If $T$ represents the points that define the perimeter of a triangle and $A = cl (T)$, that is, the closure of $T$, prove that $A$ is a convex set.

Hint: represent $A$ as the intersection of 3 half-planes in $\mathbb{R^2}$, verifying if a hyperplane is a convex set and checking properties relating to union and intersection of convex sets.

I'm really stuck here, I looked at the definitions of half-planes and hyperplane but I do not know where to go

  • $\begingroup$ which part of the hints are you stuck at? $\endgroup$ – Siong Thye Goh Mar 18 '18 at 17:57
  • $\begingroup$ There is something wrong with the question in its current form. If $T$ is the perimeter of the triangle, then $T$ is already closed and not convex, so $A$ wouldn't be convex either. $\endgroup$ – max_zorn Mar 18 '18 at 21:38
  • $\begingroup$ but $cl(T)$ it's the closure of T $\endgroup$ – Muradin Bronzebeard Mar 18 '18 at 22:36

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