EN
It is shown that if α,ζ are ordinals such that 1 ≤ ζ < α < ζω, then there is an operator from $C(ω^{ω^α})$ onto itself such that if Y is a subspace of $C(ω^{ω^α})$ which is isomorphic to $C(ω^{ω^α})$, then the operator is not an isomorphism on Y. This contrasts with a result of J. Bourgain that implies that there are uncountably many ordinals α for which for any operator from $C(ω^{ω^α})$ onto itself there is a subspace of $C(ω^{ω^α})$ which is isomorphic to $C(ω^{ω^α})$ on which the operator is an isomorphism.