The set of continuous $[0, 1] \to \Bbb R$ functions is a subspace of $\Bbb R^{[0, 1]}$ In example 1.35.b of Sheldon Axler's Linear Algebra Done Right, on page 19, it is said that "the set of continuous real-valued functions on the interval $[0, 1]$" is a subspace of $\Bbb R^{[0, 1]}$.
I am confused why this result is interesting to us. According to the definition of $F^S$, isn't "the set of [...] on the interval $[0, 1]$" equivalent to $\Bbb R^{[0,1]}$? What did I miss?
 A: $\mathbb{R}^{[0,1]}$ is the space of all functions $f : [0,1] \to \mathbb{R}$.
The subset $C$ of functions $f \in \mathbb{R}^{[0,1]}$ which are continuous is claimed to be a subspace. Of course, not all $f \in \mathbb{R}^{[0,1]}$ are continuous.
For instance: define
$$f : [0,1] \to \mathbb{R} \text{ where } f(x) := \begin{cases}
1 & x = 1/2 \\
0 & x \ne 1/2 \end{cases}$$
Then $f \in \mathbb{R}^{[0,1]}$ but $f \not \in C$. (The continuity you're concerned about in the comments is not meant to be about $\mathbb{R}$, but of the functions themselves.)
A: This is certainly interesting in that the claim is that the subset $C\subset \Bbb R^{[0,1]}$ of continuous functions among all functions from $[0,1]$ to $\Bbb R$ is claimed to be a subspace.
So, first you need to convince yourself that the set of all such functions forms a vector space.
Then you can check for closure of the subset of continuous functions under addition and scalar multiplication.
Of course there are functions in $\Bbb R^{[0,1]}$ that are not continuous.  For one that is very discontinuous consider the characteristic function of the rationals, restricted to the unit interval $\chi_\Bbb Q\restriction_{[0,1]}:[0,1]\to \Bbb R$ by $\chi_\Bbb Q\restriction_{[0,1]}(x)=\begin {cases}1\,,x\in\Bbb Q\\0,\,x\not\in \Bbb Q\end {cases}$.
