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I am having difficulties proving the equivalence of these statements.

Show that the following statements are equivalent

(1) $f: D \to \mathbb R^n$ is continuous.

(2) For every closed ball $B$ in $\mathbb R^n$ , the inverse image of $B$ under $f$ is closed in $D$.

(3) For every closed subset $S$ of $\mathbb R^n$, the inverse image of $S$ under $f$ is closed in $D$.

So to show equivalence I have to show the following implication chain right? $(1)\Rightarrow (2) \Rightarrow (3)\Rightarrow (1)$

I do understand the analogous proof for open balls and subsets but I don't know where to start based from this..I would appreciate any help!

Thank you

Maria

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  • $\begingroup$ You have to show (1)<=> (2)<=>(3)<=>(1) $\endgroup$
    – jnyan
    Commented Sep 17, 2016 at 9:54
  • $\begingroup$ I suppose $(1)\implies (3)\implies (2)\implies (1)$ is simpler $\endgroup$
    – Hagen von Eitzen
    Commented Sep 17, 2016 at 10:05

2 Answers 2

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You have proved the analogous statement about open sets, right?

$(1)\Rightarrow (2)$: Assume $f$ is continuous, and let $B$ be a closed ball in $\mathbb{R}^n$. Then $U=\mathbb{R}^n\setminus B$ is an open set, so its inverse image is open in $D$. How is the inverse image of $U$ connected to the inverse image of $B$?

$(2)\Rightarrow (3)$: Exactly analogous to the statement about open balls and open sets.

$(3)\Rightarrow (1)$: You have to show that $f$ is continuous. Let $U$ be open in $\mathbb{R}^n$ and show that $f^{-1}(U)$ is open in $D$, again using the complement set trick.

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  • $\begingroup$ 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? $\endgroup$
    – Maria
    Commented Sep 17, 2016 at 10:25
  • $\begingroup$ 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. $\endgroup$
    – Olivier Moschetta
    Commented Sep 17, 2016 at 10:29
  • $\begingroup$ 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 $\endgroup$
    – Maria
    Commented Sep 17, 2016 at 10:45
  • $\begingroup$ 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. $\endgroup$
    – Olivier Moschetta
    Commented Sep 17, 2016 at 12:33
  • $\begingroup$ 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? $\endgroup$
    – Maria
    Commented Sep 17, 2016 at 13:08
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Okay, so I have written down the proof for (1)=>(2)=>(3)=>(1) I will just write it in words, I am not familiar yet how to write the mathematical signs properly (in my notebook I write it in mathematical terms of course). Can someone take a look at it whether there are logical mistakes?

part 1 (1)=>(2): Assume f is continuous. Let B be a closed ball in R^n. To show: the inverse of B is closed in D. Take U=R^n\B which is open by definition. Then the inverse of U is open, because of the theorem with open balls. Then the invverse of B = the inverse of R^n\U = D\inverse of U. Because the inverse of U is open, then the inverse of B must be closed. qed.

part 2 (2)=>(3): Assume (2) holds. Let S be a closed set in R^n. To show: the inverse of S is closed. The complement of S (let's call it Sc) is open because is closed. Sc made up of the union of n open balls Bi. So Sc= union Bi. Because every open ball in R^n has an open counterpart in the domain (according to statement (2)), all the Bi's in D will be open sets. The inverse of Sc is therefore open. The inverse of S is therefore closed. qed.

part 3 (3)=>(1): Assume (3) holds. Let x be an element of D and epsilon > 0. Construct a closed ball around f(x) with a radius of epsilon, which we call eball. Then the inverse of the eball will also be closed (according to statement (3)). Because f(x) is an element of the eball, x is an element of the inverse of the eball. Hence, there must be a delta, such that a smaller ball (let's call it deltaball or dball) is a subset of the inverse of the eball (in other words we fit a small ball in the inverse of the eball). This is equivalent to saying that f(dball) is a subset of the eball. And this statement is proving that f is continuous in x (Cauchy definition of continuity). And because we chose x arbitrarily, (1) holds for any x. qed.

Thank you!

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  • $\begingroup$ (3)=>(1) It would be far easier to show that f is continuous by showing that every open set U in $\mathbb{R}^n$ is such that the inverse image of U is open (the two things are known to be equivalent). This would work as in part 1. Let U be open in R^n. Then S=R^n\U is closed. According to (3) the inverse image of S is also closed, hence the inverse image of U is open and f is continuous. $\endgroup$
    – Olivier Moschetta
    Commented Sep 17, 2016 at 14:33
  • $\begingroup$ (2)=>(3) An open set is not necessarily the union of open balls. Rather, a set U is open if for every x in U there is an open ball B around x so that $B\subset U$. $\endgroup$
    – Olivier Moschetta
    Commented Sep 17, 2016 at 14:40

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