Consider $X = [0,1] \subset \mathbb{R}$ with euclidean subspace topology. Let $R$ be a subset of $X \times X$ with all the pairs $(x,y)$ in it, such that $x = y$ , or $x,y \in \{0,1/2,1\} $. We know $R$ defines an equivalence relation and the resulting quotient space is denoted with $X/\sim$. The question know is, whether $p([0,1/4))$ and $p((1/4,3/4))$ are open subsets in $X\backslash \sim $ , where p ist the canonical projective map $p: X \rightarrow X\backslash \sim$ and $[0,1/4)$, $(1/4,3/4)$ are already open in X.
My approach was looking at $0,1/2$ and $1$, since the projective maps assigns those elements all to the same equivalence class ( $[0] = [1] = [1/2]$) ,and if, say $p([0,1/4))$ was an open set, $p^{-1}(p([0,1/4))$ must have been open, since $p$ is continuous. Calculating $p^{-1}(p([0,1/4))$ results in $p^{-1}(p([0,1/4)) = \{0,1/2,1\} \cup (0,1/4)$ which is not open in $X$. Same was concluded for $p((1/4,3,4)$.
Now my question is, was my approach right and if not, what did I miss?
Best wishes, M.Uon
Yes, you are right. Basically identifying the points 0, 1/2, 1, that gives you an infinity sign (folding together the line segment along these three points). If either of 0, 1/2, 1 are in U, then the intersection point is in p(U). In this case p(U), can be only be open if it contains a whole neighborhood of the intersection. This is not the case for your sets, so they cannot be open. Even more, they are actually neither open, nor closed.
Apart from this visual heuristics, your proof is the correct, rigorous one.