Need help in metric spaces proving this statement! Please if someone could help me prove this rather annoying statement.
Let $C(0,1)$ be the set of continuous functions on the open interval $(0,1) \subset \mathbb R$. Fro any two functions $x(t), y(t) \in C(0,1)$ define the set $E(x,y)=\{t \in (0,1) | x(t) \neq y(t)\}$.
Show that $E(x,y)$ is a union of disjoint open intervals.
I hope I've been clear enough. Thanks. 
 A: $E(x,y)$ is open in $(0,1)$ (by continuity of $x$ and $y$) so open in $\mathbb{R}$. Every open subset of the reals is a disjoint union of open intervals (namely the connected components of the open set).
To see the openness: consider $f: (0,1) \rightarrow \mathbb{R}^2$ defined by $f(t) = (x(t), y(t))$. Then $f$ is continuous when $x,y$ are. Then $\Delta = \{(x,y): x = y \}$ is closed in $\mathbb{R}^2$, and $E(x,y) = (0,1) \setminus f^{-1}[\Delta]$, which is the complement of a closed set. This works with all codomains. You can also consider $d(t) = x(t) - y(t)$ and $E(x,y) = d^{-1}[\mathbb{R}\setminus \{0\}]$ where $\mathbb{R} \setminus \{0\}$ is open.
A: Addition to the answer by Henno Brandsma: Take the function $h:(0,1)\rightarrow \mathbb R$ with $h(t) = x(t)-y(t)$, which is continuously (because $x$ and $y$ are continously). You have $$E(x,y) = h^{-1}(\mathbb R\setminus\{0\})$$ and thus $E(x,y)$ is open ($E(x,y)$ is the inverse image of the open set $\mathbb R \setminus \{0\}$ under the continuous function $h$).
