Prove the set of continuous real-valued functions on the interval $[0,1]$ is a subspace of $\mathbb{R}^{[0,1]}$
Note I asked this question earlier here: Prove the set of continuous real-valued functions on the interval $[0,1]$ is a subspace of $\mathbb{R}^{[0,1]}$, in which I provided a horrendously wrong proof (it's what lack of sleep does). Here I've written a new proof, which I believe should be correct. If it's not please feel free to tear it apart.
Main definition used in proof:
If $S$ is a set, then $F^S$ denotes the set of functions from $S$ to $F$
My Attempted Proof:
$$\mathbb{R}^{[0,1]} := \{ f \ | \ f : [0,1] \to \mathbb{R}\}$$
Put $V = \{f \ | \ f: [0,1] \to \mathbb{R} \ \ \text{such that $f$ is continous}\}$
Part 1 :Take $f_0 : [0,1] \to 0$ (i.e. $f_o(x) = 0 \ \ \ \forall x \in [0,1]$). Clearly $f_0$ is continuous and $f_0 \in V$. Now $f_0 + g = g$ for all $g \in V$, and thus $f_0$ is the additive inverse of $V$.
Part 2 : Now take $f,g \in V$.
Now recall for $f, g \in F^S$, the sum $f+g \in F^S$ is the function defined by $(f+g)(x) = f(x) + g(x)$.
Since $V \subset \mathbb{R^{[0,1]}}$ we have $f,g \in \mathbb{R^{[0,1]}}$, and thus the above definition of the addition operation holds, and $f(x) + g(x) = (f+g)(x)$.
And since $f$ and $g$ are both continuous, $f+g$ is also continuous, and combining with what we've shown above we can see that $f+g \in V$. Thus $V$ is closed under addition.
Part 3 :Finally we take $\gamma \in \mathbb{R}$ and $f \in V$.
Recall for $\lambda \in F$ and $f \in F^S$, the product $\lambda f \in F^S$ is the function defined by $(\lambda f)(x) = \lambda f(x)$. So all we have to do is verify that $f \in \mathbb{R^{[0,1]}}$, and scalar multiplication as defined above for $F^S$ (arbitrary function spaces) will hold.
Since $V \subset \mathbb{R^{[0,1]}}$ we have $f \in \mathbb{R^{[0,1]}}$, and $(\gamma f)(x) = \gamma f(x)$
Since $f$ is continuous, $\gamma f$ is also continuous and combining with what we've shown above we can see that $\gamma f \in V$, and thus $V$ is closed under scalar multiplication. $\square$
Is my proof correct and logical/rigorous? And if what I've written is nonsense, please say so.
Best Answer
Part 1 is o.k. But in Part 2 and 3 you only have considered functions which are constant!