

We shall use the axiom of choice to prove an extremely wimpy version of the Banach Tarski paradox, to wit:



Proof:

Use the axiom of choice to find a subset X of [0,1] which contains exactly

one element of each equivalence class. (Intuitively speaking, we are taking each

equivalence class in turn and picking one element from each, and throwing it into the

set X). This set is what logicians call "non-constructive"; there is no recipe

or formula for this set, because we can't specify exactly how we pick an element from

each equivalence class. All we know is that such a set exists by this abstract axiom.

Since X does not contain two or more elements from any equivalence class, we see

that the translates X + q are all disjoint for all rational numbers q.

In particular, the translates X + q for all q \in Q \cap [0,1] form a disjoint

partition of some subset Y of [0,2]. You can think of Y as all the "small"

rational translations of X.

Now the sets Q \cap [0,1] and Q are both countably infinite, and so there is some

(constructive!) one-to-one correspondence q --> f(q) between them.

Now what we do is that we take our set Y, which is the union of translates X + q

of X, and shift each translate so that X+q shifts to X + f(q). If we do this,

we get the union of all the rational translations of X (not just the small ones).

But this set is all of R, since every element of R is the member of some equivalence

class, and is therefore in some rational translate of X, since X contains one element

from each equivalence class. This proves the theorem.