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?

  • 2
    $\begingroup$ You missed "continuous". $\endgroup$ Aug 15 at 3:53
  • $\begingroup$ are you suggesting $R$ is NOT continuous in this context? $\endgroup$
    – Roy Huang
    Aug 15 at 3:59
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    $\begingroup$ 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.".) $\endgroup$ Aug 15 at 13:16

2 Answers 2


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


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