Normed Quotient Space

Question

If $X$ is a normed vector space and $M$ is a proper closed subspace, I want to show that for any $\epsilon>0$ there exists an $x\in X$ such that $\|x\|=1$ and $\|x+M\|\geq 1-\epsilon$. Is there anything wrong with arguing as follows:

Suppose by contradiction there exists some $\epsilon >0$ such that for all $x\in X$ with $\|x\|=1$, $\|x+M\| <1-\epsilon$. Then $\|x+M\| <1-\epsilon=\|x\|-\epsilon \implies \epsilon <\|x\|-\|x+M\| \leq \|x\|-\|x+y\|$ (for all $y\in M$), $\leq \|x\|-\|x\|-\|y\|$, i.e. $\epsilon \leq -\|y\|$ which is impossible. Can verify if this makes sense? Can someone offer a more intuitive argument? Thanks

Answer 1

Your last inequality is wrong. You have $\|x+y\|\leq\|x\|+\|y\|$, but with a minus in front the inequality is reversed.

But, more importantly, $\|x+M\|\leq\|x+y\|$, so $-\|x+M\|\geq-\|x+y\|$, so your second to last inequality is also wrong.

What you want to prove is called Riesz Lemma:

If $X$ is a normed linear space, $Y\subset X$ is a closed proper subspace and $\alpha\in(0,1)$, then there exists $x\in X$ with $\|x\|=1$ and $\|x+y\|\geq\alpha$ for all $y\in Y$.