Quotient Mapping Maps Unit Open Ball in Normed Vector Space to Unit Open Ball in Normed Quotient Vector Space
Theorem
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $N$ be a closed linear subspace of $X$.
Let $\struct {X/N, \norm {\, \cdot \,} }$ be the normed quotient vector space associated with the quotient vector space $X/N$.
Let $B_X$ be the unit open ball in $\struct {X, \norm {\, \cdot \,} }$.
Let $B_{X/N}$ be the unit open ball in $\struct {X/N, \norm {\, \cdot \,} }$.
Let $\pi$ be the quotient mapping associated with $X/N$.
Then:
- $\map \pi {B_X} = B_{X/N}$
Proof
From Quotient Mapping is Bounded in Normed Quotient Vector Space, we have:
- $\norm {\map \pi x}_{X/N} \le \norm x$
So if $x \in B_X$, we have $\norm x < 1$ and hence:
- $\norm {\map \pi x}_{X/N} < 1$
So $\map \pi x \in B_{X/N}$.
So we have:
- $\map \pi {B_X} \subseteq B_{X/N}$
Conversely, let $\mathbf x \in B_{X/N}$ and pick $x \in X$ such that $\mathbf x = \map \pi x$.
Then we have:
- $\norm {\map \pi x}_{X/N} = \inf_{z \in N} \norm {x - z} < 1$
So there exists $z_\ast \in N$ such that:
- $\norm {x - z_\ast} < 1$
so that:
- $x - z_\ast \in B_X$
From Kernel of Quotient Mapping and Quotient Mapping is Linear Transformation, we have:
- $\map \pi {x - z_\ast} = \map \pi x = \mathbf x$
So we have:
- $\mathbf x \in \map \pi {B_X}$
So we have:
- $B_{X/N} \subseteq \map \pi {B_X}$
$\blacksquare$