Does there exist a continuous function separating these two sets $A$ and $B$












2












$begingroup$


True or False:




There exists a continuous function $f : mathbb{R}^2 → mathbb{R}
> $such that $f ≡ 1$ on the set ${(x, y) in mathbb{R}^2 : x ^2+y^2
=3/2 }$ and $f ≡ 0$ on the set $B∪{(x, y) in mathbb{R}^2: x^2+y^2 ≥ 2}$ where B is closed unit disk.




I think this is just a stratightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.



I hope I am not missing something. Topology can be weird sometimes!!!



Thanks in advance.






general-topology separation-axioms







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    That's true you can use Urysohn's Lemma. You just need to show that these sets are closed and disjoint.
    $endgroup$
    – Yanko
    Jan 11 at 20:26






  • 1




    $begingroup$
    You can use Urysohn's Lemma, yes. It seems kind of like using a nuke to kill ants, though, especially when it's pretty clear how to just write down what $f$ is explicitly.
    $endgroup$
    – user3482749
    Jan 11 at 20:37
















2












$begingroup$


True or False:




There exists a continuous function $f : mathbb{R}^2 → mathbb{R}
> $such that $f ≡ 1$ on the set ${(x, y) in mathbb{R}^2 : x ^2+y^2
=3/2 }$ and $f ≡ 0$ on the set $B∪{(x, y) in mathbb{R}^2: x^2+y^2 ≥ 2}$ where B is closed unit disk.




I think this is just a stratightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.



I hope I am not missing something. Topology can be weird sometimes!!!



Thanks in advance.






general-topology separation-axioms







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    That's true you can use Urysohn's Lemma. You just need to show that these sets are closed and disjoint.
    $endgroup$
    – Yanko
    Jan 11 at 20:26






  • 1




    $begingroup$
    You can use Urysohn's Lemma, yes. It seems kind of like using a nuke to kill ants, though, especially when it's pretty clear how to just write down what $f$ is explicitly.
    $endgroup$
    – user3482749
    Jan 11 at 20:37














2












2








2





$begingroup$


True or False:




There exists a continuous function $f : mathbb{R}^2 → mathbb{R}
> $such that $f ≡ 1$ on the set ${(x, y) in mathbb{R}^2 : x ^2+y^2
=3/2 }$ and $f ≡ 0$ on the set $B∪{(x, y) in mathbb{R}^2: x^2+y^2 ≥ 2}$ where B is closed unit disk.




I think this is just a stratightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.



I hope I am not missing something. Topology can be weird sometimes!!!



Thanks in advance.






general-topology separation-axioms







share|cite|improve this question











$endgroup$




True or False:




There exists a continuous function $f : mathbb{R}^2 → mathbb{R}
> $such that $f ≡ 1$ on the set ${(x, y) in mathbb{R}^2 : x ^2+y^2
=3/2 }$ and $f ≡ 0$ on the set $B∪{(x, y) in mathbb{R}^2: x^2+y^2 ≥ 2}$ where B is closed unit disk.




I think this is just a stratightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.



I hope I am not missing something. Topology can be weird sometimes!!!



Thanks in advance.






general-topology separation-axioms




general-topology separation-axioms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 20:42









6005

37.1k752127




37.1k752127










asked Jan 11 at 20:25









StammeringMathematicianStammeringMathematician

2,7951324




2,7951324








  • 3




    $begingroup$
    That's true you can use Urysohn's Lemma. You just need to show that these sets are closed and disjoint.
    $endgroup$
    – Yanko
    Jan 11 at 20:26






  • 1




    $begingroup$
    You can use Urysohn's Lemma, yes. It seems kind of like using a nuke to kill ants, though, especially when it's pretty clear how to just write down what $f$ is explicitly.
    $endgroup$
    – user3482749
    Jan 11 at 20:37














  • 3




    $begingroup$
    That's true you can use Urysohn's Lemma. You just need to show that these sets are closed and disjoint.
    $endgroup$
    – Yanko
    Jan 11 at 20:26






  • 1




    $begingroup$
    You can use Urysohn's Lemma, yes. It seems kind of like using a nuke to kill ants, though, especially when it's pretty clear how to just write down what $f$ is explicitly.
    $endgroup$
    – user3482749
    Jan 11 at 20:37








3




3




$begingroup$
That's true you can use Urysohn's Lemma. You just need to show that these sets are closed and disjoint.
$endgroup$
– Yanko
Jan 11 at 20:26




$begingroup$
That's true you can use Urysohn's Lemma. You just need to show that these sets are closed and disjoint.
$endgroup$
– Yanko
Jan 11 at 20:26




1




1




$begingroup$
You can use Urysohn's Lemma, yes. It seems kind of like using a nuke to kill ants, though, especially when it's pretty clear how to just write down what $f$ is explicitly.
$endgroup$
– user3482749
Jan 11 at 20:37




$begingroup$
You can use Urysohn's Lemma, yes. It seems kind of like using a nuke to kill ants, though, especially when it's pretty clear how to just write down what $f$ is explicitly.
$endgroup$
– user3482749
Jan 11 at 20:37










1 Answer
1






active

oldest

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3












$begingroup$


I think this is just a straightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.




Right. As you say and as Yanko agrees, you can use Urysohn's Lemma; then it just remains to show the sets are closed and disjoint.



In case you don't like Ursyohn's Lemma -- or just for fun -- we can define the continuous function ourselves. It doesn't turn out to be so hard in this particular case. Define $f: mathbb{R}^2 to mathbb{R}$ by
$$
f(x,y) = 2(2 - x^2 - y^2).
$$



Now, this is a polynomial, so it's continuous. And on the set $A$, $f(x,y) = 2(2 - tfrac32) = 2 cdot tfrac12 = 1$. And on the set $B$, $f(x,y) = 2(2 - 2) = 0$.



If $f$ has to be positive, you can instead make $f(x,y) = max(0, 2(2-x^2 - y^2))$.




Topology can be weird sometimes!!!




I agree :)






share|cite|improve this answer









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    3












    $begingroup$


    I think this is just a straightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.




    Right. As you say and as Yanko agrees, you can use Urysohn's Lemma; then it just remains to show the sets are closed and disjoint.



    In case you don't like Ursyohn's Lemma -- or just for fun -- we can define the continuous function ourselves. It doesn't turn out to be so hard in this particular case. Define $f: mathbb{R}^2 to mathbb{R}$ by
    $$
    f(x,y) = 2(2 - x^2 - y^2).
    $$



    Now, this is a polynomial, so it's continuous. And on the set $A$, $f(x,y) = 2(2 - tfrac32) = 2 cdot tfrac12 = 1$. And on the set $B$, $f(x,y) = 2(2 - 2) = 0$.



    If $f$ has to be positive, you can instead make $f(x,y) = max(0, 2(2-x^2 - y^2))$.




    Topology can be weird sometimes!!!




    I agree :)






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$


      I think this is just a straightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.




      Right. As you say and as Yanko agrees, you can use Urysohn's Lemma; then it just remains to show the sets are closed and disjoint.



      In case you don't like Ursyohn's Lemma -- or just for fun -- we can define the continuous function ourselves. It doesn't turn out to be so hard in this particular case. Define $f: mathbb{R}^2 to mathbb{R}$ by
      $$
      f(x,y) = 2(2 - x^2 - y^2).
      $$



      Now, this is a polynomial, so it's continuous. And on the set $A$, $f(x,y) = 2(2 - tfrac32) = 2 cdot tfrac12 = 1$. And on the set $B$, $f(x,y) = 2(2 - 2) = 0$.



      If $f$ has to be positive, you can instead make $f(x,y) = max(0, 2(2-x^2 - y^2))$.




      Topology can be weird sometimes!!!




      I agree :)






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$


        I think this is just a straightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.




        Right. As you say and as Yanko agrees, you can use Urysohn's Lemma; then it just remains to show the sets are closed and disjoint.



        In case you don't like Ursyohn's Lemma -- or just for fun -- we can define the continuous function ourselves. It doesn't turn out to be so hard in this particular case. Define $f: mathbb{R}^2 to mathbb{R}$ by
        $$
        f(x,y) = 2(2 - x^2 - y^2).
        $$



        Now, this is a polynomial, so it's continuous. And on the set $A$, $f(x,y) = 2(2 - tfrac32) = 2 cdot tfrac12 = 1$. And on the set $B$, $f(x,y) = 2(2 - 2) = 0$.



        If $f$ has to be positive, you can instead make $f(x,y) = max(0, 2(2-x^2 - y^2))$.




        Topology can be weird sometimes!!!




        I agree :)






        share|cite|improve this answer









        $endgroup$




        I think this is just a straightforward application of Urysohn's Lemma as metric spaces are normal so by Urysohn's Lemma says that disjoint closed subsets can be separated by continuous function.




        Right. As you say and as Yanko agrees, you can use Urysohn's Lemma; then it just remains to show the sets are closed and disjoint.



        In case you don't like Ursyohn's Lemma -- or just for fun -- we can define the continuous function ourselves. It doesn't turn out to be so hard in this particular case. Define $f: mathbb{R}^2 to mathbb{R}$ by
        $$
        f(x,y) = 2(2 - x^2 - y^2).
        $$



        Now, this is a polynomial, so it's continuous. And on the set $A$, $f(x,y) = 2(2 - tfrac32) = 2 cdot tfrac12 = 1$. And on the set $B$, $f(x,y) = 2(2 - 2) = 0$.



        If $f$ has to be positive, you can instead make $f(x,y) = max(0, 2(2-x^2 - y^2))$.




        Topology can be weird sometimes!!!




        I agree :)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 20:41









        60056005

        37.1k752127




        37.1k752127






























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