# Function 1-1 thorughout, if its 1-1 on an open set.

• Jan 18th 2010, 04:26 AM
Chandru1
Function 1-1 thorughout, if its 1-1 on an open set.
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.
• Jan 18th 2010, 05:28 AM
tonio
Quote:

Originally Posted by Chandru1
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