I recently asked this question.
Existence of x∈X such that ∥x∥=1 and ∥x+M∥=1 for a closed subspace M
And people said that when $\mathcal{X}$ is normed vector space, even if $\mathcal{M}$ is closed and $$\|x+\mathcal{M}\|=\inf_{y\in \mathcal{M}}\|x+y \|=d$$, we cannot say that there exists $y\in \mathcal{M}$ such that $\|x+y\|=d$ due to this result(Given a point $x$ and a closed subspace $Y$ of a normed space, must the distance from $x$ to $Y$ be achieved by some $y\in Y$?).
And I started to solve a problem 5.1.12 of Folland's real analysis which says
5.1.12 Let $\mathcal{X}$ be a normed vector space and $\mathcal{M}$ a proper closed subspace of $\mathcal{X}$. Then $$\|x+\mathcal{M}\|=\inf_{y\in \mathcal{M}}\|x+y \|$$ is a norm on $\mathcal{X\setminus M}$.
In order to prove it, we need to prove that $$\|x+\mathcal{M}\|=0 \iff x\in \mathcal{M}$$, which seems contradicts to the results above.
Could anyone teach me where I am thinking wrongly?
Best Answer
Suppose $x\in \mathcal{M}$ it is clear that $\Vert x+\mathcal{M} \Vert=0$. Now suppose $x\not \in \mathcal{M} $, since $\mathcal{M}$ is a closed set, and assuming we are using the normed topology, there exists an open ball centered in x, that does not intersect $\mathcal{M}$ then for every $y\in \mathcal{X}-\mathcal{B}(x)$ we have $\Vert x-y\Vert >0$.