A question on isometric isomorphism

Question

This is problem from Conway's Functional Analysis.

If $X=\{\frac{1}{n}:n\in \mathbb N\}\cup \{0\}$, then I have to show that $C(X)=\{f:X\to \mathbb K: f \text{ is continuous }\}$ is isometrically isomorphic to $c=\{(x_n): x_n\in \mathbb K, \lim\limits_{n\to \infty}x_n\in \mathbb K\}$.

I have no idea how to proceed. Only I can guess that it can be done by first principle only as it is given just after introducing isometric isomorphism. Any help is appreciated.

Answer 1

You just map a sequence $(x_n)$ to the function $f$ with $f(1/n)=x_n$ and $f(0)=\lim_{n\to\infty}x_n$.

Answer 2

Consider the function $\phi:C(X)\longrightarrow c_0$, defined as, $\phi(f)=f(\frac{1}{n})$.

Now, this function is well defined, as $f$ is continuos, so $\lim\limits_{n\to \infty}f(\frac{1}{n})=f(0)\in \mathbb K$.

Work with this.