Let $X$ be a normed space and $Y \subseteq X$. We define $ \pi: X \rightarrow X/Y $ as below: $$\pi(x) = \hat x, \quad x\in X$$ where $\hat x$ is the coset of $x$ with respect to $Y.$
Knowing that $\pi$ is linear and a contraction, I am trying to understand a proof to its being open. The proof as given in my old class notes is as follows.
$``$Let $\hat x_0 = \pi (x_0)$ for an arbitrary $x_0\in X$ and $B_X(x_0,\epsilon)$ be an open ball around $x_0$ with $\epsilon>0.$ We must show that $$\forall \hat x_0 \in \pi \big( B_X(x_0, \epsilon)\big),\,\, \exists \delta>0, \,\, s.t. \,\, B_{X/Y}(\hat x_0, \delta ) \subseteq \pi \big( B_X(x_0, \epsilon)\big). $$ Setting $\delta = \epsilon$, we claim that $$B_{X/Y}(\hat x_0, \epsilon ) \subseteq \pi \big( B_X(x_0, \epsilon)\big)\quad (\text I)$$ due to $$||\pi(x)-\pi(x_0)||=||\hat x- \hat x_0||_{X/Y} \leq ||x-x_0||_X <\epsilon, \forall x\in B_X(x_0, \epsilon). \,\square \quad \text{(II)} " $$ Now, in the last line, I understand that the inequality is due to $\pi$ being a linear contraction. But I do not see how (I) can be concluded from (II). Plus, where did $||\pi(x) - \pi(x_0)||$ come from? What term (from the lines above) does it represent?
Thanks for any explanation.
Since $\hat x=\pi(x), \|\pi(x)-\pi(x_0)\|=\|\hat x-\hat x_0\|$.
To see $I$ consider $\hat x\in B_{X/Y}(x_0,\epsilon)$. $\|\hat x-\hat x_0\|<\epsilon$. We can write $\hat x=\pi(x)$. This implies that $\|\hat x-\hat x_0\|=\|\pi(x-x_0)\|=inf_{y\in Y}\|x-x_0-y\|$. This implies that there exists $y\in Y$ such that $\|x-y-x_0\|<\epsilon$ and $x-y\in B_X(x_0,\epsilon)$, we deduce that $\pi(x-y)=\pi(x)=\hat x\in \pi(B_X(x_0,\epsilon)$.