Doesn't this theorem hold for general normed spaces My question is: does this hold for any normed space $X$ or only for Banach spaces:

If $X$ is a Banach space then $K(X)$ (space of compact operators)
  equals $B(X)$ (space of bounded operators) if and only if $X$ is
  finite dimensional.

$\color{\grey}{\text{I know that the closed unit ball in a *normed space* $X$ is compact if and only if $X$ is finite}}$
$\color{\grey}{\text{ dimensional and I also know that the identity map is not compact if the unit ball is not.}}$
 A: That the closed unit ball is compact is indeed a characterization of 'finite dimensional'. Take for instance the space $X$ of all bounded functions $f:\mathbb R\to\mathbb R$ equipped with the maximum norm $||f||_\infty:=\sup_{x\in\mathbb R}|f(x)|$. Then one can show that $(X, ||\cdot||_\infty)$ is a (infinite dimensional) Banach space. To show that the unit sphere is not compact, consider the sequence $f_n$ of functions of $X$, given by
\begin{align*}
f_n(x)=\begin{cases}1,&\mbox{if } x=n \\ 0,&\mbox{if } x\neq n\end{cases}.
\end{align*}
Then each $f_n$ is bounded such that $||f_n||_\infty=1$. But $||f_m-f_n||_\infty=1$ whenever $m\neq n$, so $(f_n)$ cannot have a convergent subsequence.
Now, take for example the operator $T:X\to X,\ f\mapsto f$ (so $T$ is the identity). Then $T$ is obviously bounded but cannot be compact, since the closed unit ball is mapped to itself, but it is not compact as shown above.
A: The assumption that the unit ball is compact is indeed the issue. In general, a metric space is compact if and only if it is complete and totally bounded. The condition of being totally bounded is stronger than being bounded (basically, it means that for any $\varepsilon >0,$ the space can be covered by finitely many open balls of that radius (if you recall, that takes some effort to prove for a closed interval in $\mathbb{R} )).$ It is indeed the case that the closed unit ball of a complete real normed linear space is compact if and only if it is finite dimensional, and this can be found in many functional analysis texts. Note that the unit ball of an infinite dimensional normed linear space $X$ contains a sequence with no convergent subsequence (whether or not $X$ is complete- if $X$ is not complete it contains a Cauchy sequence which does not converge, and the elements can be taken from the closed unit ball of $X$ if required), so even the identity operator is not compact.
