How can I show that $X\setminus Y$ is dense in $X$ if $X$ is a normed space?

Question

Let $X$ be a normed space and $Y$ a subspace s.t. $Y\neq X$. I need to show that $X\setminus Y$ is dense in $X$.

I'm a bit confused how to do this. I thought about this geometrically, so I thought that it is enough to show that for all open balls $B(x_0,r_0)$ with $x_0\in X$, $r_0>0$ there exists $x\in (X\setminus Y)\cap B(x_0,r_0)$. But now somehow I don't see how to construct such an $x$. I thought then about proving the statement by contradiction but also this does not brings me further. Could maybe someone help me?

My approach was to assume $(X\setminus Y)\cap B(x_0,r_0)=\emptyset$, this would imply that $B(x_0,r_0)\subset Y$. But then in particular $x_0\in Y$. Now I thought that since $x_0$ is arbitrary in $X$ this would mean that $X\subset Y$ but then in particular $X=Y$ which gives a contradiction.

But I don't think this is true.

Answer 1

Let $a \in X \setminus Y$. For any $x \in X$, the sequence $\{x+\frac{a}{n}\}$ converges to $x$ and belongs to $X \setminus Y$.

Answer 2

Your logic doesn't quite work. You've taken an arbitrary $x_0$. But then you impose the restriction that $B(x_0,r_0)$ is entirely contained in $Y$. This restricts the possible values $x_0$ can take.

This does not mean you have reached a contradiction. It just means that instead of picking an arbitrary $x_0\in X$, you have now picked an arbitrary $x_0$ among all the elements in $X$ such that $B(x_0,r_0)\subseteq Y$.