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I should find at least one function that satisfies this property for every $x \in \mathbb R $ $$ f(x) = \begin{cases} 0 & \text{if $x \ge 0$} \\ 1 & \text{if $x \lt 0$ } \end{cases} $$ My try: $$ f(x)= -\frac {x}{2\lvert x \rvert}+\frac 12 $$ But it doesn't satisfy the condition that $x$ could be zero.
So, can you help me with that ?

1 Answers 1

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$$ f(x) = \begin{cases} 0 & \text{if $x \ge 0$} \\ 1 & \text{if $x \lt 0$ } \end{cases} $$

Is a function. It is defined for every $x\in\mathbb R$.

One way of writing it "without cases" is to say that $f(x)=\chi_{(-\infty, 0)}(x)$, because for any set $A\subseteq \mathbb R$, the "characteristic" or "indicator" function of $A$ is defined as being $1$ on $A$ and $0$ everywhere else.

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    I wanted a general notation – 2017-01-19
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    Certainly the most straightforward solution :) – 2017-01-19
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    @CarwLucas what does "general notation" mean? – 2017-01-19
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    A function that doesn't use cases – 2017-01-19
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    @CarwLucas But... how do you define $|x|$? – 2017-01-19
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    $|x|$ is a common function. I wanted to use common functions to generalize the function I stated without using cases. Sorry if the question was unclear. – 2017-01-19
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    Umm, don't you mean $\chi_{(-\infty,0)}$? – 2017-01-20
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    @ForgotALot Yeah, thanks. – 2017-01-20