The set of continuous $[0, 1] \to \Bbb R$ functions is a subspace of $\Bbb R^{[0, 1]}$

linear algebra

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?

Best Answer

$\mathbb{R}^{[0,1]}$ 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.)

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