I just finished sheet rocking my living room and am stuck at a stand-still.  My ceiling is 12' feet high at the tallest point and slopes down to 8'.  That said, at the 12' wall, the sheetrock only goes up 8' and is then met by a lap and gap type wood, which then continues up to the ceiling. The wood then slopes (on the wall). At the 8' point of the ceiling, the sheetrock butts up to the ceiling.  

My question is...How would I install a crown moulding or transition piece at the point where the sheetrock meets the wood?  The previous owners had a painted 1x4 where the paneling (now sheetrock) met the wood. However, I am wanting something a bit nicer than just a 1x4 transition.

Either I am the only person out there with a room like this or I am searching incorrectly on Google.  Does anyone have any ideas or pictures on this? I can definitely provide pictures if needed. I am tempted to pull the wood all the way to ceiling and continue the sheetrock all the way up. Any advice would be much appreciated!


Question: On surjections, idempotence and axiom of choice
Answer: The answer is no. 

First, I argue that it is consistent with ZF that a Dedekind
finite set $A$ can map onto $A\times 2$, and much more.

To see this, begin with any infinite Dedekind finite set $B\subset 2^\omega$, which is furthermore dense in the sense that any finite binary sequence has extensions in $B$. It is consistent with ZF that such a dense set $B$ exists, since in fact the usual symmetric-model arguments produce infinite Dedekind-finite sets that are dense.

Let $A$ consist of the finite non-repeating sequences from
$B$. Note that $A$ is still Dedekind finite, since any countably
infinite subset of $A$ can be used to produce a countably infinite
subset of $B$. Moreover, I claim that $A$ surjects onto
$A\times 2$ and indeed, onto $A^{{\lt}\omega}$. 

To see that $A$ surjects onto $A\times 2$, given $a=\langle b_0,\ldots,b_n\rangle\in A$, let $j$ be the first bit of $b_0$ and define $f(a)=(\langle b_1,\ldots,b_n\rangle,j)$. This is onto, since given any $(\langle b_1,\ldots,b_n\rangle,j)$, we just adjoin $b_0$ starting with digit $j$ to form $a=\langle b_0,\ldots b_n\rangle$, which maps to the original pair. 

For fun, let me show somewhat more, namely, that $A$ actually surjects onto $A^{{\lt}\omega}$. I shall define a function $f:A\to A^{{\lt}\omega}$ as follows. Suppose that we are given $a=\langle
b_0,\ldots,b_n\rangle\in A$. In order to define $f(a)$, we look at a certain finite initial segment of $b_0$, which we take to code a number $k$ and maps $\pi_i:n_i\to n$ for $i\lt k$. This can be coded in some canonical way, whose details are not important. (For example, perhaps $b_0$ starts with $k$ many $0$s, and after this there are $k$ blocks of $1$s, with the $i^{th}$ block of length $n_i$, and after this the bits are given to define the maps $\pi_i:n_i\to n$.) We use these maps to assemble $f(a)$ from the rest of the reals $b_1,\ldots,b_n$. Specifically, let $f(a)=\langle
\vec x_0,\ldots,\vec x_k\rangle\in A$, where $\vec x_i=\langle
b_{\pi_i(0)},\ldots,b_{\pi_i(n_i-1)}\rangle$. That is, each $\vec x_i$ enumerates a subset of $b_1,\ldots b_n$ in the order specified by $\pi_i$. In summary, a finite part of $b_0$ tells us how to assemble $f(a)$ according to a definite procedure from the other reals $b_1,\ldots,b_n$ appearing in $a$. (And if $b_0$ happens not to code things correctly, then we default to some constant value.) This defines $f:A\to A^{{\lt}\omega}$ without using the axiom of choice.
Furthermore, the map is surjective, since for any finite sequence
of injective tuples $\langle \vec x_0,\dots,\vec x_k\rangle$ from $B$, we may consider the reals appearing in those tuples and enumerate them $b_1,\ldots,b_n$, deleting repetitions, and then assemble a suitable $b_0$, using the fact that $B$ is dense in order to know that our desired collection of maps $\pi_i$ for $i\lt k$, which is coded by some finite binary sequence, can be extended to an element $b_0\in B$. It follows that $f(b_0,b_1,\ldots,b_n)$ is exactly the desired sequence of tuples. This surjectivity argument does not use the axiom of choice. 

In summary, $A$ surjects onto $A\times 2$ and even $A^{{\lt}\omega}$, but it is not bijective with $A\times 2$ or indeed with any superset of $A$, since it is Dedekind finite.

Thus, one cannot deduce in ZF that $A$ is bijective with $A\times 2$, just from knowing that it surjects onto $A\times 2$. This is true even when $A$ is a set of reals, since the example provided above has a bijection to a set of reals.