Is the equality $(0,1]\cup(1,2)=(0,2)$ true? I believe this is true because the first set contains everything up to and including $1$ and the second contains everything from $1$ onwards.
 A: Um... yes?
If $a \in (0,1]$ then $a \in (0,2)$ so $(0,1] \subset (0,2)$.
If $a \in (1, 2)$ then $a \in (0,2)$ so $(1,2) \subset (0,2)$.
So $(0,1] \cup (1,2) \subset (0,2)$.
If $b \in (0,2)$ then $b > 0$ and $b < 2$ and either 1) $b > 1$ or 2) $b \le 1$.
If 1) $b \in (1,2)$ and $b \in (0,1] \cup (1,2)$
If 2) $b \in (0,1]$ and $b \in (0,1] \cup (1,2)$
So $(0,2) \subset (0,1] \cup (1,2)$.
So $(0,1] \cup (1,2) = (0,2)$.
Or...
$(0,1] = \{x| 0< x \le 1\}; (1,2) = \{x| 1< x < 2\}$ so $(0,1] \cup (1,2) = \{x| 0<x\le 1 \text{ or } 1<x < 2\} = \{x| 0< x< 2\} = (0,1)$.
A: This is only true if you work with classical logic.  If the background logic is intuitionistic then the equality $(0,1]\cup(1,2)=(0,2)$ may not hold.  This is related to the failure of the law of trichotomy.
A: It appears to be true. To prove it, you generally need two steps:
1) Let $x\in(0,1]\cup(1,2)$. Then  $0<x\le 1$ or $1<x<2$. In either case, $x\in(0,2)$.
2) Let $x\in (0,2)$. Then either $x\le 1$ or $x>1$. Thus $x\in(0,1]$ or $x\in(1,2)$, so $x\in(0,1]\cup(1,2)$.
Depending on the level or rigor, it is important to be able to prove claims like this even if they appear obvious.
A: Let $A,B\subset X$. Then $A\cup B:=\{x\in X\ \colon\  x\in A\ \vee\ x\in B\}$.
In yours case we have: $A=(0,1]$ and $B=(1,2)$.
It means that: $(0,1]\cup(1,2)=\{x\in\mathbb{R}\ \colon\ x\in (0,1]\ \vee\ x\in (1,2)\}=(0,2)$.
Note that, since $1\in (0,1]$ then $1\in (0,1]\cup(1,2)$.
