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I'd like to proof the Theorem stated in the question Title.Which is

Let $(X,\tau )$ be a hausdorff topological vectorspace and $F\subseteq X$ be a finite dimensional Subspace. Show that F must be closed.

Now i am aware of the proof which uses an Isomorphism to Rn and uses local compactness of the preimage of the unit ball. However i want to prove it in another more straight forward way.

The idea is the following: $F\quad is\quad closed\quad <->\quad X\setminus F\quad is\quad open\quad <->\quad for\quad all\quad x\quad \in \quad X\setminus F\quad there\quad exists\quad an\quad open\quad neighbourhood\quad V\quad s.t\quad V\sqsubseteq X\setminus F\quad <->\quad Every\quad Point\quad in\quad X\setminus F\quad is\quad an\quad interior\quad Point\quad <->\quad X\setminus F\quad is\quad open$

Now my problem: To show the second part in the reasoning above we could show that for $x\in X\setminus F$ the set $F'\setminus F$ forms an open neighbourhood of x (in respect to the subspace topology of F') where $F'=Lin(F\bigcup \left\{ x \right\} )$. Im struggeling to show this part.

My try: Let $x\in U$ and U be open in the subspace toplogy of F' i.e U is of the Form $U=O\bigcap F'$ for some $O\in \tau $ so than we have

$O\bigcap F'=O\bigcap (Lin(F\bigcup \left\{ x \right\} )\sqsubseteq (O\bigcap Lin(F))\quad \bigcup \quad (O\bigcap \left\{ x \right\} )$

How do i continiue from here. I hope i made my problem clear. Thanks for any Help and Sorry for the bad editing!

MasterPI
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I think the result might not be true in general, if the field $\mathbb K$ over which your linear space is defined is not complete.

Maybe rather show directly that it is closed, instead of considering its complement.

If $F$ has dimension $1$, then $F=\{\lambda e:\lambda\in \mathbb K\}$, where $e$ is any fixed nonzero element of $F$. Then, show that the function that maps any element $\lambda e$ of $F$ to $\lambda\in\mathbb K$ (this map is well defined, since $\lambda$ is uniquely determined by $\lambda e$) is continuous. Hence, $F$ will automatically be closed.

Then, if $F$ has dimension $\geq 2$, you can maybe proceed by induction.

TrivialPursuit
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  • I dont see how the result depends on completness of the field K.There is a similar post on that Theorem refering to a proof of Terence Tao in his blog. HE used almost the same arguement. https://math.stackexchange.com/questions/3673587/finite-dimensional-subspaces-of-tvses – MasterPI Nov 30 '20 at 21:28
  • And shouldnt it be possible to show that F'\F (using my notation above) is an open neighbourhood without refering to a field K – MasterPI Nov 30 '20 at 21:29
  • Take $\mathbb R^2$ over the rationals. The $\mathbb Q(1,0)\subset \mathbb R^2$ is a proper subspace of $\mathbb R^2$ but it certainly is not closed. – Matematleta Nov 30 '20 at 21:37
  • Q(0,1) is not finite i.m.o. – MasterPI Nov 30 '20 at 21:53
  • No but $\mathbb Q(0,1)$ has dimension 1 (over the field of rationals), since it is spanned by $(0,1)$. More generally, if $\mathbb K$ is not complete, then for any nonzero element $e\in X$, the one dimensional $\mathbb K e$ has dimension 1 but needs not be closed if $\mathbb K$ is not complete. – TrivialPursuit Nov 30 '20 at 23:35