We have the following statement:
Let be $f$ a continuous function. The preimage of a closed set is again a closed set.
Let be $f:(0,\infty)\times [0,\infty)\to\mathbb{R}$ where $f(x,y)=x+y$.
(Where "$[$" means that $0$ is included and "$($" means that $0$ is excluded).
If I consider the preimage $f^{-1}(1)$ or in other words $D:=\{(x,y)\in (0,\infty)\times [0,\infty)\mid f(x,y)=1\}$ then this would mean that it is closed. However, it is easy to construct a counter-example to see that not all limit points of $D$ are an element of $D$ (e.g. the limit point $(0,1)$). So $D$ can't be a closed set.
How is this possible? Where is my mistake?