subset under continuous functions

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Sep 21, 2016 at 11:37	comment	added	Olivier Moschetta		Oh (2) does imply (3). I just meant that while obviously (3)=>(2) it is not quite clear that (2)=>(3) (although it is true).
Sep 21, 2016 at 7:56	comment	added	Maria		Thanks a lot for the help, I have a last question though, do you know of an example in which we can show that (2) does not imply (3)? (referring to your second comment in this thread)
Sep 17, 2016 at 14:15	comment	added	Maria		Okay, thank you! I have written my proof below in a separate answer, would you mind take a look at it?
Sep 17, 2016 at 13:14	comment	added	Olivier Moschetta		You're right, we have proven (1)=>(2) and (1)=>(3) at the same time. You can go either (1)=>(2)=>(3)=>(1) or (1)=>(3)=>(2)=>(1), it doesn't change much.
Sep 17, 2016 at 13:08	comment	added	Maria		First of all thank you very much for taking your time to answer!! Okay, I understand your proof. But then I again don't see the difference to proving (1)=>(3). For me it seems exactly analogous whether we take the ball B or the set S. You also didn't use any definition of what the ball B actually is.. So closed ball => closed subset, but closed subset =!> closed ball, right? Can you explain why that is the case?
Sep 17, 2016 at 12:33	comment	added	Olivier Moschetta		You have already proved the statement for open sets.. Instead we shall use that f continuous $\Leftrightarrow$ $f^{-1}(U)$ is open whenever $U$ is open. Let's call this assertion "Theorem 1". To prove 1=>2: Assume $f$ is continuous on $D$. Let $B$ be a closed ball in $\mathbb{R}^n$. We need to prove that $f^{-1}(B)$ is closed in $D$. Now $U=\mathbb{R}^n\setminus B$ is open. By Theorem 1, $f^{-1}(U)$ is open in $D$. But since $f^{-1}(B)=f^{-1}(\mathbb{R}^n\setminus U)=D\setminus f^{-1}(U)$ and $f^{-1}(U)$ is open, we see that $f^{-1}(B)$ is closed since it's the complement of an open set.
Sep 17, 2016 at 10:45	comment	added	Maria		Okay,I get this. But I still cannot make the connection from open to closed. For instance in (1)=>(2), they prove it as follows: Assume f is continuous. Let B be an open ball in R^n and let x be an element of the inverse of B. Then f(x) is an element of B. Because B is open there exsits some e-ball around f(x) that is a subset of B. And by the Cauchy definition of continuity, inside this e-ball there is a d-ball which has its preimage in the inverse of B. Hence the d-ball is open and that means the inverse of B is open. I understand what you said in the first part but I cant connect it to this
Sep 17, 2016 at 10:29	comment	added	Olivier Moschetta		The inverse image of B is the complement of the inverse image of U (which is open), so it a closed set. (3) is strictly stronger than (2). (2) states that all closed balls have a certain property and (3) that all closed sets (a larger category) have the same property. So (3)=>(2) obviously, but not the other way around.
Sep 17, 2016 at 10:25	comment	added	Maria		To answer (1)=(2): The inverse image of B should then be closed and QED? But then I don't understand the difference between the statements (2) and (3), because a closed ball is a closed subset?
Sep 17, 2016 at 10:17	history	answered	Olivier Moschetta	CC BY-SA 3.0