Is the set of triangles closed? Let $\mathcal{P}$ be the set of simple polygons in $\mathbb{R}^2$ with strictly positive area.
Define the distance between any two such polygons, say $A$ and $B$, to be $d(A,B) = \mu(A \Delta B)$ where $\mu$ is the Lebesgue measure and $A \Delta B = (A - B) \cup(B-A)$ is the symmetric difference of $A$ and $B$. The distance function $d$ makes $\mathcal{P}$ a metric space.
Let $\mathcal{T}$ denote the set of all triangles in $\mathcal{P}$.
Is $\mathcal{T}$ a closed subset of $\mathcal{P}$?
My attempt
When need to show if $T_n \in \mathcal{T}$ is such that there is a $P_0$ in $\mathcal{P}$ such that $d(T_n,P_0) \to 0$ then $P_0 \in \mathcal{T}$.
Since $|\mu(T_n) - \mu(P_0)| \leq d(T_n,P_0)$ this also means $\lim_{n\to\infty} \mu(T_n) = \mu(P_0) > 0$.
Let $x_n,y_n,z_n$ denote the vertices of $T_n$ in some order, then if we can show the $T_n$'s are uniformly bounded in $\mathbb{R}^2$ then, passing to a subsequence if necessary, we can assume wlog that $x_n,y_n,z_n$ converges to (say) $x_0,y_0,z_0$. Let $T_0$ be the (possibly degenerate) triangle described by $x_0,y_0,z_0$, then it is sufficient to show $d(T_n,T_0) \to 0$ and $T_0$ is not degenerate.
We have, $d(T_n,T_0) = \mu(T_n) + \mu(T_0) - 2\mu(T_n \cap T_0)$.
Clearly $\mu(T_n) \to \mu(T_0)$. This also shows $\mu(T_0) > 0$ and so $T_0$ is not degenerate.
And we also have, $\mu(T_n \cap T_0) = \int_{x \in \mathbb{R}^2}I_{T_n}(x)I_T(x)dx$ (where $I_A()$ is the indicator function of the set $A$). We can show $I_{T_n}(x) \to 1$ whenever $x$ is an interior point of $T_0$ and $I_{T_n}(x) \to 0$ for $x \not\in T_0$, this is sufficient to show, using DCT, $\mu(T_n \cap T_0) \to \mu(T_0)$ and we are done.
So we need to show $T_n$'s are uniformly bounded in $\mathbb{R}^2$.
 A: The answer is positive. Indeed, suppose that $T_n=A_nB_nC_n$ is a sequence of triangles of unbounded size which converges (in your sense) to a polygon $P$. Then the sequence of areas of $T_n$'s converges to the area of $P$. WLOG, we can assume that the distances $|A_nB_n|$ diverge to infinity. Then the sequence of heights $h_n$ of $T_n$'s (with respect to the base $A_nB_n$) has to converge to zero (otherwise, areas will diverge to infinity). Let $S_n$ denote the parallel strip in the plane $E^2$ which is bounded by the infinite line $(A_nB_n)$ and the parallel line through $C_n$. Thus, $h_n$ is the width of $S_n$ and $T_n\subset S_n$.
Lemma. For every compact $K\subset E^2$  the areas of $T_n\cap K$ converge to zero.
Proof. It suffices to prove the claim for compacts $K$ of the form $B=B(R)$, the closed disk of radius $R$ centered at the origin in $E^2$.  Clearly, the area of $T_n\cap B$ is at most the area of $S_n\cap B$. The latter is at most $2Rh_n$, which converges to zero as $n\to\infty$. qed
Now, we can finish the proof. Take $K=P$. By your notion of convergence $T_n\to P$,  it follows that the areas of $T_n\cap P$ converge to the area of $P$  as $n\to\infty$. By Lemma, this limit equals zero. Hence, $P$ has zero area, contradicting your definition of a polygon.
