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Since square of real numbers is zero or positive, can we write set builder form of real number set as $$\{x|x^2 \geq 0\}$$

Asaf Karagila
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my stak
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    Usually in set builder notation, we signify where we are taking the elements from. So you should specify this by writing something like ${x \in \mathbb{R} \mid x^2 \geq 0}.$ – green frog Jul 10 '19 at 10:42
  • @伽罗瓦 Good point, but with your definition, you would be defining $\mathbb{R}$ using $\mathbb{R}$ itself. Maybe ${x \in \mathbb{C} \mid x^2 \geq 0, Im(x)=0}$, but this would require as to have the complex numbers defined first. And also, that order relation $\geq$ is probably undefind unless we have the real numbers first – David Jul 10 '19 at 10:44
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    @David There doesn't seem to be any way to interpret this notation other than a circular definition of the real numbers. – Peter Foreman Jul 10 '19 at 10:45
  • @david32 you mean just like 1st comment? – my stak Jul 10 '19 at 10:48

1 Answers1

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Yes, and no.

When you write a set builder notation, you have some implicit assumption on the background universe. Namely, where do these objects exist? Normally, you would say this is just the universe of "all things", but then what does it mean for $x^2$ and $0$ and $\geq$ between two arbitrary mathematical objects?

So you can limit your scope to the real numbers. In which case, yes. That is correct, all the real numbers satisfy this property. But you can also say the same about the rational numbers, $\Bbb Q=\{x\in\Bbb Q\mid x^2\geq 0\}$, or any subset of the real numbers really.

And you can also go to $\Bbb C$, where $x^2$ makes sense, but since there is no inherent order on $\Bbb C$, there's no meaning to $x^2\geq 0$. And the only reasonable way to interpret it is to say a priori that $a\geq b$ is only defined when $a$ and $b$ are real numbers. But then... well, circularity.

On the other hand, there are other field extensions of the real numbers where $x^2\geq 0$ makes sense, but then $\Bbb R$ is a proper subset of $\{x^2\mid x^2\geq 0\}$, because some (or all) of our added elements satisfy this property as well.

Therefore the answer is "depends on your implicit universe from which you take your objects, and if the answer is yes then it's kinda trivially true".

Of course, you can overcome this difficulty by specifying what is the set from where the elements are taken, e.g., as I wrote above in the case of $\Bbb Q$ or as it were suggested in the comments.

Asaf Karagila
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