Do algebraically open sets define a vector space topology?
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
asked Jan 3 at 20:53
SmileyCraft
3,176416
add a comment |
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
asked Jan 3 at 20:53
SmileyCraft
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
add a comment |
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
asked Jan 3 at 20:53
SmileyCraft
3,176416
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
asked Jan 3 at 20:53
SmileyCraft
3,176416
asked Jan 3 at 20:53
SmileyCraft
3,176416
asked Jan 3 at 20:53
SmileyCraft
3,176416
asked Jan 3 at 20:53
SmileyCraft
3,176416
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
add a comment |
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
add a comment |
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.
answered Jan 3 at 20:54
SmileyCraft
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
add a comment |
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1 Answer
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1 Answer
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oldest
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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.
answered Jan 3 at 20:54
SmileyCraft
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
add a comment |
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.
answered Jan 3 at 20:54
SmileyCraft
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
add a comment |
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.
answered Jan 3 at 20:54
SmileyCraft
3,176416
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.
answered Jan 3 at 20:54
SmileyCraft
3,176416
answered Jan 3 at 20:54
SmileyCraft
3,176416
answered Jan 3 at 20:54
SmileyCraft
3,176416
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
add a comment |
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
add a comment |
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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