Let $\displaystyle f:\mathbb{R}^{n} \to \mathbb{R}^{n}$ be a continuous function which is 1-1 on any open set in $\displaystyle \mathbb{R}^{n}$. Prove that $\displaystyle f$ is 1-1 throughout the domain.
Let $\displaystyle f:\mathbb{R}^{n} \to \mathbb{R}^{n}$ be a continuous function which is 1-1 on any open set in $\displaystyle \mathbb{R}^{n}$. Prove that $\displaystyle f$ is 1-1 throughout the domain.
Solution 1 (The wiseguy solution): $\displaystyle \mathbb{R}^n$ is open in itself and thus, according to the given data, $\displaystyle f$ is 1-1 there.
Solution 2: $\displaystyle \mathbb{R}^n-\{0\}$ is open so $\displaystyle f$ is 1-1 there. Now, if $\displaystyle f(0)=f(x_1)$ , for some $\displaystyle x_1\in\mathbb{R}^n$ ,then take $\displaystyle \mathbb{R}^n-\{x_2\}\,,\,\,x_2\neq 0\,,\,x_1$ ...
By the way , the continuity of $\displaystyle f$ isn't required.
Tonio