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Tom Goodwillie
Tom Goodwillie
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Any two $2$-dimensional rational vector spaces in $\mathbb R$ are isomorphic as groups and also order-isomorphic (I believe), hence homeomorphic when given the relative topology from $\mathbb R$. But any isomorphism as topological groups would have to be given by multiplication by a real number. So for example $\mathbb Q+\mathbb Q\pi$ and $\mathbb Q+\mathbb Q\sqrt 5$ are a counterexample.

EDIT: Or the same thing with $\mathbb Z$ instead of $\mathbb Q$.

Any two $2$-dimensional rational vector spaces in $\mathbb R$ are isomorphic as groups and also order-isomorphic (I believe), hence homeomorphic when given the relative topology from $\mathbb R$. But any isomorphism as topological groups would have to be given by multiplication by a real number. So for example $\mathbb Q+\mathbb Q\pi$ and $\mathbb Q+\mathbb Q\sqrt 5$ are a counterexample.

Any two $2$-dimensional rational vector spaces in $\mathbb R$ are isomorphic as groups and also order-isomorphic (I believe), hence homeomorphic when given the relative topology from $\mathbb R$. But any isomorphism as topological groups would have to be given by multiplication by a real number. So for example $\mathbb Q+\mathbb Q\pi$ and $\mathbb Q+\mathbb Q\sqrt 5$ are a counterexample.

EDIT: Or the same thing with $\mathbb Z$ instead of $\mathbb Q$.

Source Link
Tom Goodwillie
Tom Goodwillie
  • 55.9k
  • 7
  • 151
  • 240

Any two $2$-dimensional rational vector spaces in $\mathbb R$ are isomorphic as groups and also order-isomorphic (I believe), hence homeomorphic when given the relative topology from $\mathbb R$. But any isomorphism as topological groups would have to be given by multiplication by a real number. So for example $\mathbb Q+\mathbb Q\pi$ and $\mathbb Q+\mathbb Q\sqrt 5$ are a counterexample.