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I'm going through the topic $C^*$ algebra and facing few questions. It would be great if you people could help me to clear the doubts.

Q3. Does there exists some $X$ belonging to a $C^*$ algebra such that resolvent set is countable.

Ans. If suppose I consider the $n \times n$ matrix $M_n(C)$ as $C^*$ algebra then as the spectrum set is the set of Eigenvalues for any element in $M_n(C)$ , by definition resolvent set = C( complex number) - n ..so uncountable for all $x$ belongs to $M_n(C)$.

The question is, am I sounding correct ? If not please suggest a suitable example.

algae
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Leo
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The spectrum of any element of a Banach algebra is compact, so the resolvent set is always nonempty and open therefore uncountable.

Robert Israel
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  • Oh yes, in Euclidean topology non empty open sets are always uncountable. Get it. – Leo Apr 22 '20 at 04:39