Function with open set properties

**Pinkk** · August 31st 2010, 07:28 PM

Suppose $f: \mathbb{R}^{n} \to \mathbb{R}^{k}$ has the following property: For any open set $U \subset \mathbb{R}^{k}$ , $S = \{x : f(x)\in U\}$ is an open set in $\mathbb{R}^{n}$ . Show that $f$ is continuous.

I am completely stuck on this. I know that there is a connection between the $\epsilon - \delta$ definition of continuity and how if $S$ is an open set, then given $y\in S$ , there exists $r>0$ such that $|x - y | < r \implies x\in S$ but I am missing the connection and can't really formalize any sort of proof. Thanks in advance.

**Iondor** · September 1st 2010, 03:20 AM

The balls B(x,r)={y : |x-y|<r} are open sets.
Maybe this helps already?

**Pinkk** · September 1st 2010, 04:47 AM

Hmm, thanks, I think that does help. So here is my attempt.

Given $\epsilon > 0$ and $a\in \mathbb{R}^{n}$ , let $U=B(\epsilon , f(a))$ , which is clearly an open set $\{y : |y - f(a)| < \epsilon \}$ . Since $f(a)\in U$ , we have that $a\in S = \{x : f(x)\in U\}$ . So $a$ is an interior point of $S$ , which means there exists $\delta > 0$ such that $|x - a|< \delta$ implies $x\in S$ , which in turn implies $f(x)\in U$ . Since $f(x)\in U$ implies $|f(x)-f(a)|<\epsilon$ , we have that given $\epsilon > 0$ , there exists $\delta > 0$ such that $|x - a| < \delta$ implies $|f(x) - f(a)| < \epsilon$ , which means $f$ is continuous at $a$ . Since $a$ was arbitrary, we have that $f$ is continuous on $\mathbb{R}^{n}$ . Q.E.D.

**Isomorphism** · September 1st 2010, 08:15 AM

Originally Posted by Pinkk

Hmm, thanks, I think that does help. So here is my attempt.

Given $\epsilon > 0$ and $a\in \mathbb{R}^{n}$ , let $U=B(\epsilon , f(a))$ , which is clearly an open set $\{y : |y - f(a)| < \epsilon \}$ . Since $f(a)\in U$ , we have that $a\in S = \{x : f(x)\in U\}$ . So $a$ is an interior point of $S$ , which means there exists $\delta > 0$ such that $|x - a|< \delta$ implies $x\in S$ , which in turn implies $f(x)\in U$ . Since $f(x)\in U$ implies $|f(x)-f(a)|<\epsilon$ , we have that given $\epsilon > 0$ , there exists $\delta > 0$ such that $|x - a| < \delta$ implies $|f(x) - f(a)| < \epsilon$ , which means $f$ is continuous at $a$ . Since $a$ was arbitrary, we have that $f$ is continuous on $\mathbb{R}^{n}$ . Q.E.D.

I think it is perfectly rigorous

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