Identity Operator is Compact iff Finite-Dimensional Normed Vector Space

Theorem

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.

Let $I : X \to X$ be the identity operator.

Then $I$ is compact if and only if $X$ is finite-dimensional.

Proof

Let $\operatorname {ball} X$ be the closed unit ball in $\struct {X, \norm {\, \cdot \,} }$.

From the definition of a compact operator, we have that $I$ is compact if and only if:

$\overline {\operatorname {ball} X}$ is compact in $\struct {X, \norm \cdot_X}$.

From Closed Ball is Closed and Set is Closed iff Equals Topological Closure, this is equivalent to:

$\operatorname {ball} X$ is compact in $\struct {X, \norm {\, \cdot \,} }$.

From Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact, this is equivalent to:

$X$ is finite-dimensional.

$\blacksquare$