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Consider a continuous function $f(x)$ defined on the interval $[0,\,1]$. My question: How do you define the set of intervals such that $f(x)>V$, where $V$ is a constant? You may assume that all of the intervals can be calculated numerically and that there are finitely many.

My attempt was as follows \begin{equation} X := \big\{ x \,\, | \,\, f(x) >V\big\} = \big\{ (x_0,\,x_1),\,\dots, \, (x_{i},\,x_{i+1}), \, \cdots ,\, (x_{N-1},\,x_{N})\big\} \tag{1} \end{equation} where $x_i<x_{i+1}$ for $i=0,\,\dots,\,i,\,\dots,\,N.$

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    $\begingroup$ Not sure what you are asking. Finding the roots to $f(x)=V$ can be quite complicated, depending on the nature of $f(x)$. Is that really what you are asking about? Or are you just asking about notation? $\endgroup$
    – lulu
    Jul 29, 2022 at 18:24
  • $\begingroup$ Your notation is wrong. $\{x \mid f(x)>V\}$ is a set of NUMBERS, a subset of $[0,1]$, while $\big\{ (x_0,\,x_1),\,\dots, \, (x_{i},\,x_{i+1}), \, \cdots ,\, (x_{N-1},\,x_{N})\big\}$ is a set of SETS, since its elements are intervals. Which one is the one you are trying to define? $\endgroup$
    – jjagmath
    Jul 29, 2022 at 18:27
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    $\begingroup$ If notation is your goal, then note that there is no reason to imagine that there are only finitely many intervals. Suppose $f(x)$ is $0$ for $x=0$ or $x=\frac 1n$ for every $n\in \mathbb N$, and positive otherwise. $\endgroup$
    – lulu
    Jul 29, 2022 at 18:29
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    $\begingroup$ Can you edit your post for clarity? As it stands it is not clear what you are asking. $\endgroup$
    – lulu
    Jul 29, 2022 at 18:36
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    $\begingroup$ It's not reasonable to expect your readers to guess your intentions, nor to expect them to guess which unstated assumptions they are intended to make. $\endgroup$
    – lulu
    Jul 29, 2022 at 18:43

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You're defining $X =\{x \in [0,1]: f(x)>V\}$, which is exactly the $f$-preimage of the interval $(V, \infty)$. You've said $f$ is continuous, so as the preimage of an open set, $X$ is open. Open sets are the union of disjoint open intervals, where since we're working in $[0,1]$ we consider $[0,a)$ and $(a,1]$ to be open sets since we're working in a subspace of $\mathbb{R}$ If all you need is a way to justify that this set of intervals exists, then you're done.

If you actually want to compute the intervals then we can potentially leverage continuity. Say you can find the set $A = \{x \in [0,1]: f(x)=V\}$. You can then partition $[0,1]$ into intervals or points that are equal to $V$, and intervals that are either strictly greater or strictly less. For each unknown interval, just pick an interior point to determine whether $f>V$ or $f<V$ on that interval; you're guaranteed one or the other by continuity.

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  • $\begingroup$ The OP has $f$ defined on $[0,1]$, not $\Bbb R$, so $X$ is open IN $[0,1]$, and then the intervals are not necessarily open. $\endgroup$
    – jjagmath
    Jul 29, 2022 at 19:08
  • $\begingroup$ Right I missed that. They're still open in $[0,1]$ with the subspace topology, which is to say that $[0,a)$ and $(a,1]$ are "honorary" open intervals. $\endgroup$
    – DanishChef
    Jul 29, 2022 at 20:02

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