i}$, then there exists a subsequence of intervals $[a_{i_{k}},b_{i_{k}}]$ to the left of $[a_{i},b_{i}]$ such that the right endpoints converge to $x$. The first assertion follows from $[0,1]-I_{i}$ being open, so that there is an interior interval of positive length between $(0,a_{i})$ in which to place a countably infinite number of closed intervals, and the second assertion is a simple consequence of the fact that two compact sets are always separated by a positive distance. Hence, if $[a_{i_{1}},b_{i_{1}}]$ is to the left of $[a_{i},b_{i}]$, then there exists intervals $[a_{i_{2}},b_{i_{2}}],\ldots,[a_{i_{N}},b_{i_{N}}]$ and positive numbers $\delta_{1},\delta_{2},\ldots,\delta_{N}$ such that $$a_{i_{1}}<b_{i_{1}}<a_{i_{2}}<b_{i_{2}}<\ldots<a_{i_{N}}<b_{i_{N}}<a_{i}<b_{i}$$ and $$0<a_{i}-b_{i{N}}=\delta_{N}<\ldots<a_{i}-b_{i_{2}}=\delta_{2}<a_{i}-b_{i_{1}}=\delta_{1}.$$ This process is clearly continuable, but more importantly, since $\delta_{k}>0$ for all $k=1,2,\ldots$, this process must continue indefinitely in order that $\{I_{i}\}_{i=1}^{\infty}$ cover $(0,1)$. And so, we obtain a sequence of distances $\delta_{k}\to0$ as $n\to\infty$ which implies $b_{i_{k}}\to a_{i}=x$ as $k\to\infty$. The same argument applies for if $x$ is a right endpoint $b_{i}$ of some $I_{i}$. Since every $x$ is the limit of a sequence in $S$, we see that $S$ is perfect. An application of the Baire Category theorem then shows $S$ is uncountable, for otherwise $S$ (being a closed perfect subset of a complete metric space, hence itself complete) is the countable union of singletons, which are no where dense, and therefore cannot be all of $S$. This is the desired contradiction, since it implies there are an uncountable number of $x\in(0,1)$ which are not covered by $\{I_{i}\}_{i=1}^{\infty}$. This completes the proof.

The above argument can be carried out without using the Baire Category theorem to show that there is at least one $x\in(0,1)$ which is not in the cover of $\{I_{i}\}_{i=1}^{\infty}$; this has the advantage of being straight-forward without having to construct an auxiliary set $S$ and then appeal to a sophisticated theorem on perfect subsets of complete metric spaces. On the other hand, it does not illuminate just how far short a countable disjoint cover of closed sets falls in attempting to cover $(0,1)$ (the Baire Category theorem shows it does not come anywhere close!). A completely different approach would rest on the observation that our auxiliary set $S$ has an identical structure to the Cantor set $C$, which is similarly the countable intersection of finite unions of open intervals removed from $[0,1]$. Therefore, if we can establish an onto map from $S$ to $C$, then we will have obtained the desired uncountability of $S$. We could also do a work-around from these facts, and obtain a contradiction by showing $(0,1)$ is topologically equivalent to $C$, but this seems unnecessary and pointless.

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

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.