Showing that a subset is compact (polynomials) I am trying to figure out how $K_{n}$ was written as a product? I am trying to prove compactness by showing the subset is compact and therefore $f(C)$ is compact. Could someone please explain what the method here is? Much appreciated!


 A: Consider the map $f:\mathbb{R}^{N+1}\to C[0,1]$ by
$$f(a_0,a_1, \dots, a_N)= \sum\limits_{i=0}^N a_i x_i.$$
It is not difficult to prove that $f$ is continuous. Note that $K_N$ is the image of $\prod\limits_{i=0}^N [-\frac{1}{1+i}, \frac{1}{1+i}]$ under $f$. By Tychonoff Theorem, $\prod\limits_{i=0}^N [-\frac{1}{1+i}, \frac{1}{1+i}]$ is compact. Being continuous image of a compact set, $K_N$ is compact.
EDIT $$K_N=f(\prod\limits_{i=0}^N [-\frac{1}{1+i}, \frac{1}{1+i}]).$$
Consider any $a_0+a_1x_1+\dots +a_Nx_N\in K_N$. Then $a_i\in [-\frac{1}{i+1},\frac{1}{i+1}]$ for each $0\leq i \leq N$. Thus, $(a_0,a_1, \dots, a_n)\in \prod\limits_{i=0}^N [-\frac{1}{1+i}, \frac{1}{1+i}]$ and $f (a_0,a_1, \dots, a_n)=a_0+a_1x_1+\dots +a_Nx_N\in f(\prod\limits_{i=0}^N [-\frac{1}{1+i}, \frac{1}{1+i}])$. Thus, $K_N \subseteq f(\prod\limits_{i=0}^N [-\frac{1}{1+i}, \frac{1}{1+i}])$. On the other hand, consider any member $p$ of $f(\prod\limits_{i=0}^N [-\frac{1}{1+i}, \frac{1}{1+i}])$. Then $p$ is of the form $f(b_0,b_1,\dots,b_N)=b_0+b_1x_1+\dots+b_Nx^N$, where $b_i\in [-\frac{1}{i+1},\frac{1}{i+1}]$. However, this shows that $p\in K_N$ and $f(\prod\limits_{i=0}^N [-\frac{1}{1+i}, \frac{1}{1+i}])\subseteq K_N$.
Hope this helps!
