# 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?

• You missed "continuous". Aug 15 at 3:53
• are you suggesting $R$ is NOT continuous in this context? Aug 15 at 3:59
• Page 19 of what? What edition? A book with the title "Linear Algebra" or "Linear Algebra Done Right"? What author? Candidate: "Linear Algebra Done Right (Undergraduate Texts in Mathematics) 3rd ed. 2015 Edition" by Sheldon Axler. ("Undergraduate Texts in Mathematics (ISSN 0172-6056) is a series of undergraduate-level textbooks in mathematics published by Springer-Verlag.".) Aug 15 at 13:16

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

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 $$\chi_\Bbb Q\restriction_{[0,1]}(x)=\begin {cases}1\,,x\in\Bbb Q\\0,\,x\not\in \Bbb Q\end {cases}$$.