Is the convex hull of union of a line and a point an open set?

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?

• It is neither open nor closed. – Rigel Mar 1 at 18:15
• @Rigel could you elaborate on this? – PulseJet Mar 2 at 4:21
• 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. – 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.