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I have this statement:

If $\mathbb{A} = (-\infty, \infty)$ write the interval in conjunctionist notation.

First I thought:: $\forall{x} \in \mathbb{Z}:x \in \mathbb{A}$

However, I am deleting irrational numbers, so after I did this:

Second answer: $\forall{x} \in \mathbb{R}:x \in \mathbb{A}$

However, I am deleting complex numbers, so finally is this:

$\forall{x} \in \mathbb{C}:x \in \mathbb{A}$, since the set of real numbers is a subset of the set of complex numbers.

But according to the guide my answer is incorrect and must be the second answer, and I have no idea why. Thanks in advance.

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  • $\begingroup$ $(-\infty, \infty)$ generally means all real numbers, not all complex numbers. $\endgroup$ – wgrenard May 29 at 0:51
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The complex numbers, while orderable (since in bijection with the reals, which are canonically ordered), are not orderable in a way that is compatible with the operations there. Thus, since $(-\infty,\infty)$ indicates an interval, and thus implies some order, we must assume (unless given more information) that it indicates a subset of the reals--typically, the set of all real numbers.

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