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I am given a question and am having a hard time understanding how to read part of a question, it reads

let $ C^{1}(0,1):= \{f:(0,1) \rightarrow \mathbb{R} \mid f\text{ is differentiable and $f'$ is continuous}\}$

so we are given some set $C^1(0,1)$ what does the ordered pair mean, does it mean our set $C^1$ is just an ordered pair? Then we see the following $\{f:(0,1) \rightarrow \mathbb{R}\}$ How is this read? I'm reading it as "The set of all functions $f$ such that the ordered pair $(0,1)$ maps to some real number. Is this how it's supposed to be read and if so what does it mean for an ordered pair to map to a real number?

Rodrigo de Azevedo
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Diehardwalnut
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1 Answers1

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$(0,1)$ does not stand for an ordered pair, but for “the interval $(0,1)$”: $$ (0,1)=\{x\in\mathbb{R}\mid 0<x<1\} $$ so you are defining the set of all real functions $f$ defined over the interval $(0,1)$ which are differentiable, with continuous derivative.

The notation $f\colon (0,1)\to\mathbb{R}$ denotes a function with real values defined over the interval $(0,1)$.

Unfortunately the notation $(a,b)$ is overloaded. Following some authors, I usually write $(0..1)$ to denote the interval. However, in this case it should be clear what the meaning is.

egreg
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  • In the same way we have $C^1(-3, 5)$ or any other interval you'd like. When you want to talk about functions that have all of the real numbers as domain, then writing $C^1(-\infty, \infty)$ is a lot more common than writing $C^1\Bbb R$. – Arthur Jun 21 '16 at 23:02