Notation (or Name) of the function which is $0$ at $x=0$ and $1$ otherwise 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.
 A: The Kronecker delta function $\delta_{ij}$ is $1$ when $i=j$ and $0$ otherwise, so $\epsilon(x)=1-\delta_{0,x}$.
A: In the Iverson bracket notation you could write
$$\epsilon(x)=[x\ne0].$$
A: You can write $\epsilon(x)=\mathrm{sgn}^2(x)$ where $\mathrm{sgn}$ is the signum function.
A: 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
$$
H(x)+ H (-x)-1  = \begin{cases}
0 & \textrm{ if }x < 0 \\
1 & \textrm{ if }x = 0 \\
0 & \textrm{ if }x >0
\end{cases}
$$
A: 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.
A: You could use the Heaviside step function $H(x)$. Your function is 
$\epsilon(x)=|H(x)-H(-x)|$
A: 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 
$$\epsilon(x)=x^0$$
