Is $\mathbb{R}^2$ a polyhedron?

Question

I'll appreciate if anyone can explain to me if $\mathbb{R}^n$ is a polyhedron? I saw this link here which says $\mathbb{R}^n$ is not a polyhedron, but could one think of $\mathbb{R}^n$ as an intersection of halfspaces in $\mathbb{R}^{n+1}$? Thanks

Answer 1

If the ambient space is $\Bbb R^3$, then by your definition, the $x,y$-plane $P$ is a polyhedron, since it is the intersection of two half-spaces:

$$P = \{(x,y,z) : z\ge0\} \cap \{(x,y,z) : z\le0\}$$

Technically, this isn't the same as $\Bbb R^2$, but there is a natural bijection between $P$ and $\Bbb R^2$.

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Explanation of technicality: $\Bbb R^2$ consists of ordered pairs $(x,y)$. On the other hand, $P$ consists of an ordered triplet $(x,y,0)$. You can map from $P$ to $\Bbb R^2$ in a natural way by defining $$f(x,y,0) = (x,y),$$ and similar for the reverse direction, so in many respects they "look the same".

Now consider a function like $g(x,y,z) = z$. When restricted to $P$, we have $g(x,y,z) = 0$ everywhere. On the other hand, $g$ cannot be applied to $\Bbb R^2$, since there simply is no $z$-coordinate.

Finally, even if you very informally speak of $\Bbb R^2$ as a plane within $\Bbb R^3$, keep in mind that there's an ambiguity: are you referring to the $x,y$-plane? The $x,z$-plane? A plane on a diagonal? (Case in point: in my answer, I wasn't sure which plane you were referring to, so I arbitrarily used the $x,y$-plane for simplicity.)

Answer 2

$\mathbb{R}^2$ may indeed be constructed as the set of tangent planes and, in general $\mathbb{R}^n$ may be constructed as the set of tangent $\mathbb{R}^(n-1)$ spaces. But these constructions are not polyhedra. The set of subspaces is known as the tangent bundle.

In the plane, the 1-sphere or circle is an example of a conic section. In 3-space the sphere is an example of a quadric surface. And so on. These can all be constructed as what are known as the envelopes of their tangent bundles.

One may obtain such an envelope as the limit in a series of $n$-sided polyhedra where $n$ tends to infinity, i.e. as an infinite-faced polyhedron or apeirohedron. But the faces degenerate to points, so one may wish to regard the apeirohedron as degenerate and therefore not a valid understanding of the sphere.