**Proof that a sequence of set has a set dense somewhere in $[a,b]$**without wasting too much if your time.

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Is the following proof correct?

**Proposition**: if we have a sequence of set $U_i$ such as $bigcup_{iin mathbb{N}} U_i=[a,b]$ 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 $forall ]u,v[$

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

Let’s define

$$leftlbrace begin{array} .E_{c,h}^i= ]c-h,c+h[ & text{ if } & U_i cap ]c-h,c+h[ =emptyset \ E_{c,h}^i= emptyset &text{ if } & U_i cap ]c-h,c+h[ neq emptyset end{array}right.$$

and

$$E_i = bigcup_{h>0} bigcup_{cin ]u,v[} E^i_{c,h}$$

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

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

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

$$forall epsilon >0 forall xin ]u,v[, exists cin E_i,quad cin ]x-epsilon,x+epsilon[$$

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

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

Contradiction.

## Answer

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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