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Let $a \in \mathbb{R}$, and consider the half open interval $(a,a]$.

Is it correct to write this half open interval as $(a,a]=\{\emptyset \}$? Or $(a,a]=\{a \}$?

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    $\begingroup$ $(x,y]:=\{r \in \mathbf{R}: x<r\le y\}$. Hence it is empty. $\endgroup$ – Paolo Leonetti Aug 4 '15 at 14:25
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    $\begingroup$ It is not the set containing the empty set, it is just the empty set $\endgroup$ – BadAtMaths Aug 4 '15 at 14:27
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    $\begingroup$ $\varnothing\neq\{\varnothing\}$. $\endgroup$ – Asaf Karagila Aug 4 '15 at 14:32
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    $\begingroup$ Possible duplicate of Is $[a, a)$ equal to $\{a\}$ or $\varnothing$? $\endgroup$ – BCLC Nov 22 '15 at 22:42
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No, $(a,a]$ has no elements. It is empty. There is no real number $x$ such that $a < x \le a$. But you do not write that as $\{\varnothing\}$. You write it as $\varnothing$. See the difference?

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  • $\begingroup$ I did not know this difference was important, so thanks. So I can clarify, $\{ \emptyset \}$ is the set containing the empty set; is this any set containing the empty set? $\endgroup$ – möbius Aug 4 '15 at 14:37
  • $\begingroup$ @möbius THE set containing the empty set. There "is" only one, since if two sets have the same elements, they are the same set. $\endgroup$ – 5xum Aug 4 '15 at 14:39
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    $\begingroup$ @möbius No. it is the set with one element, and this element is the empty set. $\endgroup$ – Crostul Aug 4 '15 at 14:39
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No. Because $\emptyset$, the empty set, is not a real number, i.e. it is not an element of $\mathbb R$, the set $\{\emptyset\}$ is not a subset of $\mathbb R$. ON the other hand, $(a,a]$ is a subset of $\mathbb R$, so there can be no equality.

You are close, though, since $(a,a]$ is the empty set, so $(a,a]=\emptyset=\{\}$.

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  • $\begingroup$ I agree that $\varnothing$ is not a real number. But (in some formulations) it may be. Those weird model theorists start with zero as $\varnothing$ and then build up the number system from that... $\endgroup$ – GEdgar Aug 4 '15 at 14:44
  • $\begingroup$ Only the natural number zero is the empty set; the real number zero is either a Dedekind cut or an equivalence class of Cauchy sequences, and in either case the real number zero is not the empty set. @GEdgar $\endgroup$ – Carl Mummert Aug 9 '15 at 16:11
  • $\begingroup$ @CarlMummert In the standard model of real numbers. It is possible to construct a model of reals that has empty set as zero. $\endgroup$ – 5xum Aug 9 '15 at 18:14
  • $\begingroup$ @5xum: well, we can always arbitrarily make any particular number be any particular set. But the usual two constructions of the reals - via Dedekind cuts or via equivalence classes of Cauchy sequences - do not make any real number be represented by the empty set. The real number 0 is not represented in the same way as the natural number 0. $\endgroup$ – Carl Mummert Aug 9 '15 at 18:30

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