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Problem: Let $X$ be a normed space with $\dim{X} \geq 2$ and $(x_n)$ a sequence of non-zero points in $X$. Show that there exists a non-zero closed subspace $Y \leq X$, such that $Y$ does not contain any of the points of the sequence $(x_n)$.

My thought about the proof is that we need to define a subspace $Y= <y>$ (of dimension $1$ ) which doesn't contain any of the points $x_n$, of course. Can you give me any ideas?

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To prove this it's enough to show that $\cup_{n\in \mathbb{N}} \mathbb{R}x_n \neq X$. Then there exists $y \in X$ such that $<y>=Y$ as you want.

Since $dim(X)\geq2$, for each $n \in \mathbb{N}$ the set $\mathbb{R} x_n$ is a nowhere dense set. Then the set $A:= \cup_{n \in \mathbb{N}} \mathbb{R}x_n$ is a set whose interior is the empty set, according to Baire's category theorem.

Therefore $A\neq X$.

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  • $\begingroup$ Thank you for your answer! can you please make it more clear why the union $A$ of nowhere dense sets $\mathbb{R}x_n$ has an empty interior? Shouldn't $X$ be a complete space in order to be true? $\endgroup$ – dimvolt Mar 11 '17 at 11:19
  • $\begingroup$ You can embedd your space in a complete space for the theorem to work. $\endgroup$ – Keen-ameteur Mar 11 '17 at 11:25
  • $\begingroup$ I am at the beginning of "Introduction to Functional Analysis" course, so I don't think I already have the tools to do that. My only tools are some knowledge from Real Analysis (Metric spaces inculding Baire category theorem). But thank you again! $\endgroup$ – dimvolt Mar 11 '17 at 11:33
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    $\begingroup$ If your space is of a finite dimension it's already complete. If not I don't currently see an alternative way without the theorem, but maybe someone else will. $\endgroup$ – Keen-ameteur Mar 11 '17 at 11:41
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    $\begingroup$ It suffices to work in $\mathbb{R}^2$: Since $\dim X\geq 2$, $X$ contains a two dimensional subspace $Z$. Then apply the previous reasoning for $Z$ instead of $X$. An alternative approach without Baire's theorem, is to consider $y, z\in X$ linearly independent and for every $t>0$ set $Y_t=\langle y+tz\rangle.$ Then the collection $\{Y_t\}_{t>0}$ consists of uncountably many closed subspaces with the property that $Y_t\cap Y_s=\{0\}$, for $t\neq s$, so the members of the sequence $(x_n)$ can only belong to at most countably many of them. $\endgroup$ – tree detective Mar 12 '17 at 15:59

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