Given a line $l$ and a point $A$ not on the line, is the convex hull of the two an open set? As per my understanding, the hull will be all points between lines $l$ and line $m$ which passes through $A$ and is parallel to $l$, including line $l$ but not line $m$, except the point $A$. Is this a closed set?

  • 3
    $\begingroup$ It is neither open nor closed. $\endgroup$ – Rigel Mar 1 at 18:15
  • $\begingroup$ @Rigel could you elaborate on this? $\endgroup$ – PulseJet Mar 2 at 4:21
  • $\begingroup$ It's not open, because the line $l$ is in the set, and on its boundary as well. It's not closed, because the line $m$ is not in the set (save $A$), but it is on its boundary. $\endgroup$ – Michael Grant Mar 5 at 1:14

No, in the plane, take say the line $y=0$ and the point $(0,1)$, then the point $(1,1)$ would on your line $m$. $(1,1)$ can be written as the limit of points in the convex hull but does not belong itself to the convex hull, i.e.,

$$(1,1) = \lim_n (1-\frac{1}{n}) (0,1) + \frac{1}{n} (n,0),$$

where $(1-\frac{1}{n}) (0,1) + \frac{1}{n} (n,0)$ belongs to the convex hull.


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