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?

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2019-05-18 22:39:17
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Answers: 3

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 $\frac{1}{2}$.

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2019-05-30 12:38:11
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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/

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2019-05-21 09:57:31
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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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2019-05-21 07:24:44
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