Proof that $\phi:C[0;1] \rightarrow \mathbb{R}^\mathbb{N}$ is isomorphism. I have problem with proof that $\phi:C[0;1] \rightarrow \mathbb{R}^\mathbb{N}$ is isomorphism.
$ C[0;1], \mathbb{R}^\mathbb{N}$ - vector spaces.
Definition of $ \phi $: $ \phi(\varphi) = (\varphi(r_1),...,\varphi(r_n),...)$ where {$ r_i\in [0;1] $ $ i\in \mathbb{N}$ } some fixed set.
I don't know how to explain that $\phi$ is bijection, at least using only definition of bijection.
And I don't what property of continuous function I need to use for proof there equals:
$\phi(\varphi_1+\varphi_1) = \phi(\varphi_1)+ \phi(\varphi_2)$ $ \varphi_1,\varphi_2 \in C[0;1]$
$\phi(\alpha\varphi_1) = \alpha\phi(\varphi_1)$ $ \alpha\in\mathbb{R}, \varphi_1\in C[0;1]$
Could you please give any ideas how to prove that $\phi$ is isomorphism ?
 A: $\phi$ is never a bijection.
Let $R = \{r_i\}$ which is an infinite set. More precisely we have a bijection $r : \mathbb N \to R, r(i) = r_i$.
Note that $\phi(f) = \phi(g)$ means that $f \mid_R = g \mid_R$.

*

*$\phi$ is injective if and only if $R$ is dense in $[0,1]$.


*

*Let $R$ be dense. Let $\phi(f) = \phi(g)$. Then $f$ and $g$ agree on the dense subset $R \subset [0,1]$. For each $x \in [0,1]$ there exists a sequence $(x_n)$ in $R$ such that $\lim_{n \to \infty} x_n = x$. By continuity of $f,g$ we get $f(x) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} g(x_n) = g(x)$. Thus $f = g$.


*Let $R$ not be dense. Then there exists $\xi \in [0,1]$ and $\epsilon > 0$ such that $(\xi-\epsilon,\xi+\epsilon) \cap R = \emptyset$. Note that $(\xi-\epsilon,\xi+\epsilon)$ is not necessarily a subset of $[0,1]$, but this is irrelavnt here. The above condition means that each point of $R$ has  distance $\ge \epsilon$ from $\xi$. Define $G :\mathbb R \to \mathbb R, G(x) = 0$ for $x \notin  (\xi-\epsilon,\xi+\epsilon)$, $G(x) = 1 - \frac{1}{\epsilon^2}(x - \xi)^2$ for $x \in  [\xi-\epsilon,\xi+\epsilon]$. Now consider $f(x) = 0$ and $g = G \mid_{[0,1]}$. Then $f \ne g$, but $\phi(f) = \phi(g)$.



*

*$\phi$ is not surjective.

Since $R$ is infinite, it has an accumulation point $\xi \in [0,1]$. Let $(r_{i_k})_{k \in \mathbb N}$ be a subsequence of $(r_i)$ such that $\lim_{k \to \infty} r_{i_k} = \xi$. Define a sequence $(s_i) \in \mathbb R^{\mathbb N}$ by $s_i = 0$ for $i \in \{i_1,i_3,i_5,i_7,\ldots \}$ and $s_i = 1$ for  $i \notin \{i_1,i_3,i_5,i_7,\ldots \}$. Assume that $(s_i) = \phi(f)$ for a continuous $f : [0,1] \to \mathbb R$. This means $f(r_i)  = s_i$ for all $i$. But we have $f(\xi) = f(\lim_{l \to \infty} r_{i_{2l+1}}) = \lim_{l \to \infty}f(r_{i_{2l+1}}) =  \lim_{l \to \infty}s_{i_{2l+1}} = 0$ and $f(\xi) = f(\lim_{l \to \infty} r_{i_{2l}}) = ) = \lim_{l \to \infty}f(r_{i_{2l}}) =  \lim_{l \to \infty}s_{i_{2l}} = 1$, a contradicton.
