Let $T$ be a triangular region in $\mathbb{R}^2$ defined by $x\geq 0, y\geq 0, x+y\leq 1$:
a) Describe the boundary of $T$
b) Prove that the point $(0.0001, 0.9998)$ is an interior point of $T$
Attempt:
For a), the boundary of $T$ is formed by all the points on the $y-$axis and $x-$axis.
For b), by inspection $0.0001+0.9998=0.9999<1$. Then if it is an interior point. Note that there exists an open ball of radius $0.0001$ around $(0.0001,0.9998)$, $B_{0.0001}(0.0001,0.9998)$; such that $B_{0.0001}(0.0001,0.9998)\subset T$. Therefore $(0.0001,0.9998)$ is an interior point.
I think more generally, it would be to take $\varepsilon > 0$ such that $\varepsilon \leq 0.0002$.
The three bounding lines meet at three points. \begin{align*} &x=0, y=0 : & &(0,0) \\ &x=0, x+y=1 : & &(0,1) \\ &y=0, x+y=1 : & &(1,0) \\ \end{align*} The region $T$ is the interior of the triangle formed by those three vertices.
The boundary of $T$ is the union of three line segments, one from $(0,0)$ to $(0,1)$, one from $(0,1)$ to $(1,0)$, and one from $(1,0)$ to $(0,0)$.
We can also specify these symbolically. $$ \{ (0,y) : y \in [0,1]\} \cup \{(t,1-t):t \in [0,1]\} \cup \{(x,0) : x \in [0,1]\} $$
(b) We test that the given point satisfies the strict version of the bounding inequalities (so is in the interior): \begin{align*} x &= 0.0001 > 0 \text{ is true,} \\ y &= 0.9998 > 0 \text{ is true, and} \\ x+y &= 0.9999 < 1 \text{ is true.} \end{align*} The given point satisfies all three strict constraints, so is an interior point of $T$.