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:


 * $B_{X/N} \subseteq \map \pi {B_X}$