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$(X, \lVert \cdot \rVert)$ is a normed space. Let $x \in X \setminus \{0\}$ and $Y \subset X$ is a subspace. Prove that if $Y$ is open then $Y=X$.

Which technique is more useful? We know that if a set is open then complement is closed. Or what else, directly use the definition of open set?

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Use that $Y$ is a neighbourhood of $0$. –  Daniel Fischer Jul 2 at 15:52
    
Yes. Use the fact than open set contains an open ball around each point it contains. Then try to scale any other point of $X$ to fit inside that ball. –  Prahlad Vaidyanathan Jul 2 at 15:52
    
@DanielFischer I'm probably forgetting something, but how do you know that $Y$ is a neighborhood of $0$? –  DanZimm Jul 2 at 15:59
    
We can show that any subspace of a topological vector space with non-empty interior must be the entire space. But what does $x$ have to do with your question? –  Stefan Hamcke Jul 2 at 16:14
    
@DanZimm: You know that $Y$ contains the origin because it is closed under scalar multiplication. –  Matt Jul 2 at 16:15

1 Answer 1

$Y$ is a (linear!) subspace, so $0 \in Y$. Then there is an open ball $B(0,r) \subset Y$, where $B(0,r) = \{ x: \|x\| < r \}$, because by assumption $Y$ is open so $0$ is an interior point of $Y$. If $x \neq 0$ is in $X$, then $y = \frac{r}{2 \|x\|}\cdot x$ has norm $\|y\| = \frac{r}{2\|x\|}\|x\| = \frac{r}{2} < r$, so $y \in Y$. As $Y$ is a linear subspace, $x = \frac{2\|x\|}{r} \cdot y$ is in $Y$ as well. So $Y = X$.

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