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Hi i´m reading a pdf about linear programming and i´m having trouble understanding the difference between a polyhedron and polytope between those two definitions

A polyhedron P ⊆ $R^{n}$ is the set of all points x ∈ $R^{n}$ that satisfy a finite set of linear inequalities. Mathematically, P = {x ∈ $R^{n}$: Ax ≤ b} for some matrix A ∈ R m×n and a vector b ∈ $R^{m}$.

And the definition of polytope is the next one

A polyhedron is called a polytope if it is bounded, i.e., can be enclosed in a ball of finite radius.

So my problem isn´t the definition of polyhedron already bounded , and also are all polyhedron in these context convex?

And for example

{x ∈ $R^{n}$: Ax ≤ b} if i guide myself from the given definitions i can only assure this is a polyhedron but not a polytipe since it isn't technically bounded but is convex and {x|$x\in\mathbb{R}$} isn't any of those two.

Any insight would be helpful

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    $\begingroup$ One might say that a polyhedron is the intersection of finitely many half-spaces, while a polytope is the bounded intersection of finitely many half-spaces. $\endgroup$
    – While I Am
    Commented Feb 11, 2021 at 5:04
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    $\begingroup$ It should be noted however, that those same terms get used differently by other authors! Then polytope becomes the general dimensional term, convex or non-convex, and polyhedron would be just the 3D instance, just like polygon is the 2D instance, and polychoron is the 4D instance. - So be careful to which definitions someone adheres! $\endgroup$
    – Dr. Richard Klitzing
    Commented Feb 11, 2021 at 20:21

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A polyhedron is not bounded in the sense that we might not be able to find a ball of finite radius to find it.

For example consider, $\{x \in \mathbb{R}^n : x \ge 0\}$, the first octant polyhedron, it is unbounded, it is a polyhedron but it is not a polytope. If you claim that it is bounded by a ball of radius $r$, consider the point $(r+1, 0, \ldots, 0)$, it is insider the polyhedron, but it is not contained inside the ball of radius $r$.

$\{x | x \in \mathbb{R}\}$ is a polyhedron since we can take $A=b=0$.

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Consider a set of linear inequalities whose solution set extends to infinity, forming what is called a polyhedral bowl.

This bowl is a polyhedron because it is the solution set. But it is not a polytope because it cannot be enclosed in a sphere of finite radius, i.e. the equalities do not fully bound it.

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  • $\begingroup$ I don't understand your definition of polyhedral bowl. Is it just any unbounded polyhedron? $\endgroup$
    – M. Winter
    Commented Feb 16, 2021 at 8:42
  • $\begingroup$ @M.Winter I didn't define a polyhedral bowl, I just gave it as an example of a solution set. Technically the bowl is the finite (part of the) limit of the solution set. If you think of a polytope as a ball, then a bowl is an infinitely tall dome. $\endgroup$
    – Guy Inchbald
    Commented Feb 16, 2021 at 8:54

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