Subspaces of $\Bbb R^n$ containing vectors whose coordinates satisfy prescribed inequalities For any integer $n\ge2$, the vector space $\Bbb R^n$ is divided into $n!$ "wedges" by prescribing which coordinate is largest, second-largest, etc. One such wedge is $$\{(x_1,\dots,x_n)\in\Bbb R^n\colon x_1>x_2>\cdots>x_n\},$$ and the other wedges have similar definitions but with the $x_j$ permuted in the string of inequalities.
Note that each wedge is invariant under translation by multiples of the vector $(1,\dots,1)$. Therefore, the subspace $\{(x_1,\dots,x_n)\in\Bbb R^n\colon x_1+x_2+\cdots+x_n=0\}$, consisting of all vectors orthogonal to $(1,\dots,1)$, intersects (the interior of) every wedge.
Question: does any smaller subspace (that is, of dimension $n-2$ or less) intersect every wedge?
Equivalently, does there exist a proper subspace of $\Bbb R^n$ that contains $(1,\dots,1)$ yet manages to intersect every wedge? (This is equivalent because if $W$ is such a subspace, then the orthogonal complement of $(1,\dots,1)$ inside $W$ is a subspace of dimension at most $n-2$ that still intersects every wedge.)
From drawing pictures and animal cunning, I know the answer is "no" for $n=2$ and $n=3$. But even for $n=4$, I haven't been able to show that a $2$-dimensional subspace can't intersect every wedge. I would love a proof that the answer is "no" for all $n$ (or, only slightly less, a proof that the answer is "yes").
 A: I believe I found a proof that the answer is indeed "no", so I'll post it here to avoid misleading future readers. The idea is that no wedge is big enough to contain vectors that are perpendicular to each other, if we aren't allowed to use the $(1,...,1)$ direction.
We prove the following equivalent statement: if $V$ is a subspace of $\Bbb R^r$ not containing $(1,\dots,1)$ and $\dim V \le r-2$, then $V$ does not intersect some wedge $W_\sigma$. The orthogonal complement of the subspace generated by $V$ and $(1,\dots,1)$ has dimension $r-(\dim V+1)\ge1$; therefore we may choose a nonzero vector $(y_1,\dots,y_r) \in \big(V\cup(1,\dots,1)\big)^\perp$; in particular, $y_1+\cdots+y_r=0$. By permuting the coordinates of $\Bbb R^r$, we may assume that $y_1\le y_2\le \cdots \le y_r$. We claim that $V$ does not intersect $W_{id} = \big\{ (x_1,\dots,x_r) \in \Bbb R^r \colon x_1 < x_2 < \cdots < x_r \big\}$.
Suppose for the sake of contradiction that there was some $(z_1,\dots,z_r) \in W_{id} \cap V$. Then $z_1<z_2<\cdots<z_r$, and $y_1z_1 + y_2z_2 + \cdots + y_rz_r=0$ since $(z_1,\dots,z_r) \in V$ and $(y_1,\dots,y_r) \in V^\perp$. For the moment assume that no $y_j$ equals $0$; choose $0<k<r$ such that $y_k<0<y_{k+1}$. Then, thanks to the known ordering $z_1<z_2<\cdots<z_r$, we have
\begin{align*}
0 = y_1z_1 + y_2z_2 + \cdots + y_rz_r &\ge (y_1+\cdots+y_k)z_k + (y_{k+1}+\cdots+y_r)z_{k+1} \\
&= (z_{k+1}-z_k)(y_{k+1}+\cdots+y_r) > 0,
\end{align*}
where the middle equality follows from $y_1+\cdots+y_r=0$. This contradiction establishes the claim in the case where no $y_j$ equals $0$. If $y_k < 0 = y_{k+1} = \cdots = y_{k'} < y_{k'+1}$, then the above argument goes through with $k+1$ replaced by $k'+1$ everywhere.
