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I have the function $f : \mathbb{R} \to \mathbb{R}$, $f(x) = \begin{cases} 1/x & (x \neq 0) \\ 0 & (x=0)\end{cases}$

So I would like to denote the function $f$ as more simple, 'one-line' notation,

$$ f(x) = \frac{1}{x} \mathbb{1}_{\{ x \neq 0 \}} $$

But I think this notation is impossible unless we give some priority to the indicator function.

Does it make sense to do so?

Moreblue
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    @MauroALLEGRANZA That is not the same function as in the question. – MSDG Dec 01 '18 at 15:18
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    Rigorously speaking, this notation makes sense under the convention that $$\frac00=0$$ otherwise, it does not (but everybody (uses and) understands it). – Did Dec 01 '18 at 15:39
  • @Did Your remark convinces me. – Moreblue Dec 01 '18 at 15:46
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    Nice remark from you, though... – Did Dec 01 '18 at 15:50
  • @Did this is absolutely not a universal mathematical convention; I'm sorry but that's just not true. You cannot divide by zero. This is like getting confused about e.g. $f(x) = \frac{x^2}{x}$ and saying $f(0) = 1$. – SBK Dec 01 '18 at 16:05
  • @T_M Maybe you never met the convention but it is widely used, notably in probability theory. Anyway, your comparison does not hold, actually it first and foremost seems to indicate a too quick reading from you of what others wrote: please be aware that everybody here is well aware that $\lim x_n=\lim y_n=0$ does not imply $\lim x_n/y_n=0$, but that this is not what was said. – Did Dec 01 '18 at 16:13
  • @Did I'm sorry but dividing by zero is just never done. There is something similar, which is this: Often in measure theory you will allow functions to take the value $+\infty$ or $-\infty$. And of course there exist sets of measure zero. The convention is that the integral of $+\infty$ over a set of measure zero is equal to zero. – SBK Dec 01 '18 at 18:46
  • @Did I mentioned limits nowhere so I'm not sure why you introduced them into our discussion. We can forget the comparison if you like, but I think it's important to realize the $f$ as written in the question is not well-defined because it quite clearly cannot be evaluated at zero. – SBK Dec 01 '18 at 18:49
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    @T_M Your example $x^2/x$ very much points at a limit when $x\to0$, which is not $1$ -- so if you did not mention this case to refer to the trivial remark I made in my last comment, I fail to see why you did. Anyway, as I mentioned from the start, it is a fact that mathematicians do use the notation $g(x)=f(x)\mathbf 1_{x\in A}$ to mean that $g(x)=0$ for every $x\notin A$, whether $f(x)$ is well defined at $x$ or not (and that $g(x)=f(x)$ at every $x\in A$, hence $f$ should be well defined on the whole set $A$). – Did Dec 01 '18 at 18:58
  • OK. I just think that's not what you said at the start, which is why I disagreed so strongly. You said 0/0 = 0 makes sense rigorously speaking as a convention that everybody uses and understands. This is very different from the specific situation of $g(x) = f(x)\mathbf{1}{A}(x)$ which some mathematicians use without further comment. Either way, there seems no reason to encourage something which is obviously a shorthand for a perfectly good piecewise defined function. If a collaborator wrote $g(x) = f(x)\mathbf{1}{A}(x)$ I would disagree with them and make it a clear, piecewise definition. – SBK Dec 02 '18 at 23:15

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Not really, no. If the domain is $\mathbb{R}$, then you need to tell me how to evaluate the function at $0$. The easiest way is just to say $f(0) = 0$.

SBK
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    Doesn't the indicator function $\mathbb{1}_{{x \neq 0}}$ tell you how to evaluate the function at $0$? – Moreblue Dec 01 '18 at 15:24
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    The indicator function is only one part of the product though. It's true that $0 \times , \text{any number} = 0$, but you don't have another number to multiply by. Put it this more abstract way: You've written $f(x) = g(x)h(x)$. The function $g$ is not defined for $x=0$. Therefore the function $f$ is not defined at $x=0$. – SBK Dec 01 '18 at 16:02