How many connected components does $A$ have? Consider the problem given below.
Let
\begin{equation*}
X = \{(a, b, c) \in \mathbb{C}^3 | a \neq b \neq c \neq a\}
\end{equation*}
be the set of pairwise distinct triples of points in $\mathbb{C} \cong \mathbb{R}^2$. Each triple $(a, b, c)$ determines a (possibly degenerate) triangle $\triangle a b c$ in the plane. Give $\mathbb{C}^3 \cong \mathbb{R}^6$ the usual metric topology, and give $X$ the subspace topology. Let
\begin{equation*}
A = \{(a, b, c) \in X | |b - a| \neq |c - b| \neq |a - c| \neq |b - a|\}
\end{equation*}
be the subspace of $X$ corresponding to the non-isosceles triangles $\triangle a b c$. How many connected components does A have?
I conjecture that $A$ is connected, and so has exactly 1 connected component (namely, itself); however, I do not know how to go about proving this. I do know that
\begin{equation*}
f(a, b, c) = (a, b - a, \frac{c - a}{b - a})
\end{equation*}
defines a homeomorphism $f: X \rightarrow Y$, here where
\begin{equation*}
Y = \mathbb{C} \times (\mathbb{C} - \{0\}) \times (\mathbb{C} - \{0, 1\})
\end{equation*}
viewed as a subspace of $\mathbb{C}^3$.
I also know that
\begin{equation*}
f(A) = \mathbb{C} \times (\mathbb{C} - \{0\}) \times Z,
\end{equation*}
here where $Z$ is a subspace of $\mathbb{C} - \{0, 1\}$. Does this help?
 A: $A$ has one component. Firstly, we note that $A$ is invariant under translation, rotation, scaling, and reflection, as each of these operations on a triangle produces a similar triangle. Now the first three of these can be done continuously, showing that a triangles related by these operations are connected by a path in $A$.
WLOG, let $(0,1,x+iy) \in A$ be a triangle, with $x,y\in\mathbb{R}$ the side from $0$ to $1$ the longest, and the side from $0$ to $x+iy$ the second longest (this implies $x>1/2$). Then $(0,1,x+iy)$ and its reflection $(0,1,x-iy)\in A$ are connected by a path in $A$ given by $$\gamma(t)=(0,1,x+(1-2t)iy).$$To show this, we note that $$1>\sqrt{x^2+y^2}>\sqrt{x^2+(1-2t)^2y^2}>\sqrt{(x-1)^2+(1-2t)^2y^2},$$
or equivalently,
$$1>|x+(1-2t)iy|>|x+(1-2t)iy-1|$$
The order of side length remains the same throughout, so the path lies in $A$.
Now we can move on to the main problem. WLOG, let $(0,1,x+iy),(0,1,x'+iy')\in A$ be two triangles with $x,y,x',y'\in\mathbb{R}$, the side from $0$ to $1$ the longest, and the side from $0$ to $x+iy$ or $x'+iy'$ the second longest for each triangle. Then the two triangles are connected by a path in $A$ given by
$$\gamma(t)=(0,1,(1-t)(x+iy)+t(x'+iy')).$$
It suffices to check that the order of the side lengths remains the same throughout the path. We can see that
$$1>\max\left(\sqrt{x^2+y^2},\sqrt{x'^2+y'^2}\right)\geq\sqrt{((1-t)x+tx')^2+((1-t)y+ty')}>\sqrt{((1-t)x+tx'-1)^2+((1-t)y+ty')}$$
which finishes the proof.
