Prove the set of continuous real-valued functions on the interval $[0,1]$ is a subspace of $\mathbb{R}^{[0,1]}$
My Attempted Proof:
$$\mathbb{R}^{[0,1]} := \{ f \ | \ f : [0,1] \to \mathbb{R}\}$$
Part 1 :Take $f_0 : [0,1] \to 0$ (i.e. $f_o(x) = 0 \ \ \ \forall x \in [0,1]$). Clearly $f_0 \in \mathbb{R}^{[0,1]}$, and thus we have our additive identity for $\mathbb{R}^{[0,1]}$.
Part 2 : Now take $\alpha : [0,1] \to a \in \mathbb{R}$ and $\beta : [0,1] \to b \in \mathbb{R}$. Fix $x \in [0,1]$, then
$$\alpha(x) + \beta(x) : (x \in [0,1] \to a \in \mathbb{R}) + (x \in [0,1] \to b \in \mathbb{R})$$
which is the same as $x \in [0,1] \to (a \in \mathbb{R} + b \in \mathbb{R}) = (a+b) \in \mathbb{R}$, and thus $$\alpha(x) + \beta(x) \in \mathbb{R}^{[0,1]}$$
and $\mathbb{R}^{[0,1]}$ is closed under addition.
Part 3 :Finally we take $\gamma \in [0,1]$ and $f \in \mathbb{R}^{[0,1]}$. $f(x) = a \in \mathbb{R}$ for some $x \in [0,1]$, then $$\gamma \cdot f(x) = \gamma \cdot a \in \mathbb{R}$$
and $\mathbb{R}^{[0,1]}$ is closed under scalar multiplication. $\square$
Is my proof correct and logical/rigorous?
Best Answer
You need to emphasize the continuity. That is, you need to show that the set of continuous functions is nonempty and that it is closed under the operations of $\mathbb{R}^{[0,1]}$.
When you take the zero function, you need to note that this function is in fact continuous. Similarly with addition, you need to take two continuous functions $f$ and $g$ and show that $f+g$ is continuous.
Also, in the end you should take an arbitrary $a\in\mathbb{R}$ and prove that $(a\cdot f)\colon [0,1]\to\mathbb{R}$ is continuous for all $f\colon [0,1]\to \mathbb{R}$ continuos, where $(a\cdot f)(x):=a\cdot f(x)$ for all $x\in [0,1]$.