I'm looking for a hint to help me solve the following problem:
"For a Normed Vector Space $X$ and a closed proper subspace $Y\subset X$, show that $\forall\,\epsilon\gt0,\,\exists \,x\in X:\|x\|=1,d(x,Y)\ge1-\epsilon.$"
I noticed that this problem seemed very similar to Riesz' lemma:
"Suppose that $X$ is a NVS, $Y$ a closed linear subspace of $X$ such that $Y\neq X$ and $\alpha\in\mathbb R:0\lt\alpha\lt1.$ Then, there exists $x_\alpha\in X:\|x_\alpha\|=1$ and $\|x_\alpha-y\|\gt\alpha,\,\forall y\in Y$"
I had a go using the proof to Riesz' lemma:
Since $Y$ is a proper closed subspace, $\exists \,x\in X/Y$. By the closedness of $Y$ we can define $d:=\inf\{\ d(x,y):y\in Y\}\gt0$. Then, taking $\epsilon\gt0$, in particular for $0\lt\epsilon\lt1$, we notice that $d\lt \frac{d}{\epsilon}$.
However, continuing along this vein I never got much farther than the actual proof to Riesz' lemma itself.
Is using the proof to Riesz' lemma the right way to go about solving this problem? Or is there another avenue which would be better to take?
Take $a\in X$ \ $Y.$ For brevity let $D=d(a,Y).$ Then $D>0$ because $a\not \in Y =\bar Y.$ Take $\delta >0$ such that $\frac {1}{1+\delta}>\max (0,1-\epsilon).$ There exists $b\in Y$ with $\|a-b\|<D(1+\delta)$ by def'n of $D.$ Then $a-b\ne 0$ because $a\in X$ \ $Y$ and $b\in Y.$
Note that $\frac {1}{\|a-b\|}>\frac {1}{D(1+\delta)}>\frac {(1-\epsilon)}{D},$ which will be used later.
Let $x=\frac {a-b}{\|a-b\|}.$ Then $\|x\|=1. $
$$\text {We have }\quad d(x,Y)=\inf_{y\in Y} \|x-y\|=\inf_{y\in Y}\left \|\frac {a-b}{\|a-b\|}-y\right \|=$$ $$=\frac {1}{\|a-b\|}\cdot \inf_{y\in Y}\|(a-b)-\|a-b\|y\;\|=$$ $$=\frac {1}{\|a-b\|}\cdot\inf_{y\in Y}\|a-(b+\|a-b\|y)\;\|.\quad (\bullet)$$ Since $\|a-b\|\ne 0$ and $b\in Y$ and $Y$ is a vector space, we have $\{b+\|a-b\|y:y\in Y\}=Y.$
So the expression in $(\bullet)$ is equal to $$\frac {1}{\|a-b\|}\inf_{y'\in Y}\|a-y'\|=$$ $$=\frac {1}{\|a-b\|}D>$$ $$> \frac {(1-\epsilon)}{D}D=1-\epsilon.$$
To illustrate the idea, suppose $X=\Bbb R^2$ and $Y=\Bbb R\times \{0\}$ and $a=(u,v)\in \Bbb R^2\setminus Y$ . The orthogonal projection of $a$ to $Y$ is $(u,0)$, and we could take $b=(u,0).$
In general there need not exist $b\in Y$ such that $\|a-b\|=d(a,Y)$ but heuristically imagine that there is $b\in Y$ which is "almost" a "projection" of $a$ to $Y.$