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Let $X$ be a normed space and let $Y \neq X$ be its vector subspace. Prove that Y does not contain any nonempty open set in X.

Could someone help please with the proof? Thanks in advance! :)

rschwieb
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  • Any finite set in $Y$ is open. So, counterexample? – Sean Roberson Jan 20 '18 at 23:06
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    Do not deface your question by erasing the contents. (And again, after I rolled back. Do not try it again.) – rschwieb Jan 20 '18 at 23:28
  • Margarita. Do not deface the question by removing parts that are essential for others to evaluate it and the answers. That is against the site rules. You don't own the question in the sense that you would be allowed to do anything you want with it. – Jyrki Lahtonen Jan 20 '18 at 23:59

2 Answers2

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Suppose otherwise. There is then a $y\in Y$ and a $r>0$ such that $B(y,r)\subset Y$. But then $B(0,r)\subset Y$, since $B(0,r)=B(y,r)-y$ and $Y$ is a vector space.

Now, if $x\in X\setminus\{0\}$, then $\frac{rx}{2\|x\|}\in B(0,r)$, since its norm is $\frac r2$. Therefore $\frac{rx}{2\|x\|}\in Y$ and so $x\in Y$. And $0\in Y$ too. This proves that $X\subset Y$. But we were supposing that $Y$ is a proper subspace.

José Carlos Santos
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  • @Margarita $\left|\frac{rx}{2|x|}\right|=\frac{r|x|}{2|x|}=\frac r2$. Do you want me to edit my answer adding this to it? – José Carlos Santos Jan 20 '18 at 23:12
  • @Margarita I don't know how to answer your first question. That's just the approach that it occured to me. And if $\frac{rx}{2|x|}\in X$, then $x\in X$ because $x=\frac{2|x|}{r}\cdot\frac{rx}{2|x|}$. – José Carlos Santos Jan 20 '18 at 23:18
  • @Margarita If my answer was useful, perhaps that you could mark it as the accepted one. If you want help for another proof, please post it as a separate question. – José Carlos Santos Jan 20 '18 at 23:23
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Assume that the interior of $Y$ is nonempty, say, $B_{\epsilon}(y)\subseteq Y$.

For $x\in X$, choose a $\delta>0$ small enough such that $\delta\|x\|<\epsilon$, then $\|(y+\delta x)-y\|<\epsilon$, so $w:=y+\delta x\in Y$.

As $Y$ is a vector subspace, then $x=\dfrac{1}{\delta}(w-y)\in Y$.

user284331
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