Find the boundary of certain set

Question

Update:

I eventually know what my question exactly is.

Question: Let $x,y,z\in\mathbb R^2$, and $x-y$ and $x-z$ are linear independent.

Prove that the boundary of the graph of $A:=\{h\in\mathbb R^2\, | \, h=i_1x+i_2y+i_3z, i_1+i_2+i_3=1,i_j \ge0$ for $j=1,2,3\}$ in $\mathbb R^2$ is the set $B:=\{g\in\mathbb R^2\, | \, g=i_1x+i_2y+i_3z, i_1+i_2+i_3=1,i_1=0 \, $or$ \,i_2=0 \,$or$\, i_3=0 \}$. I mean, in the sense of euclidean topology.

Thanks so much.

Answer 1

Let's have $C:=\{h\in\mathbb R^2\, | \, h=i_1x+i_2y+i_3z, i_1+i_2+i_3=1,i_j >0$ for $j=1,2,3\}$. It is easily shown to be an open set. The closure of $C$ is $A$, and it is also equal to $C \cup B$, thus $B$ is the boundary of $A$ (and of $C$).

Answer 2

I want to prove the result in my question in order to prove that the set $A \cup B$ is exactly a triangle in $\mathbb R^2$.

So we are assumed that we don’t know $A \cup B$ is a triangle in $\mathbb R^2$.

Now this question is NOT trivial.

I spend several hours and finally think out a proof of it.

Proof

We may assume the segment $x,y$ is in one side of $z$. For otherwise, we just rotate the space $\mathbb R^2$ and rotations are clearly homeomorphisms.

Now for any point $m$ in $B$, and for any $\epsilon \gt 0$, consider the ball $K$ where centered at $m$ and with radius $\epsilon$.

We need to prove that $\exists$ points $p\in A$ and $q\in A^c$ and $p,q\in K.$

We first prove the existence of $q$.

W.L.O.G we consider the case that $m$ lies in the segment $y-x$, i.e. the segment joints $x$ and $y$.

We will have three cases.

Case $1$: Both $x$ and $y$ are orthogonal to $z$.

We just choose a vector $q_1$ in the direction of $x$ and lies in the ball $K$. Let $q=m+q_1$, then $q=tx+(1-t)y+\delta x=(t+\delta)x+(1-t)y$, where $t\in [0,1] $ and $ 0\lt \delta \lt r$. Since $t+\delta+1-t=\delta+1\gt1$, so $q\in A^c$.

Case $2$: only one of $x$ and $y$ are orthogonal to $z$.

W.L.O.G we assume $x$ are orthogonal to $z$.

We just choose a vector $q_1$ in the direction of $x$ and lies in the ball $K$. Let $q=q_1+m$. Similar to case $1$, we can prove $q\in A^c$.

Case 3: None of $x$ and $y$ are orthogonal to $z$.(Note that it cannot happen that both $x$ and $y$ are parallel to $z$, it will contradicts the assumption that $x-y$ and $x-z$ are linear independent.)

Let $n$ be a vector which is orthogonal to $z$. Clearly, $n$ can have two choices of direction. We choose $n$ in the direction s.t. $n\cdot x\gt 0$. We choose the vector $q_1$ in the direction of $n$ and lies in the ball $K$. Let $q=q_1+m$. Similar to case $1$, we can prove $q\in A^c$.

Now we deal with the existence of $p$.

In fact, we just reverse the direction of the vector $q_1$ in above cases. And let $p=q_1+m$, then it will lies in $A$, which can be shown as above.

QED.