If $(X,\| \|)$ be a Normed Linear Space.. Show that if any vector subspace $Y$ of $X$ is open, then $Y=X$

Question

If $(X,\| \|)$ be a Normed Linear Space, Show that if any vector subspace $Y$ of $X$ is open, then $Y=X$

Attempt: Subspace $Y$ of $X$ is open, $=> \exists ~~r >0 ~~\forall~~ y \in Y$ s.t $ B(y,r) \subset Y$ where $B(y,r)$ represents the open ball in $Y$ with centre $y$ and radius $r$.

Since, $B(y,r) \subset Y => \exists ~ u \in Y ~\& ~ \exists~ r>0$ s.t :

$B(y,r) = \{ u \in Y ~~|~~ d(u,y) <r \}$ where $d$ is the associated metric.

$B(y,r) = \{ u \in Y ~~|~~ \|u-y\| <r \}$ where $y \in Y ~~............(1)$

$Y$ is a subspace of $X$ . $=> \forall ~~u,y \in Y, ~ u+y \in Y ~~..........(2) $

and $\alpha ~y \in Y~ \forall ~y \in Y ~~~ ...................(3) $

How do I proceed next? Help will be appreciated. Thanks.

Answer 1

Hint: If $Y$ is open, then for each $y\in Y$, there exists $\epsilon>0$ such that $y+v\in Y$ for all $\left\|v\right\|<\epsilon$ (this is just a reformulation of the definition of "open").

Can you apply this to a suitable $y\in Y$? You will need to use the fact that $Y$ is a subspace of $X$ - e.g., that $Y$ is closed under scalar multiplication.

Hope this helps!

Answer 2

Contraposition works best here. Show that if $X\neq Y$, $Y$ is not open. To do this, construct a sequence in $X\setminus Y$converging to $0$. Note that if $x\in X\setminus Y, \lambda x\in X\setminus Y$ for all $\lambda \in \mathbb{K}$.

Edit: To show that $Y$ is not open, show that $B_{\epsilon}(0)\not\subset Y$ for all $\epsilon>0$. So given $\epsilon>0$, you must find $x\in X\setminus Y$ such that $\|x\|<\epsilon$.