WHat's wrong in my argument?

  • Thread starter mansi
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In summary, the argument for proving that every punctured open set in R^2 is path connected involves taking points x and y and a point z taken off from U. If x, y, and z do not lie on a straight line, then there is a segment between x and y. If they do lie on a straight line, another point p can be chosen outside of the line, and segments can be drawn between x and p and p and y, creating a path between x and y. However, this argument is incorrect as it involves using straight lines. The challenge is to find a path that circles the puncture in a non-convex set.
  • #1
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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  • #2
What if your set is not convex?
 
  • #3
Can you show that there's a 'path' that circles the puncture?
 

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