# Is the standard structure of a topological vector space on reals unique?

The standard stucture of a topological vector space on reals is this given by the metric d(x,y)=|x-y| on the vector space $\mathbb{R},$ with the field of scalars $\mathbb R$ with standard topology.

I would like to know if we can change the topology on the reals (considered as a vector space, not a field) and still obtain the structure of a topological vector space. To be clear, the only thing I want to change is the topology of the topological vector space. I don't want to change any operations, or the field of scalars, or the topology on the field of scalars.

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If you want the topology to have a norm, then in the finite dimensional case there is only one topology. Otherwise, there are many normed topology over a real vector space which has not finite dimension. – Josué Tonelli-Cueto Jan 21 '12 at 12:23
I don't require the topology to even be metrizable if this is what you maen. – Bartek Jan 21 '12 at 12:32
Ok, normed means only to be metrizable with a normed metric, i.e. a metric that arise from $d(x,y)=|x-y|$, where $|\cdot|$ is a norm. Nevertheless, if it unique it has to be in the finite dimensional case, since in the infinite dimensional case there are norms that give not rise to the same topology. I hope this helped in some manner. – Josué Tonelli-Cueto Jan 21 '12 at 12:40
So, if we denote $\mathcal T$ the topology we are looking for and $\mathcal T_u$ the usual topology on $\mathbb R$, $\mathcal T$ has to satisfy: $M:(\alpha,x)\mapsto \alpha x$ continuous with the topology $\mathcal T\otimes\mathcal T_u$ on $\mathbb R^2$ and $\mathcal T$ on $\mathbb R$ and $S:(x,y)\to\mathcal x+y$ with the topology $\mathcal T\otimes\mathcal T$ on $\mathbb R^2$ and $\mathcal T$ on $\mathbb R$. – Davide Giraudo Jan 21 '12 at 14:52
Since for all $x_0$, $\{x_0\}$ is compact, we have $T([0,1]\times \{x_0\})=[0,x_0]$ is compact for $\mathcal T_u$ and since this topology is Hausdorff, it's closed. Hence for all $a<b$, $[a,b]$ is closed for $\mathcal T_u$. – Davide Giraudo Jan 21 '12 at 14:52

If you don't require the topology to be $T_0$, you can simply take the trivial topology on $\mathbb{R}$.
If you do require it to be $T_0$, then it must be $T_2$ as we know from the general theory. The nice argument of Davide Giraudo (see comments to the original post) shows in this case that any such topology $\mathcal{T}$ on $\mathbb{R}$ must contain every usual open set, that is $\mathcal{T_\mathrm{usual}}\subseteq\mathcal{T}$. Now you can show, if I haven't made any mistakes, that the usual topology is the only one that works, by reductio ad absurdum:
Suppose there is some $U\in\mathcal{T}$ that is not open in the usual topology. Let $F = U^C$ be its complement and let $M:\mathbb{R}_\mathrm{usual}\times\mathbb{R}\to\mathbb{R}$ be the scalar multiplication, where $\mathbb{R}$ is equiped with the topology $\mathcal{T}$ and $\mathbb{R}_\mathrm{usual}$ has the usual topology. Since $F$ is closed, $M^{-1}(F) =\lbrace(x,y)|\hbox{ }xy\in F\rbrace$ is closed and therefore its intersection with $\mathbb{R}\times\lbrace1\rbrace$ is also closed in the product topology. But $M^{-1}(F)\cap\mathbb{R}\times\lbrace1\rbrace = \lbrace(x,1)|\hbox{ }x\in F\rbrace = F\times\lbrace 1\rbrace$, so $F$ would have to be closed in the usual topology. This is not the case, so this is a contradiction.
Conclusion: there is only one such Hausdorff topology on $\mathbb{R}$, the usual one.