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The $\emptyset$ must be in $\mathbb{R}$ because every non-empty set has $\emptyset$. But if $\emptyset \in\mathbb{R}$ that that implies that there are empty holes in $\mathbb{R}$ contradicts the completeness property of real numbers.

Can you see $\emptyset$ in the real line?

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    The empty set is a subset of every set, not an element of every set. – José Carlos Santos Oct 17 '19 at 08:46
  • Thank you. That clarifies things. –  Oct 17 '19 at 08:58
  • Standard nitpicking: $\emptyset=0\in\mathbb R$. – Michael Hoppe Oct 17 '19 at 09:47
  • But is it?... An $\emptyset$ is a set that contains nothing. The $0$ is an element, not a set. Hence an element can't equal a set, or can it? –  Oct 17 '19 at 10:55
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    @MichaelHoppe nitpicking nitpicking: no, $0\in\omega$, but we define $\Bbb R$ to be(one of the ways) $\mathbb Q^\omega/\sim$ with $\Bbb Q=\omega^4$, so $\emptyset\notin\Bbb R$. You can decide to define $\Bbb R$ with $0$ in it(or with $\omega$ as the natural of the reals) but if you nitpick, nitpick correctly – ℋolo Oct 17 '19 at 12:15
  • It is possible to construct a field with the least upper bound property such that $\emptyset$ is an element. But $\mathbb R$ is assumed to be a set whose elements are distinct from other mathematical objects, even though it may be isomorphic to that structure you constructed. –  Oct 17 '19 at 15:04

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