Complex variables : open connected sets

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SUMMARY

The set S, defined as the union of the open unit disk |z|<1 and the open disk |z-2|<1, is not connected due to the spatial separation between the two regions. The definition of a connected set, as stated in "Complex Variables and Applications" by Brown, requires that any two points within the set can be joined by a polygonal line entirely contained in S. Since the two disks do not intersect, there are no polygonal paths connecting points from |z|<1 to points from |z-2|<1, confirming that S is disconnected.

PREREQUISITES
  • Understanding of complex numbers and their geometric representation
  • Familiarity with the concept of open sets in topology
  • Knowledge of connectedness in mathematical analysis
  • Basic understanding of circular discs in the complex plane
NEXT STEPS
  • Study the properties of open and closed sets in topology
  • Learn about connected components in metric spaces
  • Explore the implications of disconnected sets in complex analysis
  • Investigate the concept of homotopy and its relation to connectedness
USEFUL FOR

Students and educators in mathematics, particularly those studying complex variables and topology, will benefit from this discussion. It is also relevant for anyone interested in understanding the properties of connected and disconnected sets in mathematical analysis.

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Homework Statement



Let S be the open set consisting of all points such that |z|<1 or |z-2|<1 . State why S is not connected.


Homework Equations





The Attempt at a Solution



According to my complex variables book the definition of a connected set are pairs of points that can be joined by a polygonal line, consisting of a finite number of line segements joined end to end, that lies entirely in S. (Complex variables and applications, Brown).

I guess S is not connected is because both |z| and |z-2| have the same slope and therefore are parallel to each other . Therefore , since both |z| and |z-2| are parallel to each other, line segments are not connected , since parallel lines will not touch each other.
 
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Those inequalities don't describe lines. They describe circular discs. |z|<1 is the open unit disc. Want to try rephrasing that explanation?
 
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