2
$\begingroup$

I'm wondering if a compact set $A\subset\mathbb{C}$ satisfying the properties that

• $A$ and its complement have finitely many connected components

• every connected component of $\partial A$ is the image of a path (say piecewise analytic, or $C^1$)

has a standard name? I'm not sure "set with rectifiable boundary" qualifies, but maybe I'm mistaken. In any case I'm looking for a shorter term if at all possible.

Thank you in advance for any pointer/suggestion.

Improve this question
$\endgroup$
5
  • $\begingroup$ How about just a "piecewise smooth set"? $\endgroup$
    – Lee Mosher
    Commented May 16, 2013 at 11:58
  • $\begingroup$ @LoïcTeyssier Very good type of subsets of $\mathbb{C}$. An operator theoretical version of your question:"To what extent all $C^{*}$ algebras with the property that the spectrum of all its element,as a subset of $\mathbb{C}$, satisfies 2 properties you mentioned, are classified? What are examples of infinite dimensional $C^{*}$ algebra with this property? what would be an algebraic (and spectral) interpretation for this topological property which you mentioned? On the other hand every compact set of $\mathbb{C}$ determines an ideal $J$ in $C_{0}(\mathbb{C}$.so the next question could be : $\endgroup$
    – Ali Taghavi
    Commented May 14, 2014 at 8:05
  • $\begingroup$ What can be said about the extension $0\to J\to C_{0}(\mathbb{C})\to C_{0}(\mathbb{C})/J\to 0$, provided your topological property is hold? Does this introduce a new $C^{*}$ algebraic invariant? $\endgroup$
    – Ali Taghavi
    Commented May 14, 2014 at 8:08
  • $\begingroup$ What can be said about the Busby invariant of such type of Extension? Morover, The Brouwn Douglas Filmore theory is about the classification of essential normal extension of compact operators in $B(H)$. If I am not mistaken it is closely related to the relative position of compact subsets of the plane. So it would be interesting to obtain some interpretation for your property, in term of BDF theory. $\endgroup$
    – Ali Taghavi
    Commented May 14, 2014 at 8:26
  • $\begingroup$ @AliTaghavi : I don't know anaything about $C^*$-algebras, sorry... $\endgroup$
    – Loïc Teyssier
    Commented May 14, 2014 at 10:51

2 Answers 2

Reset to default
1
$\begingroup$

I'll finaly settle for "a cutout compact set" for want of a better term. But I think this word expresses well the "finitely many" (connected components, non-smooth points) side of the object, which is what I wanted to highlight.

Improve this answer
$\endgroup$
0
$\begingroup$

What about "set with piecewise smooth boundary"?

Improve this answer
$\endgroup$
3
  • $\begingroup$ Alright, but I was looking for something shorter and "more catchy". It is a term I intend to use a lot in an article and therefore I would like to avoid cumbersome expressions as this one. Yet, if nothing better comes up I'll have to face the truth ;) $\endgroup$
    – Loïc Teyssier
    Commented May 16, 2013 at 10:11
  • 1
    $\begingroup$ You can define these sets at the beginning of your paper as "good sets", or "allowable sets", or even "regular sets", or something similar. I don't think there is a specific standard name in literature. $\endgroup$
    – Daniele Zuddas
    Commented May 16, 2013 at 10:14
  • $\begingroup$ Yeah, that's what I meant by "the truth" ... Something more poetic would have done, though! $\endgroup$
    – Loïc Teyssier
    Commented May 16, 2013 at 11:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .