# Is $[0,1]$ a countable disjoint union of shut collections?

Can you share $[0,1]$ as a countable disjoint union of shut collections, apart from the unimportant means of doing this?

The response to the inquiry as mentioned is no, as others have actually clarified. Nonetheless, if we loosen up the theory from disjoint to non - overlapping, after that the solution is of course.

2 periods $I_1$ and also $I_2$ are non - overlapping if $I_1^{\circ}\cap I_2^{\circ}=\emptyset$ ; that is, if their *insides * are disjoint. If the periods are shut and also non - overlapping, after that they converge at the majority of in their borders. As an example, in $\mathbb{R}$, the periods $\left[0,\frac{1}{2}\right]$ and also $\left[\frac{1}{2},1\right]$ are non - overlapping, yet plainly not disarrange as they share the factor $\frac{1}{2}$.

No. $[0,1]$ can not be created as the union of countable disjoint shut periods. Attempt to make use of the Baire Category theory. You might additionally refer this blog post http://terrytao.wordpress.com/2010/10/04/covering-a-non-closed-interval-by-disjoint-closed-intervals/

The solution is no. Actually, as Steve D claimed, we have a theory that holds for a vast class of rooms, that includes shut periods, circles, rounds and also dices. It was confirmed by Sierpiński in $1918$ $[1]$. You can locate the evidence in guide "General Topology" by Ryszard Engelking, yet I'll upload below given that it is hard to locate it online. First a definition: a topological room is called a continuum if it is a portable linked Hausdorff room. The specific declaration is the following:

**Theorem (Sierpiński). ** If a continuum $X$ has a countable cover $\{X_i\}_{i=1}^{\infty}$ by pairwise disjoint shut parts, after that at the majority of among the collections $X_i$ is non - vacant.

In order to confirm this we'll require the adhering to lemmas:

**Lemma $1$. ** Allow $X$ be a continuum. If $F$ is a non - unimportant shut part of $X$, after that for every single part $C$ of $F$ we have that $\text{Bd}(F) \cap C$ is non - vacant.

*Evidence *. Allow $x_0$ remain in $C$. Given that $X$ is Hausdorff portable, quasicomponents accompany parts, so $C$ is the junction of all open - shut embed in $F$ which have $x_0$. Intend that $C$ is disjoint from $\text{Bd}(F)$. After that, by density of $\text{Bd}(F)$, there is one open - shut set $A$ in $F$ having $x_0$ and also disjoint from $\text{Bd}(F)$. Take an open set $U$ such that $A = U \cap F$. Hence the equal rights $A \cap \text{Bd}(F) = \emptyset$ indicates that $A = U \cap \text{Int}(F)$, so $A$ is open in $X$. Yet $A$ is additionally enclosed $X$, and also has $x_0$, so $A=X$. Yet after that $\text{Bd}(F) = \emptyset$, which is not feasible given that $F$ would certainly be non - unimportant open - enclosed $X$. $\bullet$

**Lemma $2$. ** If a continuum $X$ is covered by pairwise disjoint shut collections $X_1, X_2, \ldots$ of which at the very least 2 are non - vacant, after that for every single $i$ there exists a continuum $C \subseteq X$ such that $ C \cap X_i = \emptyset$ and also at the very least 2 embed in the series $C \cap X_1, C \cap X_2, \ldots$ are non - vacant.

*Evidence *. If $X_i$ is vacant after that we can take $C = X$ ; hence we can think that $X_i$ is non - vacant. Take $j \ne i$ such that $X_j \ne \emptyset$. Given that $X$ is Hausdorff portable, there are disjoint open collections $U,V \subseteq X$ pleasing $X_i \subseteq U$ and also $X_j \subseteq V$. Allow $x$ be a factor of $X_j$ and also $C$ the part of $x$ in the subspace $\overline{V}$. Plainly, $C$ is a continuum, $ C \cap X_i = \emptyset$ and also $ C \cap X_j \ne \emptyset$. By the previous lemma, $C \cap \text{Bd}( \overline{V}) \ne \emptyset$ and also given that $X_j \subseteq \text{Int}(\overline{V})$, there exist a $k \ne j$ such that $C \cap X_k \ne \emptyset$. $\bullet$

Now we can confirm the theory:

*Proof *. Think that at the very least 2 of the collections $X_i$ are non - vacant. From lemma $2$ it adheres to that there exists a lowering series $C_1 \supseteq C_2 \ \supseteq \ldots$ of continua had in $X$ such that $C_i \cap X_i = \emptyset$ and also $C_i \ne \emptyset$ for $i=1,2, \ldots$ The first component indicates that $\bigcap_{i=1}^{\infty} C_i = \emptyset$ and also from the 2nd component and also density of $X$ it adheres to that $\bigcap_{i=1}^{\infty} C_i \ne \emptyset$. $\bullet$

The Hausdorff theory is basic. As an example, take into consideration $X$ a countable boundless set with the cofinite geography. After that $X$ is portable, linked and also a $T_1$ - room. Nonetheless, we can write $X$ as a disjoint union of countable singletons, which are shut.

$[1]$ Sierpiński, W: *Un théorème sur les continus *, Tôhoku Math. J. **13 ** (1918 ), 300-- 305.

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