I was proving a theorem stated below.
Theorem.
Suppose that $(\mathcal{X},\|\cdot\|)$ is a normed vector space and $\mathcal{M}\leq \mathcal{X}$ is a closed proper subspace. Then for any $\epsilon>0$ there exists $x\in \mathcal{X}$ such that $\|x+\mathcal{M}\|\geq 1-\epsilon$ where $$\|x+\mathcal{M}\|=\inf_{y\in \mathcal{M}}\|x+y\|.$$
After proving, I found that, actually, we can find $x\in\mathcal{X}$ such that $\|x+\mathcal{M}\|=1$. But if it is true, then there is no need to consider $\epsilon$ in the theorem. Thus, I am not confident about my proof.
Here is my proof!
Let $z\in \mathcal{X}\setminus \mathcal{M}$ be given. Then $$d=\|z+\mathcal{M}\|=\inf_{y\in \mathcal{M}}\|z+y\|>0$$
since $\mathcal{M}$ is closed.
(If $\|z+\mathcal{M}\|=0$, there exists $\{y_n\}\subset \mathcal{M}$ such that $\|y_n+z\|\rightarrow 0$ and there exists a subsequence $\{y_{n_k}\}$ such that $y_{n_k}\rightarrow -z$ which means $z\in \mathcal{M}$ since $\mathcal{M}$ is closed.)
Then, there exists $y\in \mathcal{M}$ such that $\|z+y\|=d$.
Now let $x=\frac{z+y}{\|z+y\|}$ then $\|x\|=1.$ And note that
$$\|x+\mathcal{M}\|=\frac{\inf_{a\in \mathcal{M}}\|z+y+a\|}{\|z+y\|}=1$$
since $$\|z+y\|=\inf_{b\in \mathcal{M}}\|x+b\|$$.
Thus, our $x$ has been discovered.
*Is there any defect in my proof? Any comment would be helpful for me 🙂 *
Best Answer
This sentence is essentially asserting the truth of the statement that was to be proved. In general, no such $y$ exists. A counterexample is given in Given a point $x$ and a closed subspace $Y$ of a normed space, must the distance from $x$ to $Y$ be achieved by some $y\in Y$?