Prove that a polynomial of $k$ variables is continuous Let $k \ge 1$, let $I$ be a finite subset of $\mathbb{N}^k$, and let $c: I \to \mathbb{R}$ be a function. Form the function $P : \mathbb{R}^k \to \mathbb{R}$ defined by 
$$P(x_1, ... , x_k) : = \sum_{(i_1, ... , i_k)\in I} c(i_1, ..., i_k) x_1^{i_1}, ... , x_k^{i_k}.$$ 
Show that $P$ is continuous. 
I have already proven the case for a polynomial of two variables (i.e., $P(x, y) = \sum_{i=0}^n \sum_{j=0}^m c_{ij} x^iy^j$ is continuous.) In addition, if $f : X \to \mathbb{R}^m$ and $g: X \to \mathbb{R}^n$ are continuous functions, $f \oplus g : X \to R^{m+n}$ is also continuous. The author suggest to use induction on $k$ as well as above results. 
The base case has been already shown (the case for two variables). Suppose that the polynomial of $k$ variables is continuous. Now we need to consider the case for the polynomial of $k+1$ variables. Let $f : X \to \mathbb{R}^k$ defined by $f(x) = (x_1, ..., x_k)$. This implies that $P\circ f (x) = P(x_1, ..., x_k)$. Define $g: X \to \mathbb{R}$ by $g(x) = x_{k+1}$. Then, $f \oplus g (x) = (x_1, ..., x_{k+1})$. Then, $P(x_1, ..., x_{k+1}) = P \circ (f\oplus g) (x)$. However, this is not enough because I don't know whether $f$ and $g$ are continuous. 
Am I on the right track? how can I proceed from here? 
 A: I'll prove how $P$ is continuous for $k=3$ given that it is continuous for $k=2$, but you'll see that the idea is the same for the general induction.
Let $P:\mathbb{R}^3\to\mathbb{R}$ be $$P(x,y,z)=\sum_{i,j,k}c_{ijk}x^iy^jz^k=\sum_k\left[z^k\sum_{i,j}c_{ijk}x^iy^j\right]=\sum_kz^k\cdot Q_k(x,y)$$
where the $Q_k:\mathbb{R}^2\to\mathbb{R}$ is continuous because of our hypothesis. Hence, $P$ is the finite sum of the $z^k\cdot Q_k(x,y)$, which are continuous because $z^k$ is continuous (the identity function is continuous, hence the k-product of identity functions is continuous) and $Q$ is continuous. In order to proof the general induction, just factor the $k+1$ term and do the same thing.
Coming back to what you proposed. Let's call $P_n$ to the function you've defined, where $n$ marks the dimension of its domain, that is, $\mathbb{R}^n$. We want to prove that $P_{n+1}$ is continuous if $P_n$ is continuous. Let $X=\mathbb{R}^{n+1}$, and let $f$ and $g$ be defined in $X$ as $$f(x_1,\dots,x_{n+1})=(x_1,\dots,x_n), \ \ g(x_1,\dots,x_{n+1})=x_{n+1}$$
$f$ and $g$ are continuous because each of their projections is continuous (they are in fact identity functions in $\mathbb{R}$. In this case, we could write $$P_{n+1}(x_1,\dots,x_{n+1})=P_{n+1}\circ(f\oplus g)(\mathbf x)=(P_n\circ f(\mathbf x)) \cdot (P_1\circ g(\mathbf x))$$
again continuous because the product of continuous functions is continuous.
A: The base case for $k = 1$ is obvious. Suppose inductively that all polynomials of $k$ variables are continuous. Let $I \subseteq \mathbb{N}^{k+1}$ be finite. By the inductive hypothesis, $\forall i = (i_{1}, ... ,i_{k+1}) \in I$, the function $f_{i}: \mathbb{R}^{k+1} \to \mathbb{R}$ given by $(x_{1}, ... ,x_{k+1}) \to x_{1}^{i_{1}}...x_{k}^{i_k}$ is continuous. Clearly, the function $g: \mathbb{R}^{k+1} \to \mathbb{R}$ given by $(x_{1}, ... ,x_{k+1}) \to x_{k+1}$ is also continuous.
The function $P_{i} \circ (f_{i} \oplus g): \mathbb{R}^{k+1} \to \mathbb{R}$ given by $x \to c(i)f_{i}(x)g(x)^{i_{k+1}}$ is thus continuous, where $P_{i}: \mathbb{R}^{2} \to \mathbb{R}$ defined by $P_{i}(x, y) = c(i)xy^{i_{k+1}}$ is a polynomial of two variables.
Thus we have $\sum_{(i_1, ... ,i_{k+1}) \in I}c(i_1, ... ,i_{k+1})x_{1}^{i_1} ... x_{k}^{i_k+1} = \sum_{i \in I}P_{i} \circ (f_{i} \oplus g)(x)$ is continuous since it is the sum of finitely many continuous functions.
