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Say we got this simple example

$$f(x) = \begin{cases} p_1(x),\text{ for }5 \leq x \leq 8,\\ p_2(x),\text{ for }8 \leq x \leq 10 \end{cases}$$

Assuming $p_1(8)=p_2(8)$, is this permissable?

Arguments for and against:

  • Yes, because $f(x)$ maps each x-value to exactly one y-value.
  • No, because it becomes unclear whether to choose $p_1$ or $p_2$ when $x=8$.
    • Counter: well, it doesn't matter what function you choose, they yield identical y-values.
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  • $\begingroup$ The second bullet point doesn't make sense because $p1(8)=p_2(8)$ so the supposed "choice" is not actually there. $\endgroup$ – rubikscube09 Feb 25 at 19:48
  • $\begingroup$ This is acceptable. $\endgroup$ – Ethan Bolker Feb 25 at 20:05
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I would say it is an acceptable definition, in the sense its formalization describes a function. However, some people may be displeased, but again some people will always be displeased...

A function $f:X\rightarrow Y$ is a subset $F$ of $X\times Y$ such that:

  • for every $x\in X$, there exists a $y\in Y$ for which $(x,y)\in F$;
  • and, if $(x, y_{1}), (x, y_{2})\in F$, then $y_{1}=y_{2}$.

When we define a function by cases, that is not absolutely precise in the sense of this definition, but the correspondent idea can always be formalized: in your case, the set $F$ corresponding to the function may be defined as the subset of $[5, 10]\times\mathbb{R}$ (I am assuming here that your function is real-valued and not defined outside of $[5,10]$) such that $$F=\{(x,y)\in[5,10]\times\mathbb{R} : \text{if $x\in[5,8]$, $y=p_{1}(x)$, and if $x\in [8,10]$, $y=p_{2}(x)$}\}.$$ Does that define a function?

  • For every $x\in [5,10]$, there indeed exists an $y\in\mathbb{R}$ such that $(x,y)\in F$, equal to $p_{1}(x)$ if $x\in[5,8)$, equal to $p_{2}(x)$ if $x\in(8,10]$, and equal to $p_{1}(x)=p_{2}(x)$ if $x=8$.
  • If $(x, y_{1}), (x, y_{2})\in F$, there are three cases: if $x\in[5,8)$, since $p_{1}$ is a function $y_{1}=y_{2}$; if $x\in (8,10]$, since $p_{2}$ is a function $y_{1}=y_{2}$; and if $x=8$, given both $p_{1}$ and $p_{2}$ are functions and $p_{1}(8)=p_{2}(8)$, $y_{1}=y_{2}$.

So $F$ indeed describes a function, and your definition is indeed acceptable.

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