Back-and-Forth Argument vs. "One-Way" Argument The wikipedia article on the Back and Forth Argument claims at the end:

If we iterated only step $(1)$, rather than going back and forth, then in some cases the resulting function from A to B would fail to be surjective. In the easy case of unbounded dense totally ordered sets it is possible to avoid step $2$ by choosing the element $b_j$ more carefully (by choosing $j$ as small as possible), but this does not work for more complicated examples such as atomless Boolean algebras where steps $1$ and $2$ are both needed.

The emphasis is my own. This "easy case" is exactly the same thing stipulated in the suppositions:

  
*
  
*$(A, ≤_A)$ and $(B, ≤_B)$ are linearly ordered sets;
  
*They are both unbounded, in other words neither A nor B has either a maximum or a minimum;
  
*They are densely ordered, i.e. between any two members there is another;
  
*They are countably infinite.
  

I haven't learnt anything about atomless boolean algebras. So can someone explain why the latter method cannot be applied to them?
More info: Silver, Charles L. "Who invented Cantor's back-and-forth argument?"
 A: Let me first explain why the argument using only step 1 does work for dense linear orders.  Given two countable dense linear orders $A=\{a_n\}$ and $B=\{b_n\}$, we define $f:A\to B$ as follows.  Having defined $f(a_0),\dots,f(a_{n-1})$, the rest of the set $B$ is divided into $n+1$ intervals by comparing with these elements $f(a_0),\dots,f(a_{n-1})$.  The set $A$ is also split into $n+1$ intervals, by comparing with the elements $a_0,\dots,a_{n-1}$.  The element $a_n$ lies in one of these intervals in $A$, which corresponds to one of the intervals in $B$.  We define $f(a_n)$ to be $b_j$ for the least $b_j$ that is in that corresponding interval of $B$.
Now, why is this surjective?  Well, we clearly have $f(a_0)=b_0$.  Now say $b_1>b_0$.  Then as soon as we reach an element of $A$ which is greater than $a_0$, that element will map to $b_1$.  Now say $b_0<b_2<b_1$.  As soon as we reach an element of $A$ which is between $a_0$ and $f^{-1}(b_1)$, that element will map to $b_2$.  And so on.  Once we have already found that $b_0,\dots,b_{j-1}$ are in the image of $f$, $b_j$ is in one of the intervals determined by the values of $f$ defined so far.  Eventually we must find some element of $A$ which is in the corresponding interval of $A$ (since $A$ is dense), and that element will map to $b_j$.
The key thing that makes this work is that the condition which guarantees an element of $A$ will map to $b_j$ doesn't change.  We have an entire interval in $A$ which will map to $b_j$ as soon as we pick an element of that interval.  And even if we define $f$ on other values in the meantime, the interval that needs to map to $b_j$ doesn't change.  When you define a new value $f(a_n)$, that only splits up one of the intervals of $A$ (namely the one that $a_n$ was in), and the rest don't change.
When you make the analogous argument for atomless Boolean algebras, however, the condition for an element of $A$ to map to any particular $b_j$ is constantly changing.  That is, with each new value $f(a_n)$ which we define, every "interval" of $A$ determined by the values we have defined so far gets split up.  This means that we might go through the entire process of defining $f$ without ever picking an element of $A$ which is in the correct "interval" for $b_j$, since that correct "interval" shrinks at every step.  As a result, the version of the argument using only step 1 won't work for atomless Boolean algebras, since you can't guarantee that $f$ will be surjective.
[Explicitly, the elements $a_0,\dots,a_{n-1}$ split a Boolean algebra $A$ into "intervals" as follows.  Given an element $x\in A$, consider the set $S$ of elements $s$ of the subalgebra generated by $a_0,\dots,a_{n-1}$ such that $x\geq s$ and the set $T$ of $t$ such that $x\leq t$.  We then partition $A$ into pieces, where in each piece the values of these sets $S$ and $T$ are fixed: these are what I referred to as the "intervals".]
