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I asked proving isomorphism between the next structures: $$M_1=<\left(2,\infty\right),\cdot>,\ M_2=<\left({\sqrt2}^{\sqrt2},\infty\right),\cdot>$$ I have tried to call a function: $H(x)=x^{\sqrt{\frac{1}{x}}}$, however, I don't know how to prove that for every real number in the range, there is a real number in the domain. Thanks for the help!

  • @NoahSchweber No. Have a look at the structures, this function is exactly like: $H(x) = {\sqrt x}^{\sqrt x}$. –  Dec 16 '20 at 16:38
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    @AtticusStonestrom lol it is asked before. Thank you! –  Dec 16 '20 at 16:40
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    As noted above, the particular isomorphism question has been asked before. However, you ask a technical sub-question, namely how to show that a complicated function is in fact surjective. The key point is the intermediate value theorem, or more snappily the connectedness of $\mathbb{R}$. It's easy to check that $H$ takes on arbitrarily large values, and $H(2)={\sqrt{2}}^{\sqrt{2}}$; IVT then shows that $(\sqrt{2}^{\sqrt{2}},\infty)\subseteq ran(H)$. – Noah Schweber Dec 16 '20 at 16:42
  • @Noah Schweber Can I prove that for every real number in the range, there is a real number in the domain, by variables? if so, can you show me how? –  Dec 16 '20 at 16:45
  • What do you mean "prove by variables?" – Noah Schweber Dec 16 '20 at 16:46
  • @Noah Schweber I meant to take a general element of the discussion world and prove by substitution that every real number in the other structure's discussion world holds that. –  Dec 16 '20 at 16:50
  • I'm still not really sure what you mean, but if you're asking for a closed form expression of $H^{-1}$, I doubt such a thing exists. – Noah Schweber Dec 16 '20 at 16:50
  • @Noah Schweber I mean that: suppose there is t, now how can I show that $t = {\sqrt x}^{\sqrt x}$ and find to what $x$ equals... –  Dec 16 '20 at 17:32
  • I think that my comment above applies then: as far as I know, $H^{-1}$ has no closed form. You could ask this as a separate question though, I'm not an expert in this area. – Noah Schweber Dec 16 '20 at 17:33
  • @NoahSchweber Ok, but if she has no closed-form, then she isn't isomorphism. No? –  Dec 16 '20 at 17:37
  • Nope - not having a closed form has nothing to do with being an isomorphism. Some isomorphisms are just really complicated. – Noah Schweber Dec 16 '20 at 17:37
  • @NoahSchweber oh I see. Thanks! –  Dec 16 '20 at 17:38

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