# How to prove that there is no $\mathbb{E} : \mathbb{R}\subset\mathbb{E}$ which would have all those properties that $\mathbb{R}$ has? [closed]

That's a theorem from my calculus course: $\neg\exists\mathbb{E}\supset\mathbb{R}$ which would have all the $\mathbb{R}$ properties.

First part: $\neg\exists n > +\infty$ (that contradicts Archimedes axiom), same with $-\infty$.

But how about proving this: $\neg\exists e\in\mathbb{E} \hspace{3mm}e \notin \mathbb{R}:\exists x,y\in\mathbb{R}:x<e<y$.

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## closed as unclear what you're asking by Jonas Meyer, Rebecca J. Stones, Mark Fantini, Adam Hughes, JohnMar 24 at 4:00

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

You should clarify what you mean by 'properties'. For example, if such a set $\mathbb{E}$ exists and properly contains $\mathbb{R}$, then it has an element which isn't contained in $\mathbb{R}$, which means that $\mathbb{E}$ doesn't have the property of not containing any non-real elements. Does this count? –  Clive Newstead Jan 15 '12 at 15:11
You really need to make your question self-contained. What are "the $\mathbb R$ properties." –  Thomas Andrews Jan 15 '12 at 15:11
@Henning: If you make that an answer, I’ll upvote it. –  Brian M. Scott Jan 30 '12 at 7:01
@Henning: That’s too bad; it would have disposed of the question pretty thoroughly. –  Brian M. Scott Jan 30 '12 at 21:33