Difference between convex set, closed convex set, polyhedron and polytope? I'm having a hard time differentiating between a  convex set, a closed convex set, a polyhedron and a polytope? I have a good understanding of what a convex set is, but I can't seem to understand was further conditions are placed if it is 'closed'.
Similarly, I know a polyhedron is the intersection of half-spaces(which are convex sets) so its a convex set too. But is it a closed convex set or not?
And what is meant by 'bounded', when we say that a polytope is a bounded polyhedron?
I'd really appreciate a reply, thank you!
 A: A convex set is closed if the limit of any convergent sequence of points from the set is contained in the set. As a simple example in one dimension, $[a,b]$ is a closed convex set but $(a,b)$ is not. The latter is convex, but you can easily find a sequence of points in the interval converging to $a$, and $a$ does not belong to the set.
A polyhedron is a set (not necessarily closed, not necessarily bounded and not necessarily convex) with a finite number of flat sides. As the Wikipedia entry notes, there is a lack of uniformity in how people interpret the word "polyhedron". The "toroidal polyhedron" diagram in that entry depicts a polyhedron that is clearly not convex (and is not an intersection of half-spaces).
A convex polyhedron is the intersection of a finite number of half-spaces, where a half-space is basically one side of a hyperplane. So, for instance, $X=\lbrace x\in \Re^2 : x_1 \ge 0, x_2 \ge 0, x_1 - x_2 \ge -1\rbrace$ is a convex polyhedron, and is closed.
Consider $Y=\lbrace x\in \Re^2 : x_1 \ge 0, x_2 \ge 0, x_1 - x_2 > -1\rbrace$ (where the last inequality is weak). It is again the intersection of half-spaces, including two of the three bounding hyperplanes but not the third one, so it is a convex polyhedron but not closed. $(0,1)$ belongs to $X$ but not $Y$, but you can find a sequence of points in $Y$ converging to it.
"Bounded" just means extending finitely far in any direction. A polytope is a closed, bounded polyhedron, and a convex polytope is a closed, bounded, convex polyhedron. The polyhedron $X$ above is not bounded (because $(M, M)$ is in the set for arbitrarily large $M$). If we intersect $X$ with the half-space $x_1 \le 10$, we get a closed, bounded, convex polyhedron, hence a polytope. In an optimization context, being closed and bounded guarantees that the maximum or minimum of a continuous objective function over the set is achieved somewhere in the set. If the set is either not closed or not bounded, it may be that the objective function $f(x)$ has no minimum (or maximum, whichever matters), but instead there is a lower (upper) bound that is approached but never reached by a sequence of points in the set.
