Given a compact subset K of C, there exists a bounded operator T on a Hilbert space such that sp(T) = K and the set of eigenvalues of T is dense in K.

******************************

Suppose that H = l^{2}, (e_{n}) is the standard orthonormal basis for H and (λ_{n}) is a dense sequence in K. Set T (∑_{n=1}^{∞}α_{n}e_{n}) = ∑_{n=1}^{∞}λ_{n}α_{n}e_{n} where (α_{n}) ∈ l^{2}. Obviously K ⊆ sp(T). If λ ∉ K, then inf{|λ − μ|; μ ∈ K} > 0 and so S(∑_{n=1}^{∞}α_{n}e_{n}) = ∑_{n=1}^{∞}(λ − λ_{n})^{−1}α_{n}e_{n} is a well-defined operator on H. S is the inverse of λI − T. Therefore λ ∉ sp(T). Thus K = sp(T). For every n, T e_{n} = λ_{n}e_{n}. In fact {λ_{1}, λ_{2}, ⋯} is the set of all eigenvalues of T that is dense in sp(T).