Bernstein's Theorem

Sometimes called the Cantor-Bernstein Theorem, this is a very useful result in set theory. We first define two sets A and B to be equivalent, similar, or equipollent if there exists a one to one map from set A onto set B; this implies that the two sets have the same cardinality. Notationally, we say A ∼ B. This can easily be seen to satisfy all the properties of an equivalence relation, it is reflexive (obviously a set is equivalent to itself), symmetric (if a one-to-one onto map exists from one set to another, an inverse map exists that maps the other way) and transitive (by performing composition of maps).

Bernstein's Theorem states that if A ∼ B', where B' ⊂ B, and B ∼ A', where A' ⊂ A, then A ∼ B.

Proof: Let the one to one mapping of A onto B' be φ and let the map of B onto A' be ψ. Then we construct subsets of A and B which cover the sets:

A₀ = A - A',

A_r = ψ(B_r-1),

        ∞
A = A - ∪  A
 ∞     r=0  r

and

B₀ = B - B',

B_r = φ(A_r-1),

        ∞
B = B - ∪  B
 ∞     r=0  r

The union of all the A_r's plus A_∞ is clearly A, while the union of all the B_r and B_∞ is B. All of the even indexed A_r's can be made to correspond to the odd indexed B_r's with the map φ, while in the same way the odd indexed A_r's can be made to correspond to the even indexed B_r's with ψ^-1. A_∞ can be converted into B_∞ with the map φ. In this way, a one-to-one map of A onto B can be constructed.

Using this theorem it can easily be shown that the open unit square S in R² is a continuum. Of course, the unit open interval (0,1) (equivalence to which we have taken as the definition of a continuum) is equivalent to a proper subset of S. Now, we can construct a mapping of S onto (0,1) as follows. Every element (x,y) of S can be expressed as (0.x₁x₂x₃..., 0.y₁y₂y₃...) where the x_n's and y_n's are the digits of the number in some base. This can be mapped onto an element of a proper subset of (0,1) by taking the element 0.x₁y₁x₂y₂... No two elements of S can produce the same element of (0,1) in this way. As (0,1) is similar to a subset of S, and S is similar to a subset of (0,1), it follows from Bernstein's theorem that S is a continuum.

In the same way, the unit cube can also be shown to be a continuum, as can the unit n-dimensional hypercube, even an infinite dimensional unit hypercube is also a continuum. Because the whole real line R has also been shown to be a continuum, (see the continuum writeup for the mapping), it also follows by repeatedly applying this theorem that R² is also a continuum, and so on to an infinite-dimensional space.