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From Conway, A course in functional analysis, page 104.

If $H$ is a finite dimensional vector space and $F_{1},F_{2}$ are two topologies on $H$, that makes $H$ into a TVS. Then $F_{1}=F_{2}$.

I do not really know how to prove this because $H$ may not be normable if all its non-empty open sets are unbounded. If $H$ is normable, then a subtle argument showed any norm on a finite dimensional space is equivalent to each other, hence $F_{1}=F_{2}$. (See Conway page 69). But I do not really know what to do when $H$ is just a finite dimensional TVS with no additional structure given. Even in the one dimensional case, it seems to me that $H$ may have different topologies. However, I also do not know how to construct one which is continuous with respect to addition and multiplication but not coincide withthe usual topology (in one dimensional case).

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    $\begingroup$ I suppose Conway demands that a TVS is Hausdorff? (Otherwise it's trivially wrong.) $\endgroup$ – Daniel Fischer Jul 17 '13 at 5:22
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    $\begingroup$ @QiaochuYuan, Hausdorffness is probably in the definition of TVS in Conway. The possibility that the field be some other or that it is one of those two but with a different topology are more or less pathologies: I honestly do not think belaboring those points would help anyone here. $\endgroup$ – Mariano Suárez-Álvarez Jul 17 '13 at 5:26
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    $\begingroup$ (Conway says in the book that all his topological spaces are Hausdorff, in fact) $\endgroup$ – Mariano Suárez-Álvarez Jul 17 '13 at 5:28
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    $\begingroup$ Page 103. The example on page 67 is probably where the convention is instated. $\endgroup$ – Mariano Suárez-Álvarez Jul 17 '13 at 5:30
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    $\begingroup$ terrytao.wordpress.com/2011/05/24/… seems relevant $\endgroup$ – kahen Jul 17 '13 at 5:33

With any bijection between two sets, if one of the sets carries a topology, you can transport the topology to the other. The bijection then becomes a homeomorphism.

For topological vector spaces, of course not every bijection will transport a TVS topology on the one space to a TVS topology on the other, but a linear isomorphism will (elementary but tedious verifications left as an exercise).

So given any finite-dimensional $\mathbb{K}$ vector space $V$, with any linear isomorphism, we can transport its topology to $\mathbb{K}^n$.

The claim is: If the topology $\mathcal{T}$ on $V$ was a Hausdorff TVS topology, then the topology $\mathcal{T}_V$ on $\mathbb{K}^n$ induced by the linear isomorphism is the standard (product) topology $\mathcal{S}_n$ on $\mathbb{K}^n$.

Let $(e_i)_{1 \leqslant i \leqslant n}$ the standard basis in $\mathbb{K}^n$.

First, we note that the map $\lambda \mapsto \lambda\cdot e_i$ is continuous as a map $\bigl(\mathbb{K},\,\mathcal{S}_1\bigr) \to \bigl(\mathbb{K}^n,\, \mathcal{T}_V\bigr)$, since $\mathcal{T}_V$ is a TVS topology.

Hence the map $\mathbb{K}^n \to \bigl(\mathbb{K}^n\bigr)^n$ given by $(\lambda_1,\, \dotsc,\,\lambda_n) \mapsto (\lambda_1\cdot e_1,\, \dotsc,\, \lambda_n\cdot e_n)$ is continuous for the product topologies obtained from $\mathcal{S}_1$ resp. $\mathcal{T}_V$.

Now, $\mathcal{T}_V$ being a TVS topology, addition is continuous, therefore

$$\begin{gather} \Phi \colon \bigl(\mathbb{K}^n,\, \mathcal{S}_n\bigr) \to \bigl(\mathbb{K}^n,\, \mathcal{T}_V\bigr)\\ (\lambda_1,\, \dotsc,\, \lambda_n) \mapsto \sum_{i = 1}^n \lambda_i \cdot e_i = (\lambda_1,\, \dotsc,\, \lambda_n) \end{gather}$$

is continuous ($\mathcal{S}_n$ is the product topology of $\mathcal{S}_1$).


The standard topology is the finest TVS topology that $\mathbb{K}^n$ can carry.

Note that we did not assume the Hausdorff property, so that result holds even if Hausdorffness is not required.

Next, we remark

In a topological $\mathbb{K}$ vector space $E$, $0$ has a neighbourhood basis consisting of balanced sets.

(A set $B \subset E$ is balanced if $(\forall \lambda \in \mathbb{K})(\lvert \lambda\rvert \leqslant 1 \Rightarrow \lambda\cdot B \subset B)$.)

That follows from the continuity of scalar multiplication $\mathbb{K} \times E \to E$ in $(0,\,0)$, since that demands that for each neighbourhood $N$ of $0$ in $E$, there is a neighbourhood $D\times M$ of $(0,\,0)$ such that $D\cdot M \subset N$. $D$ contains a disk $D_\varepsilon = \{z \colon \lvert z\rvert < \varepsilon\}$ and $D_\varepsilon \cdot M$ is easily verified to be balanced.

Now, let's finish the proof.

If $\mathcal{T}$ is Hausdorff, so is $\mathcal{T}_V$ (trivial verification).

The closed unit ball $B$ in $\mathbb{K}^n$ is compact in $\mathcal{S}_n$, and so is its boundary $S$. Since $\Phi$ is continuous, and $\mathcal{T}_V$ Hausdorff, $\Phi(B)$ and $\Phi(S)$ are compact (in $\mathcal{T}_V$).

Since $0 \notin \Phi(S)$, for each $x \in \Phi(S)$, there is a balanced neighbourhood $U_x$ of $0$ and a neighbourhood $W_x$ of $x$ such that $U_x \cap W_x = \varnothing$.

Since $\Phi(S)$ is compact, there exist finitely many $x_1,\, \dotsc,\, x_k$ such that $\Phi(S) \subset \bigcup\limits_{i = 1}^k W_{x_i}$. Then $U := \bigcap\limits_{i=1}^k U_{x_i}$ is a balanced ($\mathcal{T}_V$) neighbourhood of $0$ that doesn't intersect $\Phi(S)$.

$\Phi^{-1}(U)$ is a balanced ($\mathcal{S}_n$) neighbourhood of $0$ that doesn't intersect $S$, hence $\Phi^{-1}(U) \subset \overset{\circ}{B}$, whence $U \subset \Phi(\overset{\circ}{B})$.

By linearity, it follows that for every $\mathcal{S}_n$-neighbourhood $N$ of $0$, $\Phi(N)$ is a $\mathcal{T}_V$-neighbourhood of $0$.

That in turn implies that $\Phi$ is open.

An open and continuous bijection is a homeomorphism, hence $\mathcal{T}_V = \mathcal{S}_n$. $\qquad$ c.q.f.d.

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    $\begingroup$ Now reading this carefully, it is much more detailed (and long winded) in comparison with Tao's proof, but it is a proof nevertheless. So thanks. $\endgroup$ – Bombyx mori Jul 22 '13 at 6:00

To turn $V$ into a TVS:

  • Pick any linear isomorphism $V\to\mathbb{R}^n$ for some $n$ and put on $V$ the unique topology which makes it a homeo.

  • Show that the topology you get this way does not depend on the linear isomorphism you started with.

  • $\begingroup$ You still need to prove that $\mathbb{K}^n$ carries only one Hausdorff TVS topology. $\endgroup$ – Daniel Fischer Jul 17 '13 at 5:31
  • $\begingroup$ @DanielFischer, I did no claim to have done that... $\endgroup$ – Mariano Suárez-Álvarez Jul 17 '13 at 5:32
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    $\begingroup$ The second bullet may be interpreted as though it claimed to be a proof of the uniqueness of the topology for finite dimensional spaces. I thought one should make explicit that it isn't. $\endgroup$ – Daniel Fischer Jul 17 '13 at 5:35
  • $\begingroup$ The second bullet point tells us that the topology you get in the way explained in the first bullet does not depend on the isomorphism. $\endgroup$ – Mariano Suárez-Álvarez Jul 17 '13 at 5:37
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    $\begingroup$ You are confused, methinks. $\endgroup$ – Mariano Suárez-Álvarez Jul 17 '13 at 5:45

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