Is Every Punctured Open Set in R^2 Path Connected?

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SUMMARY

Every punctured open set in R² is path connected. The discussion revolves around proving this property by considering points x and y within the set and a punctured point z. The initial argument incorrectly relies on straight lines, which is not permissible. Instead, one must demonstrate the existence of a continuous path that circumvents the puncture, potentially using paths that circle around the removed point.

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  • Understanding of topological concepts, specifically open sets and path connectedness.
  • Familiarity with R² and its geometric properties.
  • Knowledge of continuous functions and their role in topology.
  • Basic principles of convexity and non-convex sets in topology.
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  • Study the definition and properties of path connectedness in topology.
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  • Investigate the concept of winding numbers and their relevance to paths around punctures.
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Mathematicians, topology students, and educators seeking to deepen their understanding of path connectedness in punctured spaces.

mansi
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I was asked to prove, every punctured open set in R^2 is path connected.

My argument : take points x and y. let z be the point we've taken off from U (open).
if x, y,z do not pass through a staright line, we have a segment between a and y.
Now if the 3, i.e. x,y,z lie on a straight line...then pick another point ,say p, not lying on the staright line.
we have segments joining x and p and p and y. hence, we've found a path between x and y ,as required.

apparently, this is wrong...my prof told me not bring in straight lines anywhere! any hints?? thanks!
 
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What if your set is not convex?
 
Can you show that there's a 'path' that circles the puncture?
 

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