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Revision difference : topological vector space |
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| A topological \PMlinkescapetext{vector space} is a pair $(V,\mathcal{T})$ where $V$ is a vector space over a topological field $K$, and $\mathcal{T}$ is a topology on $V$ such that under $\mathcal{T}$, the vector space operations $v\mapsto\lambda v$ and $(v,w)\mapsto v+w$ are continuous for all $\lambda\in K$ and all $v,w\in V$. |
A topological \PMlinkescapetext{vector space} is a pair $(V,\mathcal{T})$ where $V$ is a vector space over a topological field $K$, and $\mathcal{T}$ is a topology on $V$ such that under $\mathcal{T}$, the vector space operations $v\mapsto\lambda v$ and $(v,w)\mapsto v+w$ are continuous for all $\lambda\in K$ and all $v,w\in V$. |
| A finite dimensional vector space inherits a natural topology. For if $V$ is a finite dimensional vectos space, then V is isomorphic to $K^n$ for some $N$; then let $f:V\rightarrow K^n$ be such an isomorphism, and suppose $K^n$ has the product topology. Give $V$ the topology where a subset $A$ of $V$ is open in $V$ if and only if $f(A)$ is open in $K^n$. This topology is independent of the choice of isomorphism $f$. |
A finite dimensional vector space inherits a natural topology. For if $V$ is a finite dimensional vectos space, then V is isomorphic to $K^n$ for some $N$; then let $f:V\rightarrow K^n$ be such an isomorphism, and suppose $K^n$ has the product topology. Give $V$ the topology where a subset $A$ of $V$ is open in $V$ if and only if $f(A)$ is open in $K^n$. This topology is independent of the choice of isomorphism $f$. |
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