If $I$ is an interval of $Bbb{R}$, $f: I to Bbb{R}$ connected and $forall y in I; f^{-1}({y})$ closed in $I$...











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I have found this theorem in a calculus book




We say a function $f: I to Bbb{R}$, $I$ interval of $Bbb{R}$ is connected if $forall J subseteq I$, with $J$ interval, $f(J)$ is an interval. Prove that if $f$ is connected and if $forall y in Bbb{R} , f^{-1}({y})$ is a closed set in the relative topology of $I$, then $f$ is continuous.




I have no idea where I can start, can you give me some hints?










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  • 1




    To the OP: I would recommend adding the topology tag. This is a question that seems more suited toward a topological answer than a calculus-type of answer. (I have ideas from topology on how to approach this problem, but I personally am not sure how I’d be using strictly calculus topics to approach this problem. Hence my first question; you could be taking the two courses concurrently and mixed up the titles.)
    – Clayton
    Nov 14 at 17:10










  • @Clayton if you can solve it using topology it's good too. Perhaps the problem is that I cannot distinguish between calculus and mathematical analysis since in my main language the latter is used to indicate both; but having seen some foreign books of both, I'm quite confident the one I use is a calculus book. I'm sorry for any misunderstandings due to this semantical problem...
    – LuxGiammi
    Nov 14 at 17:10










  • Note: anyway, I'm still a student and I've not begun studying mathematical analysis yet...
    – LuxGiammi
    Nov 14 at 17:12










  • Can you state a few of the topological definitions/theorems that you have available? I assume you know the “inverse image of an open set is open” characterization for continuous functions?
    – Clayton
    Nov 14 at 17:16






  • 1




    @Clayton I know that $f: D rightarrow Bbb{R}$ is continuous $iff$ $forall X,; text{X open set of } Bbb{R} ; f^{-1}(X) = f^leftarrow(X)$ is opened in the relative topology of $D$ and that holds also if we replace "opened" with "closed". Anyway, I have some elementary general topology knowledge and the book states some of them (actually, three chapters of it are dedicated to topology...).
    – LuxGiammi
    Nov 14 at 19:27

















up vote
4
down vote

favorite












I have found this theorem in a calculus book




We say a function $f: I to Bbb{R}$, $I$ interval of $Bbb{R}$ is connected if $forall J subseteq I$, with $J$ interval, $f(J)$ is an interval. Prove that if $f$ is connected and if $forall y in Bbb{R} , f^{-1}({y})$ is a closed set in the relative topology of $I$, then $f$ is continuous.




I have no idea where I can start, can you give me some hints?










share|cite|improve this question




















  • 1




    To the OP: I would recommend adding the topology tag. This is a question that seems more suited toward a topological answer than a calculus-type of answer. (I have ideas from topology on how to approach this problem, but I personally am not sure how I’d be using strictly calculus topics to approach this problem. Hence my first question; you could be taking the two courses concurrently and mixed up the titles.)
    – Clayton
    Nov 14 at 17:10










  • @Clayton if you can solve it using topology it's good too. Perhaps the problem is that I cannot distinguish between calculus and mathematical analysis since in my main language the latter is used to indicate both; but having seen some foreign books of both, I'm quite confident the one I use is a calculus book. I'm sorry for any misunderstandings due to this semantical problem...
    – LuxGiammi
    Nov 14 at 17:10










  • Note: anyway, I'm still a student and I've not begun studying mathematical analysis yet...
    – LuxGiammi
    Nov 14 at 17:12










  • Can you state a few of the topological definitions/theorems that you have available? I assume you know the “inverse image of an open set is open” characterization for continuous functions?
    – Clayton
    Nov 14 at 17:16






  • 1




    @Clayton I know that $f: D rightarrow Bbb{R}$ is continuous $iff$ $forall X,; text{X open set of } Bbb{R} ; f^{-1}(X) = f^leftarrow(X)$ is opened in the relative topology of $D$ and that holds also if we replace "opened" with "closed". Anyway, I have some elementary general topology knowledge and the book states some of them (actually, three chapters of it are dedicated to topology...).
    – LuxGiammi
    Nov 14 at 19:27















up vote
4
down vote

favorite









up vote
4
down vote

favorite











I have found this theorem in a calculus book




We say a function $f: I to Bbb{R}$, $I$ interval of $Bbb{R}$ is connected if $forall J subseteq I$, with $J$ interval, $f(J)$ is an interval. Prove that if $f$ is connected and if $forall y in Bbb{R} , f^{-1}({y})$ is a closed set in the relative topology of $I$, then $f$ is continuous.




I have no idea where I can start, can you give me some hints?










share|cite|improve this question















I have found this theorem in a calculus book




We say a function $f: I to Bbb{R}$, $I$ interval of $Bbb{R}$ is connected if $forall J subseteq I$, with $J$ interval, $f(J)$ is an interval. Prove that if $f$ is connected and if $forall y in Bbb{R} , f^{-1}({y})$ is a closed set in the relative topology of $I$, then $f$ is continuous.




I have no idea where I can start, can you give me some hints?







calculus general-topology continuity






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 15 at 6:20

























asked Nov 14 at 15:34









LuxGiammi

1609




1609








  • 1




    To the OP: I would recommend adding the topology tag. This is a question that seems more suited toward a topological answer than a calculus-type of answer. (I have ideas from topology on how to approach this problem, but I personally am not sure how I’d be using strictly calculus topics to approach this problem. Hence my first question; you could be taking the two courses concurrently and mixed up the titles.)
    – Clayton
    Nov 14 at 17:10










  • @Clayton if you can solve it using topology it's good too. Perhaps the problem is that I cannot distinguish between calculus and mathematical analysis since in my main language the latter is used to indicate both; but having seen some foreign books of both, I'm quite confident the one I use is a calculus book. I'm sorry for any misunderstandings due to this semantical problem...
    – LuxGiammi
    Nov 14 at 17:10










  • Note: anyway, I'm still a student and I've not begun studying mathematical analysis yet...
    – LuxGiammi
    Nov 14 at 17:12










  • Can you state a few of the topological definitions/theorems that you have available? I assume you know the “inverse image of an open set is open” characterization for continuous functions?
    – Clayton
    Nov 14 at 17:16






  • 1




    @Clayton I know that $f: D rightarrow Bbb{R}$ is continuous $iff$ $forall X,; text{X open set of } Bbb{R} ; f^{-1}(X) = f^leftarrow(X)$ is opened in the relative topology of $D$ and that holds also if we replace "opened" with "closed". Anyway, I have some elementary general topology knowledge and the book states some of them (actually, three chapters of it are dedicated to topology...).
    – LuxGiammi
    Nov 14 at 19:27
















  • 1




    To the OP: I would recommend adding the topology tag. This is a question that seems more suited toward a topological answer than a calculus-type of answer. (I have ideas from topology on how to approach this problem, but I personally am not sure how I’d be using strictly calculus topics to approach this problem. Hence my first question; you could be taking the two courses concurrently and mixed up the titles.)
    – Clayton
    Nov 14 at 17:10










  • @Clayton if you can solve it using topology it's good too. Perhaps the problem is that I cannot distinguish between calculus and mathematical analysis since in my main language the latter is used to indicate both; but having seen some foreign books of both, I'm quite confident the one I use is a calculus book. I'm sorry for any misunderstandings due to this semantical problem...
    – LuxGiammi
    Nov 14 at 17:10










  • Note: anyway, I'm still a student and I've not begun studying mathematical analysis yet...
    – LuxGiammi
    Nov 14 at 17:12










  • Can you state a few of the topological definitions/theorems that you have available? I assume you know the “inverse image of an open set is open” characterization for continuous functions?
    – Clayton
    Nov 14 at 17:16






  • 1




    @Clayton I know that $f: D rightarrow Bbb{R}$ is continuous $iff$ $forall X,; text{X open set of } Bbb{R} ; f^{-1}(X) = f^leftarrow(X)$ is opened in the relative topology of $D$ and that holds also if we replace "opened" with "closed". Anyway, I have some elementary general topology knowledge and the book states some of them (actually, three chapters of it are dedicated to topology...).
    – LuxGiammi
    Nov 14 at 19:27










1




1




To the OP: I would recommend adding the topology tag. This is a question that seems more suited toward a topological answer than a calculus-type of answer. (I have ideas from topology on how to approach this problem, but I personally am not sure how I’d be using strictly calculus topics to approach this problem. Hence my first question; you could be taking the two courses concurrently and mixed up the titles.)
– Clayton
Nov 14 at 17:10




To the OP: I would recommend adding the topology tag. This is a question that seems more suited toward a topological answer than a calculus-type of answer. (I have ideas from topology on how to approach this problem, but I personally am not sure how I’d be using strictly calculus topics to approach this problem. Hence my first question; you could be taking the two courses concurrently and mixed up the titles.)
– Clayton
Nov 14 at 17:10












@Clayton if you can solve it using topology it's good too. Perhaps the problem is that I cannot distinguish between calculus and mathematical analysis since in my main language the latter is used to indicate both; but having seen some foreign books of both, I'm quite confident the one I use is a calculus book. I'm sorry for any misunderstandings due to this semantical problem...
– LuxGiammi
Nov 14 at 17:10




@Clayton if you can solve it using topology it's good too. Perhaps the problem is that I cannot distinguish between calculus and mathematical analysis since in my main language the latter is used to indicate both; but having seen some foreign books of both, I'm quite confident the one I use is a calculus book. I'm sorry for any misunderstandings due to this semantical problem...
– LuxGiammi
Nov 14 at 17:10












Note: anyway, I'm still a student and I've not begun studying mathematical analysis yet...
– LuxGiammi
Nov 14 at 17:12




Note: anyway, I'm still a student and I've not begun studying mathematical analysis yet...
– LuxGiammi
Nov 14 at 17:12












Can you state a few of the topological definitions/theorems that you have available? I assume you know the “inverse image of an open set is open” characterization for continuous functions?
– Clayton
Nov 14 at 17:16




Can you state a few of the topological definitions/theorems that you have available? I assume you know the “inverse image of an open set is open” characterization for continuous functions?
– Clayton
Nov 14 at 17:16




1




1




@Clayton I know that $f: D rightarrow Bbb{R}$ is continuous $iff$ $forall X,; text{X open set of } Bbb{R} ; f^{-1}(X) = f^leftarrow(X)$ is opened in the relative topology of $D$ and that holds also if we replace "opened" with "closed". Anyway, I have some elementary general topology knowledge and the book states some of them (actually, three chapters of it are dedicated to topology...).
– LuxGiammi
Nov 14 at 19:27






@Clayton I know that $f: D rightarrow Bbb{R}$ is continuous $iff$ $forall X,; text{X open set of } Bbb{R} ; f^{-1}(X) = f^leftarrow(X)$ is opened in the relative topology of $D$ and that holds also if we replace "opened" with "closed". Anyway, I have some elementary general topology knowledge and the book states some of them (actually, three chapters of it are dedicated to topology...).
– LuxGiammi
Nov 14 at 19:27












1 Answer
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For $x in I$ let $I_n(x) = I cap (x - frac{1}{n},x + frac{1}{n})$. This is an open neighborhood of $x$ in $I$. We have
$$(*) phantom{xx} bigcap_{n=1}^infty f(I_n(x)) = { f(x) } .$$
"$supset$" is trivial. To verify "$subset$", let $y in bigcap_{n=1}^infty f(I_n(x))$. Then there exist $x_n in I_n(x)$ such that $f(x_n) = y$. Obviously $x_n to x$. Since $f^{-1}(y)$ is closed in $I$ , we see that $x in f^{-1}(y)$, i.e. $f(x) = y$.



Let $varepsilon > 0$. Since $I_n(x)$ is an interval, also $J_n = f(I_n(x))$ is an interval. We have $J_n = langle a_n,b_n rangle$, where $langle a_n,b_n rangle$ stands for an open, half-open or closed interval such that $a_n le f(x) le b_n$. $a_n = -infty$, $b_n = infty$ is allowed. The sequence $(a_n)$ is increasing, the sequence $(b_n)$ decreasing. But $(*)$ shows that $a_n, b_n to f(x)$, hence $f(I_n(x)) = J_n subset (f(x)- varepsilon, f(x)+ varepsilon)$ for sufficiently large $n$. This means that $f$ is continuous.






share|cite|improve this answer























  • Very good proof. I've just skimmed through it and I've understood pretty much everything. I will look in more details into it in the next few days. Thanks!
    – LuxGiammi
    Nov 16 at 15:56













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active

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up vote
3
down vote



accepted










For $x in I$ let $I_n(x) = I cap (x - frac{1}{n},x + frac{1}{n})$. This is an open neighborhood of $x$ in $I$. We have
$$(*) phantom{xx} bigcap_{n=1}^infty f(I_n(x)) = { f(x) } .$$
"$supset$" is trivial. To verify "$subset$", let $y in bigcap_{n=1}^infty f(I_n(x))$. Then there exist $x_n in I_n(x)$ such that $f(x_n) = y$. Obviously $x_n to x$. Since $f^{-1}(y)$ is closed in $I$ , we see that $x in f^{-1}(y)$, i.e. $f(x) = y$.



Let $varepsilon > 0$. Since $I_n(x)$ is an interval, also $J_n = f(I_n(x))$ is an interval. We have $J_n = langle a_n,b_n rangle$, where $langle a_n,b_n rangle$ stands for an open, half-open or closed interval such that $a_n le f(x) le b_n$. $a_n = -infty$, $b_n = infty$ is allowed. The sequence $(a_n)$ is increasing, the sequence $(b_n)$ decreasing. But $(*)$ shows that $a_n, b_n to f(x)$, hence $f(I_n(x)) = J_n subset (f(x)- varepsilon, f(x)+ varepsilon)$ for sufficiently large $n$. This means that $f$ is continuous.






share|cite|improve this answer























  • Very good proof. I've just skimmed through it and I've understood pretty much everything. I will look in more details into it in the next few days. Thanks!
    – LuxGiammi
    Nov 16 at 15:56

















up vote
3
down vote



accepted










For $x in I$ let $I_n(x) = I cap (x - frac{1}{n},x + frac{1}{n})$. This is an open neighborhood of $x$ in $I$. We have
$$(*) phantom{xx} bigcap_{n=1}^infty f(I_n(x)) = { f(x) } .$$
"$supset$" is trivial. To verify "$subset$", let $y in bigcap_{n=1}^infty f(I_n(x))$. Then there exist $x_n in I_n(x)$ such that $f(x_n) = y$. Obviously $x_n to x$. Since $f^{-1}(y)$ is closed in $I$ , we see that $x in f^{-1}(y)$, i.e. $f(x) = y$.



Let $varepsilon > 0$. Since $I_n(x)$ is an interval, also $J_n = f(I_n(x))$ is an interval. We have $J_n = langle a_n,b_n rangle$, where $langle a_n,b_n rangle$ stands for an open, half-open or closed interval such that $a_n le f(x) le b_n$. $a_n = -infty$, $b_n = infty$ is allowed. The sequence $(a_n)$ is increasing, the sequence $(b_n)$ decreasing. But $(*)$ shows that $a_n, b_n to f(x)$, hence $f(I_n(x)) = J_n subset (f(x)- varepsilon, f(x)+ varepsilon)$ for sufficiently large $n$. This means that $f$ is continuous.






share|cite|improve this answer























  • Very good proof. I've just skimmed through it and I've understood pretty much everything. I will look in more details into it in the next few days. Thanks!
    – LuxGiammi
    Nov 16 at 15:56















up vote
3
down vote



accepted







up vote
3
down vote



accepted






For $x in I$ let $I_n(x) = I cap (x - frac{1}{n},x + frac{1}{n})$. This is an open neighborhood of $x$ in $I$. We have
$$(*) phantom{xx} bigcap_{n=1}^infty f(I_n(x)) = { f(x) } .$$
"$supset$" is trivial. To verify "$subset$", let $y in bigcap_{n=1}^infty f(I_n(x))$. Then there exist $x_n in I_n(x)$ such that $f(x_n) = y$. Obviously $x_n to x$. Since $f^{-1}(y)$ is closed in $I$ , we see that $x in f^{-1}(y)$, i.e. $f(x) = y$.



Let $varepsilon > 0$. Since $I_n(x)$ is an interval, also $J_n = f(I_n(x))$ is an interval. We have $J_n = langle a_n,b_n rangle$, where $langle a_n,b_n rangle$ stands for an open, half-open or closed interval such that $a_n le f(x) le b_n$. $a_n = -infty$, $b_n = infty$ is allowed. The sequence $(a_n)$ is increasing, the sequence $(b_n)$ decreasing. But $(*)$ shows that $a_n, b_n to f(x)$, hence $f(I_n(x)) = J_n subset (f(x)- varepsilon, f(x)+ varepsilon)$ for sufficiently large $n$. This means that $f$ is continuous.






share|cite|improve this answer














For $x in I$ let $I_n(x) = I cap (x - frac{1}{n},x + frac{1}{n})$. This is an open neighborhood of $x$ in $I$. We have
$$(*) phantom{xx} bigcap_{n=1}^infty f(I_n(x)) = { f(x) } .$$
"$supset$" is trivial. To verify "$subset$", let $y in bigcap_{n=1}^infty f(I_n(x))$. Then there exist $x_n in I_n(x)$ such that $f(x_n) = y$. Obviously $x_n to x$. Since $f^{-1}(y)$ is closed in $I$ , we see that $x in f^{-1}(y)$, i.e. $f(x) = y$.



Let $varepsilon > 0$. Since $I_n(x)$ is an interval, also $J_n = f(I_n(x))$ is an interval. We have $J_n = langle a_n,b_n rangle$, where $langle a_n,b_n rangle$ stands for an open, half-open or closed interval such that $a_n le f(x) le b_n$. $a_n = -infty$, $b_n = infty$ is allowed. The sequence $(a_n)$ is increasing, the sequence $(b_n)$ decreasing. But $(*)$ shows that $a_n, b_n to f(x)$, hence $f(I_n(x)) = J_n subset (f(x)- varepsilon, f(x)+ varepsilon)$ for sufficiently large $n$. This means that $f$ is continuous.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 15 at 23:09

























answered Nov 15 at 23:01









Paul Frost

7,4341527




7,4341527












  • Very good proof. I've just skimmed through it and I've understood pretty much everything. I will look in more details into it in the next few days. Thanks!
    – LuxGiammi
    Nov 16 at 15:56




















  • Very good proof. I've just skimmed through it and I've understood pretty much everything. I will look in more details into it in the next few days. Thanks!
    – LuxGiammi
    Nov 16 at 15:56


















Very good proof. I've just skimmed through it and I've understood pretty much everything. I will look in more details into it in the next few days. Thanks!
– LuxGiammi
Nov 16 at 15:56






Very good proof. I've just skimmed through it and I've understood pretty much everything. I will look in more details into it in the next few days. Thanks!
– LuxGiammi
Nov 16 at 15:56




















 

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