Can a bijection be extended to three dimensions?

[LordCalculus]

Can the definition of a bijection be extended to three dimensions? So for example,

f(a, b) = c, where f : N X N $$\rightarrow$$ N

 

[lm_aquarius]

According to set theory, bijection can be defined between any two sets - including Rⁿ and R^m. :)

 

[pinsky]

I don't see why not. To see it in the example you ask, you have to have a function which connects two sets.

The problem might arise if you say that in the first row of the set NxN there is the same number of elements as there is in the set N, so that excludes injection.

But, if you change the way you count the elements of the NxN and count in this order:

n_11,n_21, _n12, n_31, n_22, n_13 ...

Then a bijection is a possibility :)

$$\begin{Bmatrix}<br /> n_{11}\; n_{12}\; n_{13}\; ... \\ <br /> n_{21}\; n_{22}\; n_{23}\; ... \\ <br /> n_{31}\; n_{32}\; n_{24}\; ... \\ <br /> n_{41}\; n_{42}\; n_{25}\; ... \\<br /> \vdots \;\; \; \vdots \; \; \;\; \; \vdots \; \; \;\; \; ... \\<br /> \end{Bmatrix}$$