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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',

Ar = ψ(Br-1),

```        ∞
A = A - ∪  A
∞     r=0  r
```

and

B0 = B - B',

Br = φ(Ar-1),

```        ∞
B = B - ∪  B
∞     r=0  r
```

The union of all the Ar's plus A is clearly A, while the union of all the Br and B is B. All of the even indexed Ar's can be made to correspond to the odd indexed Br's with the map φ, while in the same way the odd indexed Ar's can be made to correspond to the even indexed Br'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 xn's and yn'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.

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