# Example of continuous real valued functions

Provide an example of continuous real valued functions, so a map $f : \mathbb{R} \to \mathbb{R}$, that is open but not closed. Make sure to justify why your example subsets satisfy the desired criteria

Given that: A map $f : X \to Y$ is called open if for every open set $U$ in $X$, the set $f(U)$ is open in $Y$ and $f$ is called closed if it maps closed sets in $X$ to closed sets in $Y$.

I am not sure if there has to be one function, or a set of functions being open and not closed

I need a start on this. Please help

• You just need to construct a Single map which is open but not closed.What are your thoughts on this? – Math Lover Oct 28 '16 at 17:31
• I'm not sure such a map exists. – fleablood Oct 28 '16 at 17:37
• It is possible. Why would one think a homeomorphism is impossible? Can we avoid that. – fleablood Oct 28 '16 at 18:51
• Hint: f(R) need not be R. – fleablood Oct 28 '16 at 18:54
• – fleablood Oct 28 '16 at 18:55

## 2 Answers

Hint: It's not hard to show that if $f:\mathbb{R}\rightarrow \mathbb{R}$ is open then it has to be strictly monotone.

• How can such a map not be closed? – fleablood Oct 28 '16 at 17:53
• @fleablood That would spoil the fun for the OP, but it can be done. – Jose27 Oct 28 '16 at 18:11
• It can? ... Oh! D'oh. It can! Hint. An continuous monotone function will map limit points to limit points So to map closed B to not closed C there must be a limit point of C that is not in C which is not mapped from a limit point in B. But it's not mapped from an interior of comp(B) either. So.... – fleablood Oct 28 '16 at 18:29

Continuous functions map closed sets to closed sets. So such continuous open but not closed function doesn't exist.

• Are you sure? That's actually not true. Can f(B) have a limit point y where f^-1(y) is not a limit point of B? – fleablood Oct 28 '16 at 18:34
• Closed sets in R can be null set, R, interval (compact) or countable set. Since function is continuous, R is mapped to R for unbounded function, which is also closed. For bounded function R is mapped to a compact set. Null sets are mapped to null sets. Compact sets are mapped to compact sets. I can't think of any counterexample continuous functions. – jnyan Oct 28 '16 at 19:03
• "Since function is continuous, R is mapped to R for unbounded function" Who said f is unbounded? – fleablood Oct 28 '16 at 19:07
• I considered both possibilities. Bounded and unbounded – jnyan Oct 28 '16 at 19:09
• "For bounded function R is mapped to a compact set." Why do you think this? Indeed that is certainly false. – fleablood Oct 28 '16 at 19:10