$(X,\|.\|)$ is a normed space over $\mathbb F$ field $M \subset X$ is a vector subspace of $X.$ If $M$ is open vector subspace of $X$, then $X=M$?

Question

I know there are two questions which are same with mine, but their hints and comments are not clear enough for me.

Let $(X,\|.\|)$ is a normed space over $\mathbb F$ field $M \subset X$ is a vector subspace of $X.$

Show that if $M$ is open vector subspace of $X$ then $X=M$

First I tried obtaining a contradiction via using

Let $M$ is open and $X\neq M$. Since $M$ is open $M=int(M)$. Hence $\exists x\in X\setminus int(M)$

but I couldn't obtain from there.

Then, I have tried to prove it directly but I couldn't see why $\forall x \in X$ then $x \in M$

I've written

$M$ is open $\Rightarrow$ $\exists r\gt0$ $\forall m \in M$, $B(m,r)\subseteq M$

$B(m,r):= \{n \in M : \|n-m\| \lt r\}$

and

Since $M$ is vector subspace

-$\forall m,n \in M \subset X$, $m+n \in M$

-$\forall m \in M \subset X$,$\forall \lambda \in \mathbb F$, $\lambda m \in M$

I cannot combine being vector subspace and being open. Could someone help me about seeing why "$\forall x \in X$ then $x \in M$" or obtainig contradiction in the easiest and clearest way without topological terms please?

Thanks in advance

Answer 1

Let $\mathbb{F}\in \{ \mathbb{R}, \mathbb{C} \}$, let $(X, \ \Vert \cdot \Vert)$ be a normed space over $\mathbb{F}$ and $V\subseteq X$ an open subspace of $X$. As $V$ is open, there exists $\varepsilon>0$ such that

$$ B_\varepsilon (0) = \{ x \in X \ : \ \Vert x \Vert < \varepsilon \} \subseteq V.$$

Let $0\neq x\in X$, then we have

$$ x=\underbrace{\left(\frac{2\Vert x \Vert}{\varepsilon} \right)}_{\in \mathbb{F}} \underbrace{\left( \frac{\varepsilon x}{2\Vert x \Vert }\right)}_{\in B_\varepsilon(0)\subseteq V} \in V.$$

Where we used, that $V$ is a vector space and hence scalar multiple of a vector in $V$ lie again in $V$. As $V$ is a vector space, we also have $0\in V$. Thus, $V=X$.

Answer 2

In fact, if $M$ contains any open ball $B(x,r)$ then $M = X$.

Indeed, for any $y \in X, y \ne 0$ we have

$$x + \frac{r}{2\|y\|}\cdot y \in B(x,r) \subseteq M$$

so since $M$ is closed under addition and scalar multiplication$$y = \frac{2\|y\|}{r}\left(\underbrace{x + \frac{r}{2\|y\|}\cdot y}_{\in M} - \underbrace{x}_{\in M}\right) \in M$$

On the other hand, if $y = 0$ then $$0 = \underbrace{x}_{\in M} - \underbrace{x}_{\in M} \in M$$

We conclude $M = X$.