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Is there a traditional notation or name of this function:

$$ \epsilon (x) = \begin{cases} 0 & \textrm{ if }x = 0 \\ 1 & \textrm{ if }x \neq 0 \end{cases} $$

I know one can use Indicator function $1_{A}(x)$ for $A = \mathbb{R} - \{0\}$, but I still want to know if there is a traditional, simpler, notation for it.

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    $\begingroup$ Not that I know of. The simplest way of writing is either $1_{\mathbb R\setminus \{0\}}$ or $1 - 1_{\{0\}}$. Maybe even $1-\delta_0$, but that is already slightly ambiguous. $\endgroup$ – 5xum May 5 '15 at 7:28
  • $\begingroup$ Well... $1_{\Re \backslash \{ 0 \} }$ is not a function. Moreover, $1 - \delta_0$ cannot be correct because the value of $\delta_0$ is infinite at $x=0$. This is a bizarre function, and I suspect the question poser isn't quite clear on what he or she needs. One can write the function as a limit, but again, that seems rather awkward. $\endgroup$ – David G. Stork May 5 '15 at 7:31
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    $\begingroup$ @David Yes, it is a function. And no, it does not take the value infinity. $\endgroup$ – Tobias Kildetoft May 5 '15 at 7:42
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    $\begingroup$ @DavidG.Stork The first one is precisely defined like the function the OP is asking about. In general it is a common notation to use $1_A$ as the indicator function for the set $A$, meaning that $1_A(x)$ is $1$ if $x\in A$ and $0$ otherwise. $\delta_0$ is a common shorthand for the indicator function for the set $\{0\}$. $\endgroup$ – Tobias Kildetoft May 5 '15 at 8:01
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    $\begingroup$ @DavidG.Stork why would you write it as a limit when there is a simple explicit definition as an indicator function? And why do you doubt indicator functions are functions? $\endgroup$ – Sasho Nikolov May 6 '15 at 4:39
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The Kronecker delta function $\delta_{ij}$ is $1$ when $i=j$ and $0$ otherwise, so $\epsilon(x)=1-\delta_{0,x}$.

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In the Iverson bracket notation you could write $$\epsilon(x)=[x\ne0].$$

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You can write $\epsilon(x)=\mathrm{sgn}^2(x)$ where $\mathrm{sgn}$ is the signum function.

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  • $\begingroup$ It seems (to me) this is the best way. $\endgroup$ – user565739 May 5 '15 at 8:27
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    $\begingroup$ Well, it's compact for sure, but I wouldn't recommend using it to communicate with other humans. :) $\endgroup$ – Chris Culter May 5 '15 at 8:35
  • $\begingroup$ $\epsilon(x)=\mathrm{sgn}|x|$ $\endgroup$ – charlie Mar 24 '19 at 8:49
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The Heaviside step function is defined as

$$ H (x) = \begin{cases} 0 & \textrm{ if }x < 0 \\ 1 & \textrm{ if }x => 0 \end{cases} $$

Clearly

$$ H (-x) = \begin{cases} 1 & \textrm{ if }x <= 0 \\ 0 & \textrm{ if }x > 0 \end{cases} $$

Summing gives:

$$ H(x)+ H (-x) = \begin{cases} 1 & \textrm{ if }x < 0 \\ 2 & \textrm{ if }x = 0 \\ 1 & \textrm{ if }x >0 \end{cases} $$

Subtract 1

$$ H(x)+ H (-x)-1 = \begin{cases} 0 & \textrm{ if }x < 0 \\ 1 & \textrm{ if }x = 0 \\ 0 & \textrm{ if }x >0 \end{cases} $$

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    $\begingroup$ You could also multiply two Heaviside functions: $\epsilon (x)=H(x)H(-x)$ $\endgroup$ – tomi May 5 '15 at 7:44
  • $\begingroup$ All this H(x) + H(-x) reminds me of hyperbolic functions - is there any link? $\endgroup$ – tomi May 5 '15 at 7:45
  • $\begingroup$ Yes, I do. Thanks - but I can't fix it... $\endgroup$ – tomi May 5 '15 at 7:59
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    $\begingroup$ You can't edit a comment after 5 minutes, but you can copy, delete, and repost it. Anyway, I thought you wanted to fix your answer: by changing $H(x)+H(-x)-1$ to $2-H(x)-H(-x)$. $\endgroup$ – bof May 5 '15 at 8:07
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    $\begingroup$ How is this better than writing out the definition? $\endgroup$ – Tobias Kildetoft May 5 '15 at 8:13
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This funtion is also known as the "discrete norm", which is to say an absolute value (in the sense of ring theory) whose metric induces the discrete topology. The discrete metric is $d(x,y)=1$ if $x\ne y$ and $d(x,x)=0$, while the discrete norm on any domain $X$ is $f(x)=1$ if $x\in X\setminus\{0\}$ and $f(0)=0$. For the case of $X=\Bbb R$, this reduces to your function.

I hope this answers the question posed by some of why this would be a natural or useful function, rather than a simple modification of some other function like $\delta_{x,0}$ which is useful in other contexts.

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You could use the Heaviside step function $H(x)$. Your function is

$\epsilon(x)=|H(x)-H(-x)|$

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  • $\begingroup$ True, but see my post which avoids using absolute function. $\endgroup$ – tomi May 5 '15 at 7:43
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    $\begingroup$ How is this better than writing out the definition? $\endgroup$ – Tobias Kildetoft May 5 '15 at 8:13
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    $\begingroup$ This isn't "traditional notation" (nor simpler) like the question asked, it seems more like obfuscation than anything.... unless this definition from the Heaviside function allows one to derive nontrivial information about the function, it seems pointless. $\endgroup$ – Thomas May 5 '15 at 8:23
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Here's one possible definition, though extremely dependent on how one defines the value $0^0$.

If you happen to subscribe to the belief that $0^0=0$ (disclaimer, I personally define $0^0=1$), then $$\epsilon(x)=x^0$$

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  • $\begingroup$ Would the downvoter like explain why they did so? $\endgroup$ – Hayden May 6 '15 at 16:59

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