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I read that for a finite dimensional vector space there is only Hausdorff topology that makes addition and scalar multiplication continuous and this is the usual topology on $\mathbb{C}^n$ or $\mathbb{R}^n$, and that in infinite dimensions there are many Hausdorff topologies that will make addition and scalar multiplication continuous.

How can it be the case that only one Hausdorff topology will make addition and scalar multiplication continuous in finite dimensions? Is there some proof that shows this property for finite spaces? Is there a proof that shows the property does not hold, and in fact many Hausdorff topologies have this property, in the infinite dimensional case?

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Define a real (or complex ) sequence as a function $f:\mathbb N\to \mathbb R$ (or $f:\mathbb N\to \mathbb C$). Let $X$ be the set of all bounded real (or complex) sequences. That is, if $f\in X$ then $\sup\{|f(n)|:n\in \mathbb N\}<\infty.$

For $r\in \mathbb R$ (or $r\in \mathbb C$) and $f\in X$ define $(rf)(n)=rf(n)$ for all $n\in \mathbb N.$ For $f,g \in X$ define $(f+g)(n)=f(n)+g(n)$ for all $n\in \mathbb N.$ So we have a real (or complex) vector space.

Define two metrics on the set $X$: For $f,g \in X$ let $d_1(f,g)=\sup \{|f(n)-g(n)|:n\in \mathbb N\}$ and $d_2(f,g)=\sum_{n\in \mathbb N}2^{-n}|f(n)-g(n)|$.

Let $S$ be the set of all $f\in X$ such that Re$(f(n))\in \mathbb Q$ and Im$(f(n))\in \mathbb Q$ for all $n$, and such that $\{n: f(n)\ne 0\}$ is finite. Then $S$ is countable, and with respect to the topology induced by $d_2$, the set $S$ is dense in $X.$

But with respect to the topology induced by $d_1$, no countable set is dense in $X.$ Because the set of all subsets of $\mathbb N $ is uncountable, so for each $U\subset \mathbb N$ consider the characteristic function $\chi_U\in X.$ If $U,V$ are unequal subsets of $\mathbb N$ then $d_1(\chi_U,\chi_V)=1.$

So the family $F=\{B_{d_1}(\chi_U,1/2):U\subset \mathbb N\}$ of open $d_1$-balls is a pairwise-disjoint uncountable family, and a dense subset of $X$ in the $d_1$-topology would have to have to intersect each member of $F.$

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The statement "there is only one Hausdorff topology that makes addition and scalar multiplication continuous" means that whenever you have a Hausdorff Topological Vector Space $V$ whose dimension as a vector space is finite, then there is an isomorphism of TVS (topological vector space) between $V$ and $\mathbb{R}^n$ (or $\mathbb{C}^n$), where $n=\hbox{dim}V$. By an "isomorphism of TVS", is meant a linear isomorphism of the vector-space structure, which is also a homeomorphism of the topologies. It is this homeomorphism of topologies which causes the topological structure of a finite dimensional TVS to be essentially unique.

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