Proof that products of vector is a continuous function

Question

Let $\prod : \mathbb{R}^n \to \mathbb{R}$ where $\prod (\boldsymbol{x}) = x_1 \times \dots \times x_n$ for all $\boldsymbol{x}=(x_1, \dots, x_n) \in \mathbb{R}^n$. How to proof that $\prod$ is continuous?

I know that the product of continuous functions is continuous and $x \mapsto x$ is continuous. However, in this case $\prod$ is a vector function. Can we write:

$\prod (\boldsymbol{x}) = x_1 \times \dots \times x_n = a(x_1) \times \dots \times a(x_n)$

where $a:\mathbb{R} \to \mathbb{R}$ where $a(x)=x, \forall x\in\mathbb{R}$ such that $\prod = a \times \dots \times a$ and hence $\prod$ is continuous?

Answer 1

Fix any $\boldsymbol{x}^* = (x_1^*,\cdots,x_n^*)\in\mathbb{R}^n$ and let $\boldsymbol{x}\to\boldsymbol{x}^*$, then $x_i-x_i^*\to 0$ for each $1\leq i\leq n$. So

\begin{align} & \Pi(\boldsymbol{x^*}) - \Pi(\boldsymbol{x}) \\ ={} & x_1^*\cdots x_n^* - x_1\cdots x_n \\ ={} & (x_1^*-x_1)x_2^*\cdots x_n^* \\ & + x_1(x_2^*-x_2)x_3^*\cdots x_n^* \\ & + \cdots \\ & + x_1x_2\cdots x_{n-1}(x_n^*-x_n) \to 0. \end{align}

Answer 2

Another approach: Show that the product of two continuous functions is continuous, and projections are linear, consequently continuous.

Answer 3

If the result you refer to in your proof is that the product $fg$ of two continuous functions $f,g:\mathbb{R}\to\mathbb{R}$ is continuous, then the problem is that $fg$ is a function $\mathbb{R}\to\mathbb{R}$, not $\mathbb{R}^2\to\mathbb{R}$ (meaning $a\times a:\mathbb{R}\to\mathbb{R}\neq\Pi:\mathbb{R}^2\to\mathbb{R}$).

To prove that $\Pi$ is continuous you need to (first learn and then) use the definition of continuity from $\mathbb{R}^n$.