# Prove $\forall$ compact $M:\ M \subset C\quad \exists A:l_2\rightarrow l_2, \sigma(A)=M$ [duplicate]

Prove that for each nonempty $M$ - compact subset of $\mathbf{C}$ exists operator $A:l_2 \rightarrow l_2$, such that $\sigma(A) = M$, where $\sigma$ denotes spectrum.

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## marked as duplicate by t.b., Davide Giraudo, Asaf Karagila, Byron Schmuland, Jonas MeyerNov 19 '11 at 16:14

This question was marked as an exact duplicate of an existing question.

Since $M$ is separable, let $\{\lambda_n\}$ a dense sequence in $M$. Define $A$ by $A(e_n)=\lambda_n e_n$, where $e_n$ is the sequence whose $n$-th term is $1$ and the others $0$. Show that it's indeed a continuous operator, that the spectrum contains all the $\lambda_n$ and then use the fact that the spectrum is closed. For the reserve direction, use the fact that is $z\notin M$ then we can find a $\delta$ such that for all $n$, $|z-\lambda_n|\geq \delta$. Then you can find an inverse to $A-zI$. – Davide Giraudo Nov 19 '11 at 15:23
It's not the answer, but may be helpful: mathoverflow.net/questions/66345/… -- see there Notes and Comments. – Damian Sobota Nov 19 '11 at 15:24
You want to assume $M$ to be non-empty since the spectrum of a (bounded) operator is never empty. Davide's construction and the one using multiplication operators are both mentioned in the linked thread "Can spectrum "specify" an operator?" (that question was the motivation for the question Damian linked to). – t.b. Nov 19 '11 at 15:41