I am trying to define the Absolute function in set notation. $$ |x|= \begin{cases} x, \text{ if x $\ge$ 0} \\ -x, \text{ if x $<$ 0} \end{cases} $$

Here are my two attempts: $$f_1=\{(x,y)\in\mathbb{R \times R}: (x\ge0\implies y=x) \land (x<0\implies y=-x)\}$$ $$f_2=\{(x,y)\in\mathbb{R \times R}: (x\ge0\land y=x) \lor (x<0\land y=-x)\}$$

The first attempt is restating the definition but in set notation.

The set written should be ONLY one set. No unions are allowed.


1) Which definition is correct(or both are wrong) ? Explain why.

2) Is there any mechanical way for writing functions in set notation given their cases ?

3) Assuming one of the above are correct how can I prove that it is a function ? If none are correct then provide correct one with explanation.

4) I am not sure but I think $f_1=f_2$. Is that right ?

  • 1
    $\begingroup$ I would write $$\{(x,x)\mid x\in \mathbb R, x\ge 0\}\cup\{(-x,x)\mid x\in\mathbb R, x \ge 0 \}$$ $\endgroup$ – Henning Makholm Oct 31 '16 at 22:32
  • $\begingroup$ @HenningMakholm I wrote it that way first. But I am trying to have it all in one set notation not two sets. $\endgroup$ – Nameless Oct 31 '16 at 22:34

One possibility could be $\;\{(x,y) \in \mathbb{R} \times \mathbb{R}^+ \mid x^2 = y^2\}\;$.


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