Suppose that $(L,\leq\_L,0,1)$ is a distributive and complemented Lattice that is *dense* as an order (i.e. if $a<\_L b\in L$ then there exists $x\in L$, s.t. $a<\_L x<\_L b$)


Suppose that there exists $U,V\subset L\setminus \left\{0\right\}$ s.t. 


1)$(U,\leq\_L)$ and $(V,\leq\_L)$ are total ordered set


2)$\inf (U\cup V)=0$


3)$\forall a,b\in U\cup V,\, \inf (a,b)\ne 0$


We also suppose that $|V|$ and $|U|$ are regular cardinals. 



> 
> Does there exists $T\subset L\setminus \left\{0\right\}$ totally ordered by $\leq\_L$, s.t. $\inf T=0$ and $|T|\leq |U\cup V|$
> 
> 
> 


Motivation :


If we suppose that there exists $C=\left\{c\_i,i<|C|\right\}\subset L\setminus \left\{0\right\}$ such that $|C|$ is regular and s.t. $\inf (C)=0$ and $\forall a,b\in C,\, \inf(a,b)\ne 0$, then we can do the hypothesis that $C=U\cup V$ with $U$ and $V$ totally ordered by $\leq\_L$. To get this I suppose that I give $C$ an ordinal indexation and I "try" to built a totally ordered set $U=\left\{u\_i,i<\alpha\right\}$, by taking $u\_0=c\_0$, $u\_{i+1}=\inf(u\_i,c\_{i+1})$ and I chose $u\_i\leq\_L u\_k$ for any $k<i$ if $i$ is a *limit* ordinal. In this last case, I have to be careful that $\inf(u\_i,c)\ne 0$ for any $c\in C$, and if it is **not possible** ... then then one can build a $V'=\left\{v'\_k\in L,\,k<\alpha,\, \forall u\in U,\, v'\_k\leq\_L u \right\}$, s.t $v'\_i\leq\_L v'\_k$ iff $i<k$, and s.t no lower bound of $U$ is greater than every member of $V'$. (I don't develop details to make it short but it is an easy consequence of the **not possible** case and of the regularity of $|C|$...) Then the set of complements of members of $V'$ that is $V=\left\{(c\_i)^c,i<\alpha\right\}$ is such that $V\cup U$ satisfy 2) and 3), and of course $|\alpha|=|C|$


So the resolution of the *question* would get a single tower from the two towers get from $C$...



getting one tower from two
This is a fantastic question! I spent the whole morning thinking
about it, and I finally have a solution.


The answer is no, not necessarily.


To build a counterexample, I claim first that there is a linear
order $\langle\ell,<\rangle$ with the following properties:


* $\ell$ is a dense linear order.
* $\ell$ has no cuts of type $(\omega,\omega)$ or
$(\omega\_1,\omega\_1)$. Specifically, if $$a\_0<a\_1<a\_2<\cdots\qquad\qquad\cdots<b\_2<b\_1<b\_0$$ is an ascending $\omega$-sequence with descending $\omega$-sequence above, then there is a nontrivial interval of points strictly between, and the same with increasing/decreasing $\omega\_1$-sequences on each side.
* every point in $\ell$ has type $(\omega\_2,\omega\_2)$, meaning that the cofinality and coinitiality of approaching the point from below or from above is $\omega\_2$.
* there is a unfilled gap of type $(\omega,\omega\_1)$.
$$a\_0<a\_1<a\_2<\cdots\quad )(\quad\cdots <b\_\alpha<\cdots<b\_2<b\_1<b\_0.$$


A cut in $\ell$ (sometimes called a pre-gap) has type
$(\kappa,\lambda)$, if it is determined by an increasing $\kappa$
sequence and decreasing $\lambda$ sequence above, with at most one
element in $\ell$ filling the cut, that is, above the lower sequence and below the upper sequence. A *gap* is an unfilled cut.


We can build such a linear order $\ell$ as follows. Let
$\text{No}\_{\omega\_2}$ be the surreal line built at birthday
stages below $\omega\_2$. That is, you build a linear order in $\omega\_2$ many steps, filling every cut at every stage. In particular, every cut of type $(\alpha,\beta)$ with
$\alpha,\beta<\omega\_2$ will be filled in this order at the first stage after all of its elements are born. So every $(\omega,\omega)$ cut and every
$(\omega\_1,\omega\_1)$ cut is filled again and again, giving a nontrivial interval inside. This order has no unfilled
gaps of type $(\omega,\omega\_1)$, but it does have unfilled gaps of
type $(\omega,\omega\_2)$ and others of type $(\omega\_2,\omega\_1)$, since one can fix an increasing $\omega$ sequence and a decreasing $\omega\_1$-sequence, and then the corresponding cuts will keep getting additional points up to $\omega\_2$. By
excising the region between such cuts, and bringing the left and right sides together, we can create artificially an
unfilled gap of type $(\omega,\omega\_1)$, and there will
be exactly one. This is my order $\ell$. (We could also have built it by starting with points for the desired unfilled cut of type $(\omega,\omega\_1)$, and then proceeding as in the surreals to fill all other cuts in $\omega\_2$ many steps, but never filling that cut.)


By design, the order $\ell$ has an unfilled cut determined by an
increasing $\omega$-sequence $a\_n$ and decreasing
$\omega\_1$-sequence $b\_\alpha$. 
$$a\_0<a\_1<a\_2<\cdots\quad)(\quad\cdots<b\_\alpha<\cdots<b\_2<b\_1<b\_0.$$


Let $\mathbb{B}$ be the interval algebra of $\ell$, the atomic Boolean
algebra consisting of the finite unions of intervals in $\ell$. And consider $\mathbb{B}$ modulo atoms, so that two intervals are the same if they differ by only finitely many points. Thus, we have essentially the interval algebra, but where one does not care about endpoints of intervals, and an open interval counts as equivalent to half-open or closed. Since $\ell$ is dense, this makes the quotient modulo atoms atomless, or dense in the terminology of the question.


Let $U$ consist of the intervals $(a\_n,b\_0)$ for $n<\omega$ and let $V$ consist of the
intervals $(a\_0,b\_\alpha)$ for $\alpha<\omega\_1$. These are each
linearly ordered and compatible in the interval algebra 
$$(a\_n,b\_0)\wedge (a\_0,b\_\alpha)=(a\_n,b\_\alpha)\neq \emptyset$$ because each of these intervals jumps over to the other side of the unfilled cut, giving them a nontrivial intersection. But since the intersection of all these intersections is empty, as
the cut is unfilled, it follows that $\inf U\cup V=0$.


Meanwhile, I claim that there can be no linearly ordered descending sequence of size at most $\omega\_1$ with infimum $0$, because any descending nested
$\omega$-sequence or $\omega\_1$-sequence of intervals will have a
nonempty intersection, because of the property that every nested sequence of intervals of size $\omega$ or $\omega\_1$ has nontrivial intersection. Thus, there is no $T$ of size
$\omega\_1$ as you requested.


The essence of the argument is to make the cofinalities of the two towers different, but the single linear tower would have to use just a single cofinality, making it impossible.


