There is a well known theorem that says that if $X$ is a compact Hausdorff space, then the space $C(X)$ of the continuous functions on $X$ is a complete Banach space with the sup norm.
It's clear why the space $X$ should be compact: in this way the sup norm is well defined.
It's natural that the Hausdorff hypothesis is necessary to have a REASONABLE space, but why the Hausdorff hypothesis is necessary here?
Can you make a counterexample of the theorem failing without it?
Best Answer
You don't need it for $C(X)$ to be a Banach space. But it guarantees that enough continuous functions exist to have nice theorems about the relation between $X$ and $C(X)$, eg. For $X$ any cofinite infinite space (so only compact $T_1$), the space $C(X)$ only consists of constant functions, so is isometrically isomorphic to $\mathbb{R}$, regardless of the cardinality of $X$. For compact Hausdorff spaces we have nicer theorems, and the algebraic structure of the ring $C(X)$ determines the topology of $X$ uniquely. So the theory is nicer when we add Hausdorffness.