\(\mathbb{R}\)-variable functions

On two important occasions I have seen the need for `exotic' differentials:

1. Two years ago, in exploring Special Relativity in functions spaces: \(\int f(x,y)\sqrt{dx\ dy}\)
2. Nowadays, in `taking the square root' of the metric tensor \(g_{\mu\nu}\).

This will eventually boil down to defining functions with number of variables being in \(\mathbb{R}\), hence \(\mathbb{R}^x, \ x\in\mathbb{R}\).

To do it, observe that

$$2^\emptyset=\{\emptyset\}.$$

If we write this as

$$\left(\epsilon^{\log_\epsilon 2}\right)^\emptyset =\{\emptyset\},$$

and define

$$\log_\epsilon N=-D$$

we will have

$$\boxed{N^\emptyset=\left(\epsilon^{-D}\right)^\emptyset=\epsilon^{-D\emptyset}=\{\emptyset\}}$$

which means \(\{\emptyset\}\) can be safely regarded as the Identity Element of the Cartesian Product.

In particular, this means that the number of subsets of \(S\) in a D-dimensional N-valued logic will be

$$|\mathcal{P}(S)|=N^{|S|}=\epsilon^{-D|S|}.$$

Thus, we define

$$\textbf{1}\boldsymbol{:= }\{\emptyset\}.$$

Then using this logarithm we can transform the Cartesian product \(\mathbb{R}^x, \ x\in\mathbb{R}\) to Union: we define logarithm of a set as 

$$\ln(A\cup\textbf{1}):=\ln(A\cup \{\emptyset\}):= \bigcup^\infty_{n=1} (-1)^{n+1}\frac{A^n}n = A \cap  \frac{A^2}2 \cup  \frac{A^3}{3}\cup \mathcal{O}(x^4).$$

In particular 

$$\ln(\mathbb{R}\cup\textbf{1}):=\ln(\mathbb{R}\cup \{\emptyset\}):= \bigcup^\infty_{n=1} (-1)^{n+1}\frac{\mathbb{R}^n}n = \mathbb{R} \cap \frac{\mathbb{R}^2}2 \cup \frac{\mathbb{R}^3}{3}\cup\mathcal{O}(\mathbb{R}^4).$$

Hence, finally

$$\ln\mathbb{R}^x=x\ln\mathbb{R}.$$