What is the negation of a closed subset?












-2












$begingroup$


Closed subset:= If $(a_n)_{ninmathbb{N}}$ converges and the elements of the sequence are all in $A$ then the Limit sits in $A$.



If I could Show that there exists a convergent sequence where all the Elements of the sequence are in $A$ and the Limit is not in $A$ would it be equivalent to say that $A$ is not a closed subset?$(*)$



I.e $A$ not a closed subset $iff exists (a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$ convergent and $lim_{nrightarrowinfty}a_nnotin A$



I have the Feeling that something is wrong because the Definition says:



$A$ closed subset $iff forall(a_n)_{ninmathbb{N}},a_nin A forall_{ninmathbb{N}} convergentRightarrow lim_{nrightarrowinfty}a_nin A$



Therefor the Negation would have to be



$A$ not closed subset $iff forall(a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$ convergent and $lim_{nrightarrowinfty}a_nnotin A$



If I have $(*)$ can it help me to prove:



$A$ closed subset $iff A^C$ is open






general-topology definition







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, it should be exists, as you describe at the beginning of your post. The negation at the end of your post is incorrect.
    $endgroup$
    – angryavian
    Jan 13 at 19:21






  • 1




    $begingroup$
    The negation of $forall$ is $exists$.
    $endgroup$
    – hamam_Abdallah
    Jan 13 at 19:22










  • $begingroup$
    but $neg (ARightarrow B )= A$ and $neg B$ with $A$ being:$forall(a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$convergent and $B$ being $lim_{nrightarrowinfty}a_nin A$
    $endgroup$
    – RM777
    Jan 13 at 19:25












  • $begingroup$
    You said I have to Change the quantorsign but the Formula above says that A must remain as it is
    $endgroup$
    – RM777
    Jan 13 at 19:27
















-2












$begingroup$


Closed subset:= If $(a_n)_{ninmathbb{N}}$ converges and the elements of the sequence are all in $A$ then the Limit sits in $A$.



If I could Show that there exists a convergent sequence where all the Elements of the sequence are in $A$ and the Limit is not in $A$ would it be equivalent to say that $A$ is not a closed subset?$(*)$



I.e $A$ not a closed subset $iff exists (a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$ convergent and $lim_{nrightarrowinfty}a_nnotin A$



I have the Feeling that something is wrong because the Definition says:



$A$ closed subset $iff forall(a_n)_{ninmathbb{N}},a_nin A forall_{ninmathbb{N}} convergentRightarrow lim_{nrightarrowinfty}a_nin A$



Therefor the Negation would have to be



$A$ not closed subset $iff forall(a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$ convergent and $lim_{nrightarrowinfty}a_nnotin A$



If I have $(*)$ can it help me to prove:



$A$ closed subset $iff A^C$ is open






general-topology definition







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, it should be exists, as you describe at the beginning of your post. The negation at the end of your post is incorrect.
    $endgroup$
    – angryavian
    Jan 13 at 19:21






  • 1




    $begingroup$
    The negation of $forall$ is $exists$.
    $endgroup$
    – hamam_Abdallah
    Jan 13 at 19:22










  • $begingroup$
    but $neg (ARightarrow B )= A$ and $neg B$ with $A$ being:$forall(a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$convergent and $B$ being $lim_{nrightarrowinfty}a_nin A$
    $endgroup$
    – RM777
    Jan 13 at 19:25












  • $begingroup$
    You said I have to Change the quantorsign but the Formula above says that A must remain as it is
    $endgroup$
    – RM777
    Jan 13 at 19:27














-2












-2








-2





$begingroup$


Closed subset:= If $(a_n)_{ninmathbb{N}}$ converges and the elements of the sequence are all in $A$ then the Limit sits in $A$.



If I could Show that there exists a convergent sequence where all the Elements of the sequence are in $A$ and the Limit is not in $A$ would it be equivalent to say that $A$ is not a closed subset?$(*)$



I.e $A$ not a closed subset $iff exists (a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$ convergent and $lim_{nrightarrowinfty}a_nnotin A$



I have the Feeling that something is wrong because the Definition says:



$A$ closed subset $iff forall(a_n)_{ninmathbb{N}},a_nin A forall_{ninmathbb{N}} convergentRightarrow lim_{nrightarrowinfty}a_nin A$



Therefor the Negation would have to be



$A$ not closed subset $iff forall(a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$ convergent and $lim_{nrightarrowinfty}a_nnotin A$



If I have $(*)$ can it help me to prove:



$A$ closed subset $iff A^C$ is open






general-topology definition







share|cite|improve this question











$endgroup$




Closed subset:= If $(a_n)_{ninmathbb{N}}$ converges and the elements of the sequence are all in $A$ then the Limit sits in $A$.



If I could Show that there exists a convergent sequence where all the Elements of the sequence are in $A$ and the Limit is not in $A$ would it be equivalent to say that $A$ is not a closed subset?$(*)$



I.e $A$ not a closed subset $iff exists (a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$ convergent and $lim_{nrightarrowinfty}a_nnotin A$



I have the Feeling that something is wrong because the Definition says:



$A$ closed subset $iff forall(a_n)_{ninmathbb{N}},a_nin A forall_{ninmathbb{N}} convergentRightarrow lim_{nrightarrowinfty}a_nin A$



Therefor the Negation would have to be



$A$ not closed subset $iff forall(a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$ convergent and $lim_{nrightarrowinfty}a_nnotin A$



If I have $(*)$ can it help me to prove:



$A$ closed subset $iff A^C$ is open






general-topology definition




general-topology definition






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 19:29







RM777

















asked Jan 13 at 19:18









RM777RM777

38512




38512








  • 1




    $begingroup$
    Yes, it should be exists, as you describe at the beginning of your post. The negation at the end of your post is incorrect.
    $endgroup$
    – angryavian
    Jan 13 at 19:21






  • 1




    $begingroup$
    The negation of $forall$ is $exists$.
    $endgroup$
    – hamam_Abdallah
    Jan 13 at 19:22










  • $begingroup$
    but $neg (ARightarrow B )= A$ and $neg B$ with $A$ being:$forall(a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$convergent and $B$ being $lim_{nrightarrowinfty}a_nin A$
    $endgroup$
    – RM777
    Jan 13 at 19:25












  • $begingroup$
    You said I have to Change the quantorsign but the Formula above says that A must remain as it is
    $endgroup$
    – RM777
    Jan 13 at 19:27














  • 1




    $begingroup$
    Yes, it should be exists, as you describe at the beginning of your post. The negation at the end of your post is incorrect.
    $endgroup$
    – angryavian
    Jan 13 at 19:21






  • 1




    $begingroup$
    The negation of $forall$ is $exists$.
    $endgroup$
    – hamam_Abdallah
    Jan 13 at 19:22










  • $begingroup$
    but $neg (ARightarrow B )= A$ and $neg B$ with $A$ being:$forall(a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$convergent and $B$ being $lim_{nrightarrowinfty}a_nin A$
    $endgroup$
    – RM777
    Jan 13 at 19:25












  • $begingroup$
    You said I have to Change the quantorsign but the Formula above says that A must remain as it is
    $endgroup$
    – RM777
    Jan 13 at 19:27








1




1




$begingroup$
Yes, it should be exists, as you describe at the beginning of your post. The negation at the end of your post is incorrect.
$endgroup$
– angryavian
Jan 13 at 19:21




$begingroup$
Yes, it should be exists, as you describe at the beginning of your post. The negation at the end of your post is incorrect.
$endgroup$
– angryavian
Jan 13 at 19:21




1




1




$begingroup$
The negation of $forall$ is $exists$.
$endgroup$
– hamam_Abdallah
Jan 13 at 19:22




$begingroup$
The negation of $forall$ is $exists$.
$endgroup$
– hamam_Abdallah
Jan 13 at 19:22












$begingroup$
but $neg (ARightarrow B )= A$ and $neg B$ with $A$ being:$forall(a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$convergent and $B$ being $lim_{nrightarrowinfty}a_nin A$
$endgroup$
– RM777
Jan 13 at 19:25






$begingroup$
but $neg (ARightarrow B )= A$ and $neg B$ with $A$ being:$forall(a_n)_ninmathbb{N},a_nin A forall_{ninmathbb{N}}$convergent and $B$ being $lim_{nrightarrowinfty}a_nin A$
$endgroup$
– RM777
Jan 13 at 19:25














$begingroup$
You said I have to Change the quantorsign but the Formula above says that A must remain as it is
$endgroup$
– RM777
Jan 13 at 19:27




$begingroup$
You said I have to Change the quantorsign but the Formula above says that A must remain as it is
$endgroup$
– RM777
Jan 13 at 19:27










1 Answer
1






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oldest

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$begingroup$

If you define closed as sequentially closed (so every sequence in $A$ that converges in $X$ to $p$ has $ p in A$ too), then the logical negation of that



$A$ is not closed iff there exists some sequence $(a_n)$ converging to $p$ where all $a_n in A$ but $p notin A$. (so one such sequence is enough to disprove closedness).



I don't see how that helps you proving $A$ closed iff its complement is open if you don't also have a definition of open sets in terms of sequences.



A final word of caution: said equivalence of closed and sequentially closed does not holds in all spaces (but it does in e.g. metric spaces, a very common case).






share|cite|improve this answer









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    $begingroup$

    If you define closed as sequentially closed (so every sequence in $A$ that converges in $X$ to $p$ has $ p in A$ too), then the logical negation of that



    $A$ is not closed iff there exists some sequence $(a_n)$ converging to $p$ where all $a_n in A$ but $p notin A$. (so one such sequence is enough to disprove closedness).



    I don't see how that helps you proving $A$ closed iff its complement is open if you don't also have a definition of open sets in terms of sequences.



    A final word of caution: said equivalence of closed and sequentially closed does not holds in all spaces (but it does in e.g. metric spaces, a very common case).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If you define closed as sequentially closed (so every sequence in $A$ that converges in $X$ to $p$ has $ p in A$ too), then the logical negation of that



      $A$ is not closed iff there exists some sequence $(a_n)$ converging to $p$ where all $a_n in A$ but $p notin A$. (so one such sequence is enough to disprove closedness).



      I don't see how that helps you proving $A$ closed iff its complement is open if you don't also have a definition of open sets in terms of sequences.



      A final word of caution: said equivalence of closed and sequentially closed does not holds in all spaces (but it does in e.g. metric spaces, a very common case).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If you define closed as sequentially closed (so every sequence in $A$ that converges in $X$ to $p$ has $ p in A$ too), then the logical negation of that



        $A$ is not closed iff there exists some sequence $(a_n)$ converging to $p$ where all $a_n in A$ but $p notin A$. (so one such sequence is enough to disprove closedness).



        I don't see how that helps you proving $A$ closed iff its complement is open if you don't also have a definition of open sets in terms of sequences.



        A final word of caution: said equivalence of closed and sequentially closed does not holds in all spaces (but it does in e.g. metric spaces, a very common case).






        share|cite|improve this answer









        $endgroup$



        If you define closed as sequentially closed (so every sequence in $A$ that converges in $X$ to $p$ has $ p in A$ too), then the logical negation of that



        $A$ is not closed iff there exists some sequence $(a_n)$ converging to $p$ where all $a_n in A$ but $p notin A$. (so one such sequence is enough to disprove closedness).



        I don't see how that helps you proving $A$ closed iff its complement is open if you don't also have a definition of open sets in terms of sequences.



        A final word of caution: said equivalence of closed and sequentially closed does not holds in all spaces (but it does in e.g. metric spaces, a very common case).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 13 at 23:01









        Henno BrandsmaHenno Brandsma

        107k347114




        107k347114






























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