Question

Show the cartesian product of R^n and R^m, is isomorphic to R^(n+m)

Answer 1

Answer on Question #82545 – Math – Linear Algebra

Question

Show the cartesian product of $R^n$ and $R^m$, is isomorphic to $R^n(n+m)$.

Solution

$$R^n \times R^m = \{\langle (x_1, \ldots, x_n), (y_1, \ldots, y_m) \rangle \mid (x_1, \ldots, x_n) \in R^n, (y_1, \ldots, y_m) \in R^m\}.$$

We will define function $f \colon R^n \times R^m \to R^{n+m}$ so that

$$f(\langle (x_1, \ldots, x_n), (y_1, \ldots, y_m) \rangle) = (x_1, \ldots, x_n, y_1, \ldots, y_m),$$ $$(x_1, \ldots, x_n) \in R^n, (y_1, \ldots, y_m) \in R^m, \langle (x_1, \ldots, x_n), (y_1, \ldots, y_m) \rangle \in R^n \times R^m \text{ and } (x_1, \ldots, x_n, y_1, \ldots, y_m) \in R^{n+m}.$$

Then $f$ is isomorphism.

Let $(x_1, \ldots, x_n, x_{n+1}, \ldots, x_{n+m}) \in R^{n+m}$ then $(x_1, \ldots, x_n) \in R^n$, $(x_{n+1}, \ldots, x_{n+m}) \in R^m$, $\langle (x_1, \ldots, x_n), (x_{n+1}, \ldots, x_{n+m}) \rangle \in R^n \times R^m$ and $f(\langle (x_1, \ldots, x_n), (x_{n+1}, \ldots, x_{n+m}) \rangle) = (x_1, \ldots, x_n, x_{n+1}, \ldots, x_{n+m})$ therefore $f$ is a function onto.

Let $\langle (x_1, \ldots, x_n), (x_{n+1}, \ldots, x_{n+m}) \rangle, \langle (x'_1, \ldots, x'_n), (x'_{n+1}, \ldots, x'_{n+m}) \rangle \in R^n \times R^m$ and $\langle (x_1, \ldots, x_n), (x_{n+1}, \ldots, x_{n+m}) \rangle \neq \langle (x'_1, \ldots, x'_n), (x'_{n+1}, \ldots, x'_{n+m}) \rangle$ then $\exists i (1 \leq i \leq n + m \& x_i \neq x'_i)$ but then $(x_1, \ldots, x_n, x_{n+1}, \ldots, x_{n+m}) \neq (x'_1, \ldots, x_n, x_{n+1}, \ldots, x_{n+m})$ that is $f(\langle (x_1, \ldots, x_n), (x_{n+1}, \ldots, x_{n+m}) \rangle) \neq f(\langle (x'_1, \ldots, x'_n), (x'_{n+1}, \ldots, x'_{n+m}) \rangle)$.

So, $f$ is one-to-one function.

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