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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

Eduardo Longa
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user84324
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2 Answers2

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

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

jnyan
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  • 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
  • Bounded and continuous. – jnyan Oct 28 '16 at 19:12
  • f(x)= x/(|x| + 1) is bounded as -1 < f(x) < 1. – fleablood Oct 28 '16 at 19:12
  • f(x) = 2^x. f(R) = (0, infty). Is not compact. – fleablood Oct 28 '16 at 19:13
  • Oh. So can this be the answer. Since it maps every open set to open set and there is a closed set which is not mapped to a closed set – jnyan Oct 28 '16 at 19:19