To show that the Weierstrass Approximaton Theorem holds in $\mathbb{R}^n$ , we will use induction on $n$ .

For the sake of simplicity, consider first the case of the cubical region $0 \le x_i \le 1$ , $1 \le i \le n$ . Suppose that $f$ is a continuous, real valued function on this region. Let $\epsilon$ be an arbitrary positive constant.

Since a continuous functions on compact regions are uniformly continuous, $f$ is uniformly continuous. Hence, there exists an integer $N > 0$ such that $|f(a) - f(b)| < \epsilon/2$ whenever $|a - b| \le 1/N$ and both $a$ and $b$ lie in the cubical region.

Define $\phi \colon \mathbb{R} \to \mathbb{R}$ as follows: $$ \phi (x) = \left\{ \begin{matrix} 0 & x < -1 \cr 1 + x & -1 \le x \le 0 \cr 1 - x & 0 \le x \le 1 \cr 0 & x > 1\end{matrix} \right. $$ Consider the function ${\tilde f}$ defined as follows: $$ {\tilde f} (x_1, \ldots x_n) = \sum_{m = 0}^N \phi(N x_1 + m) f(m/N, x_2, \ldots x_n) $$

We shall now show that $|f(x_1, \ldots, x_n) - {\tilde f}(x_1, \ldots, x_n)| \le \epsilon/2$ whenever $(x_1, \ldots, x_n)$ lies in the cubical region. By way that $\phi$ was defined, only two of the terms in the sum defining ${\tilde f}$ will differ from zero for any particular value of $x_1$ , and hence $$ {\tilde f} (x_1, \ldots, x_n) = (Nx - \lfloor Nx \rfloor) f \left( {\lfloor N x_1 \rfloor \over N}, x_2, \ldots, x_n \right) + (\lfloor Nx \rfloor + 1 - x) f \left( {\lfloor N x_1 \rfloor + 1 \over N}, x_2, \ldots, x_n \right), $$ so \begin{eqnarray*} | {\tilde f} (x_1, \ldots, x_n) - f (x_1, \ldots, x_n) | &=& | {\tilde f} (x_1, \ldots, x_n) - \{ (Nx - \lfloor Nx \rfloor) + (\lfloor Nx \rfloor + 1 - x)\} f (x_1, \ldots, x_n) | \cr &\le& (Nx - \lfloor Nx \rfloor) | f \left( {\lfloor N x_1 \rfloor \over N}, x_2, \ldots, x_n \right) - f (x_1, X_2 \ldots, x_n) | + (\lfloor Nx \rfloor + 1 - x) |\left( {\lfloor N x_1 \rfloor + 1 \over N}, x_2, \ldots, x_n \right) - f (x_1, x_2 \ldots, x_n) | \cr &\le& (Nx - \lfloor Nx \rfloor) {\epsilon \over 2} + (\lfloor Nx \rfloor + 1 - x) {\epsilon \over 2} = {\epsilon \over 2}. \end{eqnarray*} Next, we will use the Weierstrass approximation theorem in $n-1$ dimensions and in one dimesnsion to approximate $\tilde f$ by a polynomial. Since $f$ is continuous and the cubical region is compact, $f$ must be bounded on this region. Let $M$ be an upper bound for the absolute value of $f$ on the cubical region. Using the Weierstrass approximation theorem in one dimension, we conclude that there exists a polynoial $\breve \phi$ such that $| \breve \phi (a) - \phi (a) | < \epsilon / (4MN)$ for all $a$ in the region. Using the Weierstrass approximation theorem in $n-1$ dimensions, we conclude that there exist polynomials $p_m$ , $0 \le m \le N$ such that $|p_m (x_2, \ldots, x_n) - f (m/N, x_2,\ldots x_n) | \le {\epsilon \over 4N}$ . Then one has the following inequality: \begin{eqnarray*} | {\breve \phi} (N x_1 + m) p_m (x_2, \ldots x_n) - \phi (N x_1 + m) f(m/N, x_2, \ldots x_n) | &= | {\breve \phi} (N x_1 + m) p_m (x_2, \ldots x_n) - {\breve \phi} (N x_1 + m) f(m/N, x_2, \ldots x_n) \\ &+& {\breve \phi} (N x_1 + m) f(m/N, x_2, \ldots x_n) - \phi (N x_1 + m) f(m/N, x_2, \ldots x_n) | \\ &\le& |{\breve \phi} (N x_1 + m)| | p_m (x_2, \ldots x_n) - f(m/N, x_2, \ldots x_n)| \\ &+& | f(m/N, x_2, \ldots x_n)| | {\breve \phi} (N x_1 + m) - \phi (N x_1 + m)| \\ &\le& {\epsilon \over 4N} + M {\epsilon \over 4MN} = {\epsilon \over 2N} \\ \end{eqnarray*}Define $$ {\breve f} (x_1, \ldots x_n) = \sum_{m = 0}^N {\breve \phi} (N x_1 + m) p_m (x_2, \ldots x_n). $$ As a finite sum of products of polynomials, this is a polynomial. From the above inequality, we conclude that $|{\breve f} (a) - {\tilde f} (a)| \le \epsilon / 2$ , hence $|f(a) - {\breve f} (a)| \le \epsilon$ .

It is a simple matter of rescaling variables to conclude the Weirestrass approximation theorem for arbitrary parallelopipeds. Any compact subset of $\mathbb{R}^n$ can be embedded in some paralleloped and any continuous function on the compact subset can be extended to a continuous function on the parallelopiped. By approximating this extended function, we conclude the Weierstrass approximation theorem for arbitrary compact subsets of $\mathbb{R}^n$ .