Proof that $S={xinmathbb{R}^2 | x_1>0, x_2>0 }$ is an open set












0












$begingroup$


We have the following criteria:



If $f colon mathbb{R}^n rightarrow mathbb{R}$ continuous, then the sets
$S_1={xinmathbb{R}^n | f(x)>0 }$, $S_2={xinmathbb{R}^n | f(x)<0 }$,
$S_3={xinmathbb{R}^n | f(x)neq0 }$ are open sets.



With those criteria I have to prove that the set in the title is open, but I have no idea how I could prove that. Could someone give me a hint?






elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Consider the projections on each factor: $;(x,y)mapsto x$ and $(x,y)mapsto y$.
    $endgroup$
    – Bernard
    Jan 20 at 14:20










  • $begingroup$
    If $f$ is a continuous function and $A=(0,infty)$ is an open set in $mathbb{R}$, then $f^{-1}(0,infty)$ is the pre image of an open set, so $S_1$ is open
    $endgroup$
    – Pablo_
    Jan 20 at 19:58
















0












$begingroup$


We have the following criteria:



If $f colon mathbb{R}^n rightarrow mathbb{R}$ continuous, then the sets
$S_1={xinmathbb{R}^n | f(x)>0 }$, $S_2={xinmathbb{R}^n | f(x)<0 }$,
$S_3={xinmathbb{R}^n | f(x)neq0 }$ are open sets.



With those criteria I have to prove that the set in the title is open, but I have no idea how I could prove that. Could someone give me a hint?






elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Consider the projections on each factor: $;(x,y)mapsto x$ and $(x,y)mapsto y$.
    $endgroup$
    – Bernard
    Jan 20 at 14:20










  • $begingroup$
    If $f$ is a continuous function and $A=(0,infty)$ is an open set in $mathbb{R}$, then $f^{-1}(0,infty)$ is the pre image of an open set, so $S_1$ is open
    $endgroup$
    – Pablo_
    Jan 20 at 19:58














0












0








0





$begingroup$


We have the following criteria:



If $f colon mathbb{R}^n rightarrow mathbb{R}$ continuous, then the sets
$S_1={xinmathbb{R}^n | f(x)>0 }$, $S_2={xinmathbb{R}^n | f(x)<0 }$,
$S_3={xinmathbb{R}^n | f(x)neq0 }$ are open sets.



With those criteria I have to prove that the set in the title is open, but I have no idea how I could prove that. Could someone give me a hint?






elementary-set-theory







share|cite|improve this question











$endgroup$




We have the following criteria:



If $f colon mathbb{R}^n rightarrow mathbb{R}$ continuous, then the sets
$S_1={xinmathbb{R}^n | f(x)>0 }$, $S_2={xinmathbb{R}^n | f(x)<0 }$,
$S_3={xinmathbb{R}^n | f(x)neq0 }$ are open sets.



With those criteria I have to prove that the set in the title is open, but I have no idea how I could prove that. Could someone give me a hint?






elementary-set-theory




elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 at 19:38









Daniele Tampieri

2,2922822




2,2922822










asked Jan 20 at 14:18









NegowukuNegowuku

11




11












  • $begingroup$
    Consider the projections on each factor: $;(x,y)mapsto x$ and $(x,y)mapsto y$.
    $endgroup$
    – Bernard
    Jan 20 at 14:20










  • $begingroup$
    If $f$ is a continuous function and $A=(0,infty)$ is an open set in $mathbb{R}$, then $f^{-1}(0,infty)$ is the pre image of an open set, so $S_1$ is open
    $endgroup$
    – Pablo_
    Jan 20 at 19:58


















  • $begingroup$
    Consider the projections on each factor: $;(x,y)mapsto x$ and $(x,y)mapsto y$.
    $endgroup$
    – Bernard
    Jan 20 at 14:20










  • $begingroup$
    If $f$ is a continuous function and $A=(0,infty)$ is an open set in $mathbb{R}$, then $f^{-1}(0,infty)$ is the pre image of an open set, so $S_1$ is open
    $endgroup$
    – Pablo_
    Jan 20 at 19:58
















$begingroup$
Consider the projections on each factor: $;(x,y)mapsto x$ and $(x,y)mapsto y$.
$endgroup$
– Bernard
Jan 20 at 14:20




$begingroup$
Consider the projections on each factor: $;(x,y)mapsto x$ and $(x,y)mapsto y$.
$endgroup$
– Bernard
Jan 20 at 14:20












$begingroup$
If $f$ is a continuous function and $A=(0,infty)$ is an open set in $mathbb{R}$, then $f^{-1}(0,infty)$ is the pre image of an open set, so $S_1$ is open
$endgroup$
– Pablo_
Jan 20 at 19:58




$begingroup$
If $f$ is a continuous function and $A=(0,infty)$ is an open set in $mathbb{R}$, then $f^{-1}(0,infty)$ is the pre image of an open set, so $S_1$ is open
$endgroup$
– Pablo_
Jan 20 at 19:58










3 Answers
3






active

oldest

votes


















0












$begingroup$

Indeed, consider the projection maps, $phi_x : (x,y)mapsto x$, and $phi_y:(x,y)mapsto y$. These maps are clearly continuous. Now, the set you are interested in is nothing but, $phi_x^{-1}(0,infty)cap phi_y^{-1}(0,infty)$, both of which are open sets (open set characterization of continuous maps). Now, (finite) intersection open sets is open, done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Mh yes, but this isn't what I asked for. I have to prof this with one of the three give criteria..
    $endgroup$
    – Negowuku
    Jan 20 at 22:34



















0












$begingroup$

Prove that $$
f(x_1,x_2)=begin{cases} x_1x_2 &text{if } x_1>0, ;x_2 >0 \ 0 &text{otherwise}end{cases}
$$
is continuous.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Note that $S_1=f^{-1}((0,infty))$ and $S_2=S_1=f^{-1}((-infty, 0)), $ and since $f$ is continuous, each is open in $mathbb R^n$. Therefore, so is $S_3=S_1cap S_2$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      And what would $f$ look like? I don't understand your proof, could you maybe explain it a bit more indepth?
      $endgroup$
      – Negowuku
      Jan 20 at 22:35










    • $begingroup$
      All you know about $f$ is that it is continuous. But that is enough, because $(0,infty)$ and $(-infty, 0)$ are open in $mathbb R$ so their inverse images by $f$ are open in $mathbb R^n$ and then so is their intersection. And you're done!
      $endgroup$
      – Matematleta
      Jan 20 at 22:39










    • $begingroup$
      But where is the connection to my give set with $x_1>0, x_2>0$?
      $endgroup$
      – Negowuku
      Jan 21 at 9:44











    Your Answer





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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Indeed, consider the projection maps, $phi_x : (x,y)mapsto x$, and $phi_y:(x,y)mapsto y$. These maps are clearly continuous. Now, the set you are interested in is nothing but, $phi_x^{-1}(0,infty)cap phi_y^{-1}(0,infty)$, both of which are open sets (open set characterization of continuous maps). Now, (finite) intersection open sets is open, done.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Mh yes, but this isn't what I asked for. I have to prof this with one of the three give criteria..
      $endgroup$
      – Negowuku
      Jan 20 at 22:34
















    0












    $begingroup$

    Indeed, consider the projection maps, $phi_x : (x,y)mapsto x$, and $phi_y:(x,y)mapsto y$. These maps are clearly continuous. Now, the set you are interested in is nothing but, $phi_x^{-1}(0,infty)cap phi_y^{-1}(0,infty)$, both of which are open sets (open set characterization of continuous maps). Now, (finite) intersection open sets is open, done.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Mh yes, but this isn't what I asked for. I have to prof this with one of the three give criteria..
      $endgroup$
      – Negowuku
      Jan 20 at 22:34














    0












    0








    0





    $begingroup$

    Indeed, consider the projection maps, $phi_x : (x,y)mapsto x$, and $phi_y:(x,y)mapsto y$. These maps are clearly continuous. Now, the set you are interested in is nothing but, $phi_x^{-1}(0,infty)cap phi_y^{-1}(0,infty)$, both of which are open sets (open set characterization of continuous maps). Now, (finite) intersection open sets is open, done.






    share|cite|improve this answer









    $endgroup$



    Indeed, consider the projection maps, $phi_x : (x,y)mapsto x$, and $phi_y:(x,y)mapsto y$. These maps are clearly continuous. Now, the set you are interested in is nothing but, $phi_x^{-1}(0,infty)cap phi_y^{-1}(0,infty)$, both of which are open sets (open set characterization of continuous maps). Now, (finite) intersection open sets is open, done.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 20 at 14:24









    AaronAaron

    1,902415




    1,902415












    • $begingroup$
      Mh yes, but this isn't what I asked for. I have to prof this with one of the three give criteria..
      $endgroup$
      – Negowuku
      Jan 20 at 22:34


















    • $begingroup$
      Mh yes, but this isn't what I asked for. I have to prof this with one of the three give criteria..
      $endgroup$
      – Negowuku
      Jan 20 at 22:34
















    $begingroup$
    Mh yes, but this isn't what I asked for. I have to prof this with one of the three give criteria..
    $endgroup$
    – Negowuku
    Jan 20 at 22:34




    $begingroup$
    Mh yes, but this isn't what I asked for. I have to prof this with one of the three give criteria..
    $endgroup$
    – Negowuku
    Jan 20 at 22:34











    0












    $begingroup$

    Prove that $$
    f(x_1,x_2)=begin{cases} x_1x_2 &text{if } x_1>0, ;x_2 >0 \ 0 &text{otherwise}end{cases}
    $$
    is continuous.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Prove that $$
      f(x_1,x_2)=begin{cases} x_1x_2 &text{if } x_1>0, ;x_2 >0 \ 0 &text{otherwise}end{cases}
      $$
      is continuous.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Prove that $$
        f(x_1,x_2)=begin{cases} x_1x_2 &text{if } x_1>0, ;x_2 >0 \ 0 &text{otherwise}end{cases}
        $$
        is continuous.






        share|cite|improve this answer









        $endgroup$



        Prove that $$
        f(x_1,x_2)=begin{cases} x_1x_2 &text{if } x_1>0, ;x_2 >0 \ 0 &text{otherwise}end{cases}
        $$
        is continuous.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 20 at 14:28









        ZeeklessZeekless

        577111




        577111























            0












            $begingroup$

            Note that $S_1=f^{-1}((0,infty))$ and $S_2=S_1=f^{-1}((-infty, 0)), $ and since $f$ is continuous, each is open in $mathbb R^n$. Therefore, so is $S_3=S_1cap S_2$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              And what would $f$ look like? I don't understand your proof, could you maybe explain it a bit more indepth?
              $endgroup$
              – Negowuku
              Jan 20 at 22:35










            • $begingroup$
              All you know about $f$ is that it is continuous. But that is enough, because $(0,infty)$ and $(-infty, 0)$ are open in $mathbb R$ so their inverse images by $f$ are open in $mathbb R^n$ and then so is their intersection. And you're done!
              $endgroup$
              – Matematleta
              Jan 20 at 22:39










            • $begingroup$
              But where is the connection to my give set with $x_1>0, x_2>0$?
              $endgroup$
              – Negowuku
              Jan 21 at 9:44
















            0












            $begingroup$

            Note that $S_1=f^{-1}((0,infty))$ and $S_2=S_1=f^{-1}((-infty, 0)), $ and since $f$ is continuous, each is open in $mathbb R^n$. Therefore, so is $S_3=S_1cap S_2$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              And what would $f$ look like? I don't understand your proof, could you maybe explain it a bit more indepth?
              $endgroup$
              – Negowuku
              Jan 20 at 22:35










            • $begingroup$
              All you know about $f$ is that it is continuous. But that is enough, because $(0,infty)$ and $(-infty, 0)$ are open in $mathbb R$ so their inverse images by $f$ are open in $mathbb R^n$ and then so is their intersection. And you're done!
              $endgroup$
              – Matematleta
              Jan 20 at 22:39










            • $begingroup$
              But where is the connection to my give set with $x_1>0, x_2>0$?
              $endgroup$
              – Negowuku
              Jan 21 at 9:44














            0












            0








            0





            $begingroup$

            Note that $S_1=f^{-1}((0,infty))$ and $S_2=S_1=f^{-1}((-infty, 0)), $ and since $f$ is continuous, each is open in $mathbb R^n$. Therefore, so is $S_3=S_1cap S_2$.






            share|cite|improve this answer









            $endgroup$



            Note that $S_1=f^{-1}((0,infty))$ and $S_2=S_1=f^{-1}((-infty, 0)), $ and since $f$ is continuous, each is open in $mathbb R^n$. Therefore, so is $S_3=S_1cap S_2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 20 at 16:46









            MatematletaMatematleta

            11.1k2918




            11.1k2918












            • $begingroup$
              And what would $f$ look like? I don't understand your proof, could you maybe explain it a bit more indepth?
              $endgroup$
              – Negowuku
              Jan 20 at 22:35










            • $begingroup$
              All you know about $f$ is that it is continuous. But that is enough, because $(0,infty)$ and $(-infty, 0)$ are open in $mathbb R$ so their inverse images by $f$ are open in $mathbb R^n$ and then so is their intersection. And you're done!
              $endgroup$
              – Matematleta
              Jan 20 at 22:39










            • $begingroup$
              But where is the connection to my give set with $x_1>0, x_2>0$?
              $endgroup$
              – Negowuku
              Jan 21 at 9:44


















            • $begingroup$
              And what would $f$ look like? I don't understand your proof, could you maybe explain it a bit more indepth?
              $endgroup$
              – Negowuku
              Jan 20 at 22:35










            • $begingroup$
              All you know about $f$ is that it is continuous. But that is enough, because $(0,infty)$ and $(-infty, 0)$ are open in $mathbb R$ so their inverse images by $f$ are open in $mathbb R^n$ and then so is their intersection. And you're done!
              $endgroup$
              – Matematleta
              Jan 20 at 22:39










            • $begingroup$
              But where is the connection to my give set with $x_1>0, x_2>0$?
              $endgroup$
              – Negowuku
              Jan 21 at 9:44
















            $begingroup$
            And what would $f$ look like? I don't understand your proof, could you maybe explain it a bit more indepth?
            $endgroup$
            – Negowuku
            Jan 20 at 22:35




            $begingroup$
            And what would $f$ look like? I don't understand your proof, could you maybe explain it a bit more indepth?
            $endgroup$
            – Negowuku
            Jan 20 at 22:35












            $begingroup$
            All you know about $f$ is that it is continuous. But that is enough, because $(0,infty)$ and $(-infty, 0)$ are open in $mathbb R$ so their inverse images by $f$ are open in $mathbb R^n$ and then so is their intersection. And you're done!
            $endgroup$
            – Matematleta
            Jan 20 at 22:39




            $begingroup$
            All you know about $f$ is that it is continuous. But that is enough, because $(0,infty)$ and $(-infty, 0)$ are open in $mathbb R$ so their inverse images by $f$ are open in $mathbb R^n$ and then so is their intersection. And you're done!
            $endgroup$
            – Matematleta
            Jan 20 at 22:39












            $begingroup$
            But where is the connection to my give set with $x_1>0, x_2>0$?
            $endgroup$
            – Negowuku
            Jan 21 at 9:44




            $begingroup$
            But where is the connection to my give set with $x_1>0, x_2>0$?
            $endgroup$
            – Negowuku
            Jan 21 at 9:44


















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