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

**Chandru1** · January 18th 2010, 04:26 AM

Let $f:\mathbb{R}^{n} \to \mathbb{R}^{n}$ be a continuous function which is 1-1 on any open set in $\mathbb{R}^{n}$ . Prove that $f$ is 1-1 throughout the domain.

**tonio** · January 18th 2010, 05:28 AM

Originally Posted by Chandru1

Let $f:\mathbb{R}^{n} \to \mathbb{R}^{n}$ be a continuous function which is 1-1 on any open set in $\mathbb{R}^{n}$ . Prove that $f$ is 1-1 throughout the domain.

Solution 1 (The wiseguy solution): $\mathbb{R}^n$ is open in itself and thus, according to the given data, $f$ is 1-1 there.

Solution 2: $\mathbb{R}^n-\{0\}$ is open so $f$ is 1-1 there. Now, if $f(0)=f(x_1)$ , for some $x_1\in\mathbb{R}^n$ ,then take $\mathbb{R}^n-\{x_2\}\,,\,\,x_2\neq 0\,,\,x_1$ ...

By the way , the continuity of $f$ isn't required.

Tonio

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