Do algebraically open sets define a vector space topology?












1














Let $X$ be a vector space. A vector space topology on $X$ is a topology such that addition and scalar multiplication are continuous. A subset $A$ of $X$ is said to be algebraically open if, for all $ain A$ and $xin X$, there exists $varepsilon>0$ such that $a+(-varepsilon,varepsilon)cdot xsubseteq A$. My question is if the algebraically open sets form a vector space topology on $X$.






functional-analysis topological-vector-spaces







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  • You might notice I posted an answer to my own question rather quickly. There was another question (math.stackexchange.com/questions/3058837/…) which was answered, but this was a related question that did not get answered. I wanted to share my answer to this question from there, but I figured it deserved its own question.
    – SmileyCraft
    Jan 3 at 20:58
















1














Let $X$ be a vector space. A vector space topology on $X$ is a topology such that addition and scalar multiplication are continuous. A subset $A$ of $X$ is said to be algebraically open if, for all $ain A$ and $xin X$, there exists $varepsilon>0$ such that $a+(-varepsilon,varepsilon)cdot xsubseteq A$. My question is if the algebraically open sets form a vector space topology on $X$.






functional-analysis topological-vector-spaces







share|cite|improve this question






















  • You might notice I posted an answer to my own question rather quickly. There was another question (math.stackexchange.com/questions/3058837/…) which was answered, but this was a related question that did not get answered. I wanted to share my answer to this question from there, but I figured it deserved its own question.
    – SmileyCraft
    Jan 3 at 20:58














1












1








1







Let $X$ be a vector space. A vector space topology on $X$ is a topology such that addition and scalar multiplication are continuous. A subset $A$ of $X$ is said to be algebraically open if, for all $ain A$ and $xin X$, there exists $varepsilon>0$ such that $a+(-varepsilon,varepsilon)cdot xsubseteq A$. My question is if the algebraically open sets form a vector space topology on $X$.






functional-analysis topological-vector-spaces







share|cite|improve this question













Let $X$ be a vector space. A vector space topology on $X$ is a topology such that addition and scalar multiplication are continuous. A subset $A$ of $X$ is said to be algebraically open if, for all $ain A$ and $xin X$, there exists $varepsilon>0$ such that $a+(-varepsilon,varepsilon)cdot xsubseteq A$. My question is if the algebraically open sets form a vector space topology on $X$.






functional-analysis topological-vector-spaces




functional-analysis topological-vector-spaces






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share|cite|improve this question











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asked Jan 3 at 20:53









SmileyCraft

3,176416




3,176416












  • You might notice I posted an answer to my own question rather quickly. There was another question (math.stackexchange.com/questions/3058837/…) which was answered, but this was a related question that did not get answered. I wanted to share my answer to this question from there, but I figured it deserved its own question.
    – SmileyCraft
    Jan 3 at 20:58


















  • You might notice I posted an answer to my own question rather quickly. There was another question (math.stackexchange.com/questions/3058837/…) which was answered, but this was a related question that did not get answered. I wanted to share my answer to this question from there, but I figured it deserved its own question.
    – SmileyCraft
    Jan 3 at 20:58
















You might notice I posted an answer to my own question rather quickly. There was another question (math.stackexchange.com/questions/3058837/…) which was answered, but this was a related question that did not get answered. I wanted to share my answer to this question from there, but I figured it deserved its own question.
– SmileyCraft
Jan 3 at 20:58




You might notice I posted an answer to my own question rather quickly. There was another question (math.stackexchange.com/questions/3058837/…) which was answered, but this was a related question that did not get answered. I wanted to share my answer to this question from there, but I figured it deserved its own question.
– SmileyCraft
Jan 3 at 20:58










1 Answer
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The answer is negative. We know that every finite dimensional topological vector space is topologically isomorphic to $mathbb{K}^n$ with the Euclidean topology, for some $ninmathbb{N}$. However, there exist algebraically open subsets of $mathbb{R}^2$ that are not open by the Euclidean topology.






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  • When you say topologically isomorphic, do you mean there exists an isomorphism of TVS between them (i.e continues operator with continues inverse), or that they are isomorphic under the identity map? If you mean the former I think this requires more explanation.
    – pitariver
    2 days ago












  • Topologically isomorphic means there exists a bijection preserving addition, scalar multiplication and topology, so a homeomorphic isomorphism. You can check the link for details.
    – SmileyCraft
    2 days ago










  • Then why does this create a contradiction? take an algebraically open subset $A$ of the plane that is not open, under the isomorphism $T:(mathbb{R}^2, algeb. open) rightarrow (mathbb{R}^2, usual)$, T(A) may not be A, thus may be open. I think you need a specific example from the post you mentioned.
    – pitariver
    2 days ago








  • 1




    @pitariver Since $T$ is a bijective linear map in finite dimensions, it is continuous with respect to the usual topology. Thus, if $T(A)$ were open, then $T^{-1}(T(A))=A$ would be open.
    – SmileyCraft
    2 days ago













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1 Answer
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The answer is negative. We know that every finite dimensional topological vector space is topologically isomorphic to $mathbb{K}^n$ with the Euclidean topology, for some $ninmathbb{N}$. However, there exist algebraically open subsets of $mathbb{R}^2$ that are not open by the Euclidean topology.






share|cite|improve this answer





















  • When you say topologically isomorphic, do you mean there exists an isomorphism of TVS between them (i.e continues operator with continues inverse), or that they are isomorphic under the identity map? If you mean the former I think this requires more explanation.
    – pitariver
    2 days ago












  • Topologically isomorphic means there exists a bijection preserving addition, scalar multiplication and topology, so a homeomorphic isomorphism. You can check the link for details.
    – SmileyCraft
    2 days ago










  • Then why does this create a contradiction? take an algebraically open subset $A$ of the plane that is not open, under the isomorphism $T:(mathbb{R}^2, algeb. open) rightarrow (mathbb{R}^2, usual)$, T(A) may not be A, thus may be open. I think you need a specific example from the post you mentioned.
    – pitariver
    2 days ago








  • 1




    @pitariver Since $T$ is a bijective linear map in finite dimensions, it is continuous with respect to the usual topology. Thus, if $T(A)$ were open, then $T^{-1}(T(A))=A$ would be open.
    – SmileyCraft
    2 days ago


















1














The answer is negative. We know that every finite dimensional topological vector space is topologically isomorphic to $mathbb{K}^n$ with the Euclidean topology, for some $ninmathbb{N}$. However, there exist algebraically open subsets of $mathbb{R}^2$ that are not open by the Euclidean topology.






share|cite|improve this answer





















  • When you say topologically isomorphic, do you mean there exists an isomorphism of TVS between them (i.e continues operator with continues inverse), or that they are isomorphic under the identity map? If you mean the former I think this requires more explanation.
    – pitariver
    2 days ago












  • Topologically isomorphic means there exists a bijection preserving addition, scalar multiplication and topology, so a homeomorphic isomorphism. You can check the link for details.
    – SmileyCraft
    2 days ago










  • Then why does this create a contradiction? take an algebraically open subset $A$ of the plane that is not open, under the isomorphism $T:(mathbb{R}^2, algeb. open) rightarrow (mathbb{R}^2, usual)$, T(A) may not be A, thus may be open. I think you need a specific example from the post you mentioned.
    – pitariver
    2 days ago








  • 1




    @pitariver Since $T$ is a bijective linear map in finite dimensions, it is continuous with respect to the usual topology. Thus, if $T(A)$ were open, then $T^{-1}(T(A))=A$ would be open.
    – SmileyCraft
    2 days ago
















1












1








1






The answer is negative. We know that every finite dimensional topological vector space is topologically isomorphic to $mathbb{K}^n$ with the Euclidean topology, for some $ninmathbb{N}$. However, there exist algebraically open subsets of $mathbb{R}^2$ that are not open by the Euclidean topology.






share|cite|improve this answer












The answer is negative. We know that every finite dimensional topological vector space is topologically isomorphic to $mathbb{K}^n$ with the Euclidean topology, for some $ninmathbb{N}$. However, there exist algebraically open subsets of $mathbb{R}^2$ that are not open by the Euclidean topology.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 3 at 20:54









SmileyCraft

3,176416




3,176416












  • When you say topologically isomorphic, do you mean there exists an isomorphism of TVS between them (i.e continues operator with continues inverse), or that they are isomorphic under the identity map? If you mean the former I think this requires more explanation.
    – pitariver
    2 days ago












  • Topologically isomorphic means there exists a bijection preserving addition, scalar multiplication and topology, so a homeomorphic isomorphism. You can check the link for details.
    – SmileyCraft
    2 days ago










  • Then why does this create a contradiction? take an algebraically open subset $A$ of the plane that is not open, under the isomorphism $T:(mathbb{R}^2, algeb. open) rightarrow (mathbb{R}^2, usual)$, T(A) may not be A, thus may be open. I think you need a specific example from the post you mentioned.
    – pitariver
    2 days ago








  • 1




    @pitariver Since $T$ is a bijective linear map in finite dimensions, it is continuous with respect to the usual topology. Thus, if $T(A)$ were open, then $T^{-1}(T(A))=A$ would be open.
    – SmileyCraft
    2 days ago




















  • When you say topologically isomorphic, do you mean there exists an isomorphism of TVS between them (i.e continues operator with continues inverse), or that they are isomorphic under the identity map? If you mean the former I think this requires more explanation.
    – pitariver
    2 days ago












  • Topologically isomorphic means there exists a bijection preserving addition, scalar multiplication and topology, so a homeomorphic isomorphism. You can check the link for details.
    – SmileyCraft
    2 days ago










  • Then why does this create a contradiction? take an algebraically open subset $A$ of the plane that is not open, under the isomorphism $T:(mathbb{R}^2, algeb. open) rightarrow (mathbb{R}^2, usual)$, T(A) may not be A, thus may be open. I think you need a specific example from the post you mentioned.
    – pitariver
    2 days ago








  • 1




    @pitariver Since $T$ is a bijective linear map in finite dimensions, it is continuous with respect to the usual topology. Thus, if $T(A)$ were open, then $T^{-1}(T(A))=A$ would be open.
    – SmileyCraft
    2 days ago


















When you say topologically isomorphic, do you mean there exists an isomorphism of TVS between them (i.e continues operator with continues inverse), or that they are isomorphic under the identity map? If you mean the former I think this requires more explanation.
– pitariver
2 days ago






When you say topologically isomorphic, do you mean there exists an isomorphism of TVS between them (i.e continues operator with continues inverse), or that they are isomorphic under the identity map? If you mean the former I think this requires more explanation.
– pitariver
2 days ago














Topologically isomorphic means there exists a bijection preserving addition, scalar multiplication and topology, so a homeomorphic isomorphism. You can check the link for details.
– SmileyCraft
2 days ago




Topologically isomorphic means there exists a bijection preserving addition, scalar multiplication and topology, so a homeomorphic isomorphism. You can check the link for details.
– SmileyCraft
2 days ago












Then why does this create a contradiction? take an algebraically open subset $A$ of the plane that is not open, under the isomorphism $T:(mathbb{R}^2, algeb. open) rightarrow (mathbb{R}^2, usual)$, T(A) may not be A, thus may be open. I think you need a specific example from the post you mentioned.
– pitariver
2 days ago






Then why does this create a contradiction? take an algebraically open subset $A$ of the plane that is not open, under the isomorphism $T:(mathbb{R}^2, algeb. open) rightarrow (mathbb{R}^2, usual)$, T(A) may not be A, thus may be open. I think you need a specific example from the post you mentioned.
– pitariver
2 days ago






1




1




@pitariver Since $T$ is a bijective linear map in finite dimensions, it is continuous with respect to the usual topology. Thus, if $T(A)$ were open, then $T^{-1}(T(A))=A$ would be open.
– SmileyCraft
2 days ago






@pitariver Since $T$ is a bijective linear map in finite dimensions, it is continuous with respect to the usual topology. Thus, if $T(A)$ were open, then $T^{-1}(T(A))=A$ would be open.
– SmileyCraft
2 days ago




















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