Is the $x in Bbb R : x ne 0$ set open, close or neither?

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Is the $x in Bbb R : x ne 0$ set open, close or neither?




The answer is open. However, I can't see why.



Set $A :=x in Bbb R : x ne 0$. Following the definition it will be open if for every $a$ from $A$ and any positive $ϵ$ the $V_ϵ(a)$ ⊆ $A$. Now, my intuition says, that if we take $a$ such that it will be really close to $0$ and $ϵ$ be big enough, then $ϵ$ neighbourhood would not be contained in $A$ since it will contain $0$ too.



Can you, please, explain it to me?




real-analysis




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  • 1




    It's not about taking $epsilon$ large enough, it's about whether there's room to fit in some $epsilon$. Take $x$ as close as you like to $0$ and choose $epsilon < x/2$; your neighbourhood then doesn't include $0$.
    – postmortes
    2 days ago






  • 2




    $xin mathbbR: xneq 0^c = 0$, which is closed
    – Rumpelstiltskin
    2 days ago











  • It's a union of two open intervals
    – JuliusL33t
    2 days ago










  • $epsilon$ is kind of "flexible" range: for any small $a$ given, you take a smaller $epsilon$ - if that $epsilon$ exists you say that the set is open. Note again: you're taking a particular $epsilon$.
    – poyea
    2 days ago











  • @TheSimpli This is not set theory, please do not add the tag.
    – Andrés E. Caicedo
    2 days ago














up vote
3
down vote

favorite













Is the $x in Bbb R : x ne 0$ set open, close or neither?




The answer is open. However, I can't see why.



Set $A :=x in Bbb R : x ne 0$. Following the definition it will be open if for every $a$ from $A$ and any positive $ϵ$ the $V_ϵ(a)$ ⊆ $A$. Now, my intuition says, that if we take $a$ such that it will be really close to $0$ and $ϵ$ be big enough, then $ϵ$ neighbourhood would not be contained in $A$ since it will contain $0$ too.



Can you, please, explain it to me?




real-analysis




share|cite|improve this question

















  • 1




    It's not about taking $epsilon$ large enough, it's about whether there's room to fit in some $epsilon$. Take $x$ as close as you like to $0$ and choose $epsilon < x/2$; your neighbourhood then doesn't include $0$.
    – postmortes
    2 days ago






  • 2




    $xin mathbbR: xneq 0^c = 0$, which is closed
    – Rumpelstiltskin
    2 days ago











  • It's a union of two open intervals
    – JuliusL33t
    2 days ago










  • $epsilon$ is kind of "flexible" range: for any small $a$ given, you take a smaller $epsilon$ - if that $epsilon$ exists you say that the set is open. Note again: you're taking a particular $epsilon$.
    – poyea
    2 days ago











  • @TheSimpli This is not set theory, please do not add the tag.
    – Andrés E. Caicedo
    2 days ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite












Is the $x in Bbb R : x ne 0$ set open, close or neither?




The answer is open. However, I can't see why.



Set $A :=x in Bbb R : x ne 0$. Following the definition it will be open if for every $a$ from $A$ and any positive $ϵ$ the $V_ϵ(a)$ ⊆ $A$. Now, my intuition says, that if we take $a$ such that it will be really close to $0$ and $ϵ$ be big enough, then $ϵ$ neighbourhood would not be contained in $A$ since it will contain $0$ too.



Can you, please, explain it to me?




real-analysis




share|cite|improve this question














Is the $x in Bbb R : x ne 0$ set open, close or neither?




The answer is open. However, I can't see why.



Set $A :=x in Bbb R : x ne 0$. Following the definition it will be open if for every $a$ from $A$ and any positive $ϵ$ the $V_ϵ(a)$ ⊆ $A$. Now, my intuition says, that if we take $a$ such that it will be really close to $0$ and $ϵ$ be big enough, then $ϵ$ neighbourhood would not be contained in $A$ since it will contain $0$ too.



Can you, please, explain it to me?





real-analysis





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edited 2 days ago









Andrés E. Caicedo

63k7150235




63k7150235









asked 2 days ago









Sargis Iskandaryan

43611




43611







  • 1




    It's not about taking $epsilon$ large enough, it's about whether there's room to fit in some $epsilon$. Take $x$ as close as you like to $0$ and choose $epsilon < x/2$; your neighbourhood then doesn't include $0$.
    – postmortes
    2 days ago






  • 2




    $xin mathbbR: xneq 0^c = 0$, which is closed
    – Rumpelstiltskin
    2 days ago











  • It's a union of two open intervals
    – JuliusL33t
    2 days ago










  • $epsilon$ is kind of "flexible" range: for any small $a$ given, you take a smaller $epsilon$ - if that $epsilon$ exists you say that the set is open. Note again: you're taking a particular $epsilon$.
    – poyea
    2 days ago











  • @TheSimpli This is not set theory, please do not add the tag.
    – Andrés E. Caicedo
    2 days ago












  • 1




    It's not about taking $epsilon$ large enough, it's about whether there's room to fit in some $epsilon$. Take $x$ as close as you like to $0$ and choose $epsilon < x/2$; your neighbourhood then doesn't include $0$.
    – postmortes
    2 days ago






  • 2




    $xin mathbbR: xneq 0^c = 0$, which is closed
    – Rumpelstiltskin
    2 days ago











  • It's a union of two open intervals
    – JuliusL33t
    2 days ago










  • $epsilon$ is kind of "flexible" range: for any small $a$ given, you take a smaller $epsilon$ - if that $epsilon$ exists you say that the set is open. Note again: you're taking a particular $epsilon$.
    – poyea
    2 days ago











  • @TheSimpli This is not set theory, please do not add the tag.
    – Andrés E. Caicedo
    2 days ago







1




1




It's not about taking $epsilon$ large enough, it's about whether there's room to fit in some $epsilon$. Take $x$ as close as you like to $0$ and choose $epsilon < x/2$; your neighbourhood then doesn't include $0$.
– postmortes
2 days ago




It's not about taking $epsilon$ large enough, it's about whether there's room to fit in some $epsilon$. Take $x$ as close as you like to $0$ and choose $epsilon < x/2$; your neighbourhood then doesn't include $0$.
– postmortes
2 days ago




2




2




$xin mathbbR: xneq 0^c = 0$, which is closed
– Rumpelstiltskin
2 days ago





$xin mathbbR: xneq 0^c = 0$, which is closed
– Rumpelstiltskin
2 days ago













It's a union of two open intervals
– JuliusL33t
2 days ago




It's a union of two open intervals
– JuliusL33t
2 days ago












$epsilon$ is kind of "flexible" range: for any small $a$ given, you take a smaller $epsilon$ - if that $epsilon$ exists you say that the set is open. Note again: you're taking a particular $epsilon$.
– poyea
2 days ago





$epsilon$ is kind of "flexible" range: for any small $a$ given, you take a smaller $epsilon$ - if that $epsilon$ exists you say that the set is open. Note again: you're taking a particular $epsilon$.
– poyea
2 days ago













@TheSimpli This is not set theory, please do not add the tag.
– Andrés E. Caicedo
2 days ago




@TheSimpli This is not set theory, please do not add the tag.
– Andrés E. Caicedo
2 days ago










4 Answers
4






active

oldest

votes

















up vote
4
down vote



accepted










You have to be very careful with the quantifiers here.



A set $A$ will be open if for every $a in A$ there is some $epsilon$ which has $V_epsilon subseteq A$, not all $epsilon$.






share|cite|improve this answer




























    up vote
    3
    down vote













    There is an error in your definition of open set. It doesn't have to hold for any positive $epsilon$ but for some positive $epsilon$. Now do you see why $A$ is open?






    share|cite|improve this answer




























      up vote
      1
      down vote













      To show that a set $A$ is open, pick an element, and you just show that there is an open ball such that the open ball resides in $A$.



      We do not have to worry about open balls with large radius, we just have to find one that is sufficiently small that resides in the set.



      In particular, if $a$ is pick, you can choose your radius to be $|a|$ and it will reside in the set.






      share|cite|improve this answer




























        up vote
        0
        down vote













        We know that this set is open since $$forall xin Aquad,quadexists epsilonquad,quad (x-epsilon,x+epsilon)in A$$for example take $epsilon=dfrac10$ , but it is not closed. Take the convergent sequence $$a_n=dfrac1n$$whose limit doesn't belong to $A$ while the terms do.






        share|cite|improve this answer





















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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          4
          down vote



          accepted










          You have to be very careful with the quantifiers here.



          A set $A$ will be open if for every $a in A$ there is some $epsilon$ which has $V_epsilon subseteq A$, not all $epsilon$.






          share|cite|improve this answer

























            up vote
            4
            down vote



            accepted










            You have to be very careful with the quantifiers here.



            A set $A$ will be open if for every $a in A$ there is some $epsilon$ which has $V_epsilon subseteq A$, not all $epsilon$.






            share|cite|improve this answer























              up vote
              4
              down vote



              accepted







              up vote
              4
              down vote



              accepted






              You have to be very careful with the quantifiers here.



              A set $A$ will be open if for every $a in A$ there is some $epsilon$ which has $V_epsilon subseteq A$, not all $epsilon$.






              share|cite|improve this answer













              You have to be very careful with the quantifiers here.



              A set $A$ will be open if for every $a in A$ there is some $epsilon$ which has $V_epsilon subseteq A$, not all $epsilon$.







              share|cite|improve this answer













              share|cite|improve this answer



              share|cite|improve this answer











              answered 2 days ago









              Daniel Mroz

              851314




              851314




















                  up vote
                  3
                  down vote













                  There is an error in your definition of open set. It doesn't have to hold for any positive $epsilon$ but for some positive $epsilon$. Now do you see why $A$ is open?






                  share|cite|improve this answer

























                    up vote
                    3
                    down vote













                    There is an error in your definition of open set. It doesn't have to hold for any positive $epsilon$ but for some positive $epsilon$. Now do you see why $A$ is open?






                    share|cite|improve this answer























                      up vote
                      3
                      down vote










                      up vote
                      3
                      down vote









                      There is an error in your definition of open set. It doesn't have to hold for any positive $epsilon$ but for some positive $epsilon$. Now do you see why $A$ is open?






                      share|cite|improve this answer













                      There is an error in your definition of open set. It doesn't have to hold for any positive $epsilon$ but for some positive $epsilon$. Now do you see why $A$ is open?







                      share|cite|improve this answer













                      share|cite|improve this answer



                      share|cite|improve this answer











                      answered 2 days ago









                      Gautam Shenoy

                      7,00011444




                      7,00011444




















                          up vote
                          1
                          down vote













                          To show that a set $A$ is open, pick an element, and you just show that there is an open ball such that the open ball resides in $A$.



                          We do not have to worry about open balls with large radius, we just have to find one that is sufficiently small that resides in the set.



                          In particular, if $a$ is pick, you can choose your radius to be $|a|$ and it will reside in the set.






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote













                            To show that a set $A$ is open, pick an element, and you just show that there is an open ball such that the open ball resides in $A$.



                            We do not have to worry about open balls with large radius, we just have to find one that is sufficiently small that resides in the set.



                            In particular, if $a$ is pick, you can choose your radius to be $|a|$ and it will reside in the set.






                            share|cite|improve this answer























                              up vote
                              1
                              down vote










                              up vote
                              1
                              down vote









                              To show that a set $A$ is open, pick an element, and you just show that there is an open ball such that the open ball resides in $A$.



                              We do not have to worry about open balls with large radius, we just have to find one that is sufficiently small that resides in the set.



                              In particular, if $a$ is pick, you can choose your radius to be $|a|$ and it will reside in the set.






                              share|cite|improve this answer













                              To show that a set $A$ is open, pick an element, and you just show that there is an open ball such that the open ball resides in $A$.



                              We do not have to worry about open balls with large radius, we just have to find one that is sufficiently small that resides in the set.



                              In particular, if $a$ is pick, you can choose your radius to be $|a|$ and it will reside in the set.







                              share|cite|improve this answer













                              share|cite|improve this answer



                              share|cite|improve this answer











                              answered 2 days ago









                              Siong Thye Goh

                              76.5k124594




                              76.5k124594




















                                  up vote
                                  0
                                  down vote













                                  We know that this set is open since $$forall xin Aquad,quadexists epsilonquad,quad (x-epsilon,x+epsilon)in A$$for example take $epsilon=dfrac10$ , but it is not closed. Take the convergent sequence $$a_n=dfrac1n$$whose limit doesn't belong to $A$ while the terms do.






                                  share|cite|improve this answer

























                                    up vote
                                    0
                                    down vote













                                    We know that this set is open since $$forall xin Aquad,quadexists epsilonquad,quad (x-epsilon,x+epsilon)in A$$for example take $epsilon=dfrac10$ , but it is not closed. Take the convergent sequence $$a_n=dfrac1n$$whose limit doesn't belong to $A$ while the terms do.






                                    share|cite|improve this answer























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      We know that this set is open since $$forall xin Aquad,quadexists epsilonquad,quad (x-epsilon,x+epsilon)in A$$for example take $epsilon=dfrac10$ , but it is not closed. Take the convergent sequence $$a_n=dfrac1n$$whose limit doesn't belong to $A$ while the terms do.






                                      share|cite|improve this answer













                                      We know that this set is open since $$forall xin Aquad,quadexists epsilonquad,quad (x-epsilon,x+epsilon)in A$$for example take $epsilon=dfrac10$ , but it is not closed. Take the convergent sequence $$a_n=dfrac1n$$whose limit doesn't belong to $A$ while the terms do.







                                      share|cite|improve this answer













                                      share|cite|improve this answer



                                      share|cite|improve this answer











                                      answered 2 days ago









                                      Mostafa Ayaz

                                      8,5203530




                                      8,5203530






















                                           

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