Find a subset such that there is a cts function whose image is {0,1}












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Want to find a subset $X$ of $mathbb{R}^3$ such that there exists a cts function $f$ on the set such that the image $f(X) = {0,1}$, that is takes both values 0 and 1.



This is part of a larger question (IVT in higher dimensions) but just wanted to help understand this portion. Should be the case that some subsets of $mathbb{R}^3$ have such a function while others don't.



For a subset that doesn't, I came up with the set that consists only of the zero vector, ${(0,0,0)}$.



I might be overthinking it but struggling to come up with one since the function that exists needs to be continuous (otherwise I'd just do a piece wise function).



Appreciate any tips










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  • Hint: the continuous image of a connected set is connected. So $X$ must be ....
    – user25959
    Nov 24 at 1:38










  • So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
    – Tommy
    Nov 24 at 1:52










  • Exactly - and check that the thing you find (use the simplest possible function) is continuous
    – user25959
    Nov 24 at 1:53
















0














Want to find a subset $X$ of $mathbb{R}^3$ such that there exists a cts function $f$ on the set such that the image $f(X) = {0,1}$, that is takes both values 0 and 1.



This is part of a larger question (IVT in higher dimensions) but just wanted to help understand this portion. Should be the case that some subsets of $mathbb{R}^3$ have such a function while others don't.



For a subset that doesn't, I came up with the set that consists only of the zero vector, ${(0,0,0)}$.



I might be overthinking it but struggling to come up with one since the function that exists needs to be continuous (otherwise I'd just do a piece wise function).



Appreciate any tips










share|cite|improve this question
























  • Hint: the continuous image of a connected set is connected. So $X$ must be ....
    – user25959
    Nov 24 at 1:38










  • So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
    – Tommy
    Nov 24 at 1:52










  • Exactly - and check that the thing you find (use the simplest possible function) is continuous
    – user25959
    Nov 24 at 1:53














0












0








0







Want to find a subset $X$ of $mathbb{R}^3$ such that there exists a cts function $f$ on the set such that the image $f(X) = {0,1}$, that is takes both values 0 and 1.



This is part of a larger question (IVT in higher dimensions) but just wanted to help understand this portion. Should be the case that some subsets of $mathbb{R}^3$ have such a function while others don't.



For a subset that doesn't, I came up with the set that consists only of the zero vector, ${(0,0,0)}$.



I might be overthinking it but struggling to come up with one since the function that exists needs to be continuous (otherwise I'd just do a piece wise function).



Appreciate any tips










share|cite|improve this question















Want to find a subset $X$ of $mathbb{R}^3$ such that there exists a cts function $f$ on the set such that the image $f(X) = {0,1}$, that is takes both values 0 and 1.



This is part of a larger question (IVT in higher dimensions) but just wanted to help understand this portion. Should be the case that some subsets of $mathbb{R}^3$ have such a function while others don't.



For a subset that doesn't, I came up with the set that consists only of the zero vector, ${(0,0,0)}$.



I might be overthinking it but struggling to come up with one since the function that exists needs to be continuous (otherwise I'd just do a piece wise function).



Appreciate any tips







real-analysis functions continuity






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edited Nov 24 at 1:36









Andrés E. Caicedo

64.7k8158246




64.7k8158246










asked Nov 24 at 1:35









Tommy

1138




1138












  • Hint: the continuous image of a connected set is connected. So $X$ must be ....
    – user25959
    Nov 24 at 1:38










  • So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
    – Tommy
    Nov 24 at 1:52










  • Exactly - and check that the thing you find (use the simplest possible function) is continuous
    – user25959
    Nov 24 at 1:53


















  • Hint: the continuous image of a connected set is connected. So $X$ must be ....
    – user25959
    Nov 24 at 1:38










  • So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
    – Tommy
    Nov 24 at 1:52










  • Exactly - and check that the thing you find (use the simplest possible function) is continuous
    – user25959
    Nov 24 at 1:53
















Hint: the continuous image of a connected set is connected. So $X$ must be ....
– user25959
Nov 24 at 1:38




Hint: the continuous image of a connected set is connected. So $X$ must be ....
– user25959
Nov 24 at 1:38












So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
– Tommy
Nov 24 at 1:52




So we didn't cover connectedness in the course that I'm taking but I'm aware of the notion. But I'm guessing $X$ must be disconnected (say union of two disjoint nonempty sets?) and can find a function that works with said $X$.
– Tommy
Nov 24 at 1:52












Exactly - and check that the thing you find (use the simplest possible function) is continuous
– user25959
Nov 24 at 1:53




Exactly - and check that the thing you find (use the simplest possible function) is continuous
– user25959
Nov 24 at 1:53










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Hint: A crucial hypothesis in IVT is connectedness. Can you think of such a function on a disconnected domain?






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    1 Answer
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    1 Answer
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    active

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    0














    Hint: A crucial hypothesis in IVT is connectedness. Can you think of such a function on a disconnected domain?






    share|cite|improve this answer


























      0














      Hint: A crucial hypothesis in IVT is connectedness. Can you think of such a function on a disconnected domain?






      share|cite|improve this answer
























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        Hint: A crucial hypothesis in IVT is connectedness. Can you think of such a function on a disconnected domain?






        share|cite|improve this answer












        Hint: A crucial hypothesis in IVT is connectedness. Can you think of such a function on a disconnected domain?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 at 1:40









        Zachary Selk

        536311




        536311






























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