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Is the following proof correct?

Proposition: if we have a sequence of set $U_i$ such as $\bigcup_{i\in \mathbb{N}} U_i=[a,b]$ then there exist a $i$ such as $U_i$ is dense somewhere in $[a,b]$

Proof: If for all $i, U_i$ is nowhere dense, we have $\forall ]u,v[$

$$\forall i, \exists c \in ]u,v[, \exists h > 0,\quad U_i \cap ]c-h,c+h[ =\emptyset\text{ and } ]c-h,c+h[ \subset ]u,v[$$

Let's define

$$\left\lbrace \begin{array} .E_{c,h}^i= ]c-h,c+h[ & \text{ if } & U_i \cap ]c-h,c+h[ =\emptyset \\ E_{c,h}^i= \emptyset &\text{ if } & U_i \cap ]c-h,c+h[ \neq \emptyset \end{array}\right.$$ and $$E_i = \bigcup_{h>0} \bigcup_{c\in ]u,v[} E^i_{c,h}$$

$E_i$ is open (as a reunion of open set), and dense in $]u,v[$. Indeed, we have as

$$\forall \epsilon >0 \forall x\in ]u,v[, \exists c \in ]x-\epsilon,x+\epsilon[,\exists h >0,\quad U_i \cap ]c-h,c+h[ = \emptyset$$

$$\forall \epsilon >0 \forall x\in ]u,v[, \exists ]c-h,c+h[ \subset U_i^C,\quad]c-h,c+h[\subset ]x-\epsilon,x+\epsilon[$$

$$\forall \epsilon >0 \forall x\in ]u,v[, \exists c\in E_i,\quad c\in ]x-\epsilon,x+\epsilon[$$

So $E_i$ is an open dense set for all i. By Baire theorem, we get that $\bigcap_{i\in\mathbb{N}} E_i$ is dense in ]u,v[. As we have $\forall i, E_i\subset U_i^C$, we get that $\bigcap_{i\in\mathbb{N}} E_i \subset \bigcap_{i\in\mathbb{N}} B^C_i$ and $\bigcap_{i\in\mathbb{N}} B_i^C$ is dense in $]u,v[$. But we also have

$$\bigcap_{i\in\mathbb{N}} \left(B_i^C \cap ]u,v[\right) = ]u,v[ \cap \left( \bigcup_{i\in\mathbb{N}} B_i \right)^C = ]u,v[^C \cap ]u,v[= \emptyset$$

Contradiction.

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  • $\begingroup$ I think you can just say that by Baire's Category Theorem, it is impossible that all $U_i$ are nowhere dense. Thus, some $U_j$ satisfies $\overline{U_j}^\circ \neq \emptyset$. $\endgroup$ – PhoemueX Feb 23 '15 at 13:38
  • $\begingroup$ Oh, yes, I didn't have this version of Baire's Theorem in mind. Thanks you ! $\endgroup$ – Tryss Feb 23 '15 at 14:46
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There is an easier proof (also by contradiction). Let us assume that [a,b] is the union of countable number of nowhere dense sets. By a version of Baire theorem, the countable union of nowhere dense sets cannot be complete. Interval [a,b] is a complete space. Contradiction.

P.S. More exactly, the sets which can be represented as a countable union of nowhere dense sets, are sometimes called the sets of the 1st category. Other sets are 2nd category. A theorem says that complete spaces are the sets of 2nd category.

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