# Cantor's Back-and-Forth method

I'm interested in Cantor's back-and-forth method so I searched online and surprisingly it was on the Wikipedia.

My question is about the paragraph that starts saying: If we iterated only step (1), rather than going back-and-forth...

For me, it is not clear why we can avoid step 2 in the case of unbounded dense totally ordered sets, why is it posible to choose $j$ "as small as posible"? It is because our sets are well-ordered? or why is it? I hope someone can help me with this :)

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Am inclined to agree. The paragraph you mention is fuzzy. The well-ordering of the two sets is arbitrary, and has nothing to do with the orderings we are proving isomorphic. My feeling is that given any fully specified purely "forth" argument, I could produce a lisitng of the sets that would make surjectivity fail. –  André Nicolas Apr 18 '12 at 23:14
Your recomendation is to use the full Back-and-Forth method? –  luciana Apr 19 '12 at 3:30
Yes, for proving isomorphism it is a very useful idea. Shows up also in proving that if $A$ has same cardinality as a subset of $B$, and vice-versa, then $A$ and $B$ have the same cardinality. Also useful in Model Theory. –  André Nicolas Apr 19 '12 at 3:31
ok I would study this method a little bit more, thanks a lot :) –  luciana Apr 19 '12 at 3:36
Depends on your field of work. You may only need it for the one isolated purpose of proving Cantor's Theorem about densely ordered countable sets. –  André Nicolas Apr 19 '12 at 3:39

Let the "only-forth" method be applied by traversing the enumeration of $A$ and pairing each $a_i$ with the $b_j$ that has the least index $j$ of all elements of $B$ compatible with the current set of constraints. This is well-defined since, as the Wikipedia article explains, there is always at least one such element of $B$, and every non-empty set of indices has a least element.
At the $n$-th stage of the construction, after $n$ pairs have been chosen, the unpaired values in $A$ are divided into $n+1$ (countably infinite) sets $A_{nm}$ ($n-1$ between the paired values and $2$ at either end), and $B$ is likewise divided into $n+1$ (countably infinite) sets $B_{nm}$. If an element of $A_{nm}$ is to be paired next, the partner can be arbitrarily chosen from the corresponding set $B_{nm}$.
A given element $b_k$ of $B$ is always in one of these sets $B_{nm}$ until it gets paired. Since there is always a corresponding set $A_{nm}$ of as yet unpaired elements of $A$ and all elements of $A$ eventually get paired, an element $b_k$ that never gets paired would experience infinitely many choices being made from its respective current $B_{nm}$. But this is impossible, since in each case the element of $B_{nm}$ with least index is chosen, and there are only finitely many indices less than $k$.