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Revision difference : topological vector space
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A \emph{topological vector space} is a pair $(V,\mathcal{T})$, A \emph{topological vector space} is a pair $(V,\mathcal{T})$,
where $V$ is a vector space over a topological field $K$, where $V$ is a vector space over a topological field $K$,
and $\mathcal{T}$ is a topology on $V$ such that under $\mathcal{T}$ and $\mathcal{T}$ is a topology on $V$ such that under $\mathcal{T}$
the scalar multiplication $(\lambda,v)\mapsto\lambda v$ the scalar multiplication $(\lambda,v)\mapsto\lambda v$
is a continuous function $K\times V\to V$ is a continuous function $K\times V\to V$
and the vector addition $(v,w)\mapsto v+w$ and the vector addition $(v,w)\mapsto v+w$
is a continuous function $V\times V\to V$, is a continuous function $V\times V\to V$,
where $K\times V$ and $V\times V$ are given the respective product topologies. where $K\times V$ and $V\times V$ are given the respective product topologies.
We will also require that $\{0\}$ is closed We will also require that $\{0\}$ is closed
(which is equivalent to requiring the topology to be Hausdorff), (which is equivalent to requiring the topology to be Hausdorff),
though some authors do not make this requirement. though some authors do not make this requirement.
Many authors require that $K$ be either $\R$ or $\C$ Many authors require that $K$ be either $\R$ or $\C$
(with their usual topologies). (with their usual topologies).
A topological vector space is necessarily a topological group: A topological vector space is necessarily a topological group:
the definition ensures that the group operation (vector addition) is continuous, the definition ensures that the group operation (vector addition) is continuous,
and the inverse operation is the same as multiplication by $-1$, and the inverse operation is the same as multiplication by $-1$,
and so is also continuous. and so is also continuous.
A finite dimensional vector space inherits a natural topology. For if $V$ is a finite dimensional vector space, then $V$ is isomorphic to $K^n$ for some $n$; then let $f\colon V\rightarrow K^n$ 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 vector space, then $V$ is isomorphic to $K^n$ for some $n$; then let $f\colon V\rightarrow K^n$ 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 $f$.