Continuity of functions of monotone sequences












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Suppose $lim_{ntoinfty} f(x_n) = f(c)$ for any monotone sequence $x_n$ approaching $c$. The prove that $f$ is continuous at $c$.



Solution: We prove it by contradiction. Assume $f$ is not continuous at $c$. Then there exists $varepsilon > 0$ such that for any $n$ belonging to $mathbb{N}$, there is $x_n$ such that $x_n$ approaches c but $|f(x_n) - f(c)| > varepsilon$. Then there is subsequence $x_{n_k}$ such that $lim_{ktoinfty} x_{n_k} = c$ and $x_{n_k}$ is monotone. Then by assumption we have $lim f(x_{n_k}) = f(c)$ which is a contradiction.



Is this right? Can someone explain why we have a contradiction?






real-analysis







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$endgroup$












  • $begingroup$
    If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
    $endgroup$
    – Offlaw
    Nov 26 '18 at 13:24


















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$begingroup$


Suppose $lim_{ntoinfty} f(x_n) = f(c)$ for any monotone sequence $x_n$ approaching $c$. The prove that $f$ is continuous at $c$.



Solution: We prove it by contradiction. Assume $f$ is not continuous at $c$. Then there exists $varepsilon > 0$ such that for any $n$ belonging to $mathbb{N}$, there is $x_n$ such that $x_n$ approaches c but $|f(x_n) - f(c)| > varepsilon$. Then there is subsequence $x_{n_k}$ such that $lim_{ktoinfty} x_{n_k} = c$ and $x_{n_k}$ is monotone. Then by assumption we have $lim f(x_{n_k}) = f(c)$ which is a contradiction.



Is this right? Can someone explain why we have a contradiction?






real-analysis







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$endgroup$












  • $begingroup$
    If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
    $endgroup$
    – Offlaw
    Nov 26 '18 at 13:24
















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$begingroup$


Suppose $lim_{ntoinfty} f(x_n) = f(c)$ for any monotone sequence $x_n$ approaching $c$. The prove that $f$ is continuous at $c$.



Solution: We prove it by contradiction. Assume $f$ is not continuous at $c$. Then there exists $varepsilon > 0$ such that for any $n$ belonging to $mathbb{N}$, there is $x_n$ such that $x_n$ approaches c but $|f(x_n) - f(c)| > varepsilon$. Then there is subsequence $x_{n_k}$ such that $lim_{ktoinfty} x_{n_k} = c$ and $x_{n_k}$ is monotone. Then by assumption we have $lim f(x_{n_k}) = f(c)$ which is a contradiction.



Is this right? Can someone explain why we have a contradiction?






real-analysis







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$endgroup$




Suppose $lim_{ntoinfty} f(x_n) = f(c)$ for any monotone sequence $x_n$ approaching $c$. The prove that $f$ is continuous at $c$.



Solution: We prove it by contradiction. Assume $f$ is not continuous at $c$. Then there exists $varepsilon > 0$ such that for any $n$ belonging to $mathbb{N}$, there is $x_n$ such that $x_n$ approaches c but $|f(x_n) - f(c)| > varepsilon$. Then there is subsequence $x_{n_k}$ such that $lim_{ktoinfty} x_{n_k} = c$ and $x_{n_k}$ is monotone. Then by assumption we have $lim f(x_{n_k}) = f(c)$ which is a contradiction.



Is this right? Can someone explain why we have a contradiction?






real-analysis




real-analysis






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edited Nov 26 '18 at 13:25









user3482749

4,047818




4,047818










asked Nov 26 '18 at 13:16









Aishwarya DeoreAishwarya Deore

324




324












  • $begingroup$
    If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
    $endgroup$
    – Offlaw
    Nov 26 '18 at 13:24




















  • $begingroup$
    If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
    $endgroup$
    – Offlaw
    Nov 26 '18 at 13:24


















$begingroup$
If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
$endgroup$
– Offlaw
Nov 26 '18 at 13:24






$begingroup$
If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
$endgroup$
– Offlaw
Nov 26 '18 at 13:24












2 Answers
2






active

oldest

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0












$begingroup$

No, it is not right. The assertion “there is $x_n$ such that $x_n$ approaches $c$ but $bigl|f(x_n)-f(c)bigr|>varepsilon$” makes no sense.



There is a $varepsilon>0$ such that, for each $delta>0$, there is a $xin(c-delta,c+delta)cap D_f$ such that $bigllvert f(x)-f(c)bigrrvertgeqslantvarepsilon$. In particular, for any natural $n$, there is a $x_ninleft(c-frac1n,c+frac1nright)cap D_f$ such that $bigllvert f(x_n)-f(c)bigrrvertgeqslantvarepsilon$. The sequence $(x_n)_{ninmathbb N}$ has a monotonic subsequence $(x_{n_k})_{kinmathbb N}$ and, since $lim_{ntoinfty}x_n=c$, $lim_{ktoinfty}x_{n_k}=c$. Therefore, we should have $lim_{ktoinfty}f(x_{n_k})=f(c)$. But we don't, since $(forall kinmathbb{N}):bigllvert f(x_{n_k})-f(c)bigrrvertgeqslantvarepsilon$. So, we have a contradiction here.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    “there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
    $endgroup$
    – Offlaw
    Nov 26 '18 at 13:30












  • $begingroup$
    "x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
    $endgroup$
    – Aishwarya Deore
    Nov 26 '18 at 13:42










  • $begingroup$
    @AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
    $endgroup$
    – José Carlos Santos
    Nov 26 '18 at 17:25





















0












$begingroup$

No. Mostly, your third sentence is mangled. Here's a corrected version:



If $f$ is not left continuous at $c$, then there is some $varepsilon > 0$ such that for any $delta > 0$, there is some $x in (c-delta,c)$ such that $|f(x)-f(c)| geq varepsilon$. Choosing, in particular, for each $n in mathbb{N}$, $delta = frac{1}{n}$, and choosing some $x_n in (c-delta,c)$, we obtain a sequence $(x_n)$ such that $(x_n)to c$ but $|f(x_n)-f(c)| geq varepsilon$ for all $n$. Now, every sequence has a monotone subsequence, so in particular $(x_n)$ has a monotone subsequence $(x_{n_k})$, and $(x_{n_k})to c$, since it's a subsequence of $(x_n)$, and $|f(x_{n_k})-f(c)|geqvarepsilon$ for all $k$. This contradicts our hypothesis about $f$, so $f$ is left continuous.



An identical proof (with $(c-delta,c)$ replaced by $(c,c+delta)$) shows that $f$ is right-continuous [or you could just patch this into the main proof if you prefer], so $f$ is both left- and right-continuous, so is continuous.






share|cite|improve this answer









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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    No, it is not right. The assertion “there is $x_n$ such that $x_n$ approaches $c$ but $bigl|f(x_n)-f(c)bigr|>varepsilon$” makes no sense.



    There is a $varepsilon>0$ such that, for each $delta>0$, there is a $xin(c-delta,c+delta)cap D_f$ such that $bigllvert f(x)-f(c)bigrrvertgeqslantvarepsilon$. In particular, for any natural $n$, there is a $x_ninleft(c-frac1n,c+frac1nright)cap D_f$ such that $bigllvert f(x_n)-f(c)bigrrvertgeqslantvarepsilon$. The sequence $(x_n)_{ninmathbb N}$ has a monotonic subsequence $(x_{n_k})_{kinmathbb N}$ and, since $lim_{ntoinfty}x_n=c$, $lim_{ktoinfty}x_{n_k}=c$. Therefore, we should have $lim_{ktoinfty}f(x_{n_k})=f(c)$. But we don't, since $(forall kinmathbb{N}):bigllvert f(x_{n_k})-f(c)bigrrvertgeqslantvarepsilon$. So, we have a contradiction here.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      “there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
      $endgroup$
      – Offlaw
      Nov 26 '18 at 13:30












    • $begingroup$
      "x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
      $endgroup$
      – Aishwarya Deore
      Nov 26 '18 at 13:42










    • $begingroup$
      @AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
      $endgroup$
      – José Carlos Santos
      Nov 26 '18 at 17:25


















    0












    $begingroup$

    No, it is not right. The assertion “there is $x_n$ such that $x_n$ approaches $c$ but $bigl|f(x_n)-f(c)bigr|>varepsilon$” makes no sense.



    There is a $varepsilon>0$ such that, for each $delta>0$, there is a $xin(c-delta,c+delta)cap D_f$ such that $bigllvert f(x)-f(c)bigrrvertgeqslantvarepsilon$. In particular, for any natural $n$, there is a $x_ninleft(c-frac1n,c+frac1nright)cap D_f$ such that $bigllvert f(x_n)-f(c)bigrrvertgeqslantvarepsilon$. The sequence $(x_n)_{ninmathbb N}$ has a monotonic subsequence $(x_{n_k})_{kinmathbb N}$ and, since $lim_{ntoinfty}x_n=c$, $lim_{ktoinfty}x_{n_k}=c$. Therefore, we should have $lim_{ktoinfty}f(x_{n_k})=f(c)$. But we don't, since $(forall kinmathbb{N}):bigllvert f(x_{n_k})-f(c)bigrrvertgeqslantvarepsilon$. So, we have a contradiction here.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      “there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
      $endgroup$
      – Offlaw
      Nov 26 '18 at 13:30












    • $begingroup$
      "x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
      $endgroup$
      – Aishwarya Deore
      Nov 26 '18 at 13:42










    • $begingroup$
      @AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
      $endgroup$
      – José Carlos Santos
      Nov 26 '18 at 17:25
















    0












    0








    0





    $begingroup$

    No, it is not right. The assertion “there is $x_n$ such that $x_n$ approaches $c$ but $bigl|f(x_n)-f(c)bigr|>varepsilon$” makes no sense.



    There is a $varepsilon>0$ such that, for each $delta>0$, there is a $xin(c-delta,c+delta)cap D_f$ such that $bigllvert f(x)-f(c)bigrrvertgeqslantvarepsilon$. In particular, for any natural $n$, there is a $x_ninleft(c-frac1n,c+frac1nright)cap D_f$ such that $bigllvert f(x_n)-f(c)bigrrvertgeqslantvarepsilon$. The sequence $(x_n)_{ninmathbb N}$ has a monotonic subsequence $(x_{n_k})_{kinmathbb N}$ and, since $lim_{ntoinfty}x_n=c$, $lim_{ktoinfty}x_{n_k}=c$. Therefore, we should have $lim_{ktoinfty}f(x_{n_k})=f(c)$. But we don't, since $(forall kinmathbb{N}):bigllvert f(x_{n_k})-f(c)bigrrvertgeqslantvarepsilon$. So, we have a contradiction here.






    share|cite|improve this answer









    $endgroup$



    No, it is not right. The assertion “there is $x_n$ such that $x_n$ approaches $c$ but $bigl|f(x_n)-f(c)bigr|>varepsilon$” makes no sense.



    There is a $varepsilon>0$ such that, for each $delta>0$, there is a $xin(c-delta,c+delta)cap D_f$ such that $bigllvert f(x)-f(c)bigrrvertgeqslantvarepsilon$. In particular, for any natural $n$, there is a $x_ninleft(c-frac1n,c+frac1nright)cap D_f$ such that $bigllvert f(x_n)-f(c)bigrrvertgeqslantvarepsilon$. The sequence $(x_n)_{ninmathbb N}$ has a monotonic subsequence $(x_{n_k})_{kinmathbb N}$ and, since $lim_{ntoinfty}x_n=c$, $lim_{ktoinfty}x_{n_k}=c$. Therefore, we should have $lim_{ktoinfty}f(x_{n_k})=f(c)$. But we don't, since $(forall kinmathbb{N}):bigllvert f(x_{n_k})-f(c)bigrrvertgeqslantvarepsilon$. So, we have a contradiction here.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 26 '18 at 13:25









    José Carlos SantosJosé Carlos Santos

    156k22125227




    156k22125227












    • $begingroup$
      “there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
      $endgroup$
      – Offlaw
      Nov 26 '18 at 13:30












    • $begingroup$
      "x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
      $endgroup$
      – Aishwarya Deore
      Nov 26 '18 at 13:42










    • $begingroup$
      @AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
      $endgroup$
      – José Carlos Santos
      Nov 26 '18 at 17:25




















    • $begingroup$
      “there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
      $endgroup$
      – Offlaw
      Nov 26 '18 at 13:30












    • $begingroup$
      "x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
      $endgroup$
      – Aishwarya Deore
      Nov 26 '18 at 13:42










    • $begingroup$
      @AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
      $endgroup$
      – José Carlos Santos
      Nov 26 '18 at 17:25


















    $begingroup$
    “there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
    $endgroup$
    – Offlaw
    Nov 26 '18 at 13:30






    $begingroup$
    “there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
    $endgroup$
    – Offlaw
    Nov 26 '18 at 13:30














    $begingroup$
    "x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
    $endgroup$
    – Aishwarya Deore
    Nov 26 '18 at 13:42




    $begingroup$
    "x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
    $endgroup$
    – Aishwarya Deore
    Nov 26 '18 at 13:42












    $begingroup$
    @AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
    $endgroup$
    – José Carlos Santos
    Nov 26 '18 at 17:25






    $begingroup$
    @AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
    $endgroup$
    – José Carlos Santos
    Nov 26 '18 at 17:25













    0












    $begingroup$

    No. Mostly, your third sentence is mangled. Here's a corrected version:



    If $f$ is not left continuous at $c$, then there is some $varepsilon > 0$ such that for any $delta > 0$, there is some $x in (c-delta,c)$ such that $|f(x)-f(c)| geq varepsilon$. Choosing, in particular, for each $n in mathbb{N}$, $delta = frac{1}{n}$, and choosing some $x_n in (c-delta,c)$, we obtain a sequence $(x_n)$ such that $(x_n)to c$ but $|f(x_n)-f(c)| geq varepsilon$ for all $n$. Now, every sequence has a monotone subsequence, so in particular $(x_n)$ has a monotone subsequence $(x_{n_k})$, and $(x_{n_k})to c$, since it's a subsequence of $(x_n)$, and $|f(x_{n_k})-f(c)|geqvarepsilon$ for all $k$. This contradicts our hypothesis about $f$, so $f$ is left continuous.



    An identical proof (with $(c-delta,c)$ replaced by $(c,c+delta)$) shows that $f$ is right-continuous [or you could just patch this into the main proof if you prefer], so $f$ is both left- and right-continuous, so is continuous.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      No. Mostly, your third sentence is mangled. Here's a corrected version:



      If $f$ is not left continuous at $c$, then there is some $varepsilon > 0$ such that for any $delta > 0$, there is some $x in (c-delta,c)$ such that $|f(x)-f(c)| geq varepsilon$. Choosing, in particular, for each $n in mathbb{N}$, $delta = frac{1}{n}$, and choosing some $x_n in (c-delta,c)$, we obtain a sequence $(x_n)$ such that $(x_n)to c$ but $|f(x_n)-f(c)| geq varepsilon$ for all $n$. Now, every sequence has a monotone subsequence, so in particular $(x_n)$ has a monotone subsequence $(x_{n_k})$, and $(x_{n_k})to c$, since it's a subsequence of $(x_n)$, and $|f(x_{n_k})-f(c)|geqvarepsilon$ for all $k$. This contradicts our hypothesis about $f$, so $f$ is left continuous.



      An identical proof (with $(c-delta,c)$ replaced by $(c,c+delta)$) shows that $f$ is right-continuous [or you could just patch this into the main proof if you prefer], so $f$ is both left- and right-continuous, so is continuous.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        No. Mostly, your third sentence is mangled. Here's a corrected version:



        If $f$ is not left continuous at $c$, then there is some $varepsilon > 0$ such that for any $delta > 0$, there is some $x in (c-delta,c)$ such that $|f(x)-f(c)| geq varepsilon$. Choosing, in particular, for each $n in mathbb{N}$, $delta = frac{1}{n}$, and choosing some $x_n in (c-delta,c)$, we obtain a sequence $(x_n)$ such that $(x_n)to c$ but $|f(x_n)-f(c)| geq varepsilon$ for all $n$. Now, every sequence has a monotone subsequence, so in particular $(x_n)$ has a monotone subsequence $(x_{n_k})$, and $(x_{n_k})to c$, since it's a subsequence of $(x_n)$, and $|f(x_{n_k})-f(c)|geqvarepsilon$ for all $k$. This contradicts our hypothesis about $f$, so $f$ is left continuous.



        An identical proof (with $(c-delta,c)$ replaced by $(c,c+delta)$) shows that $f$ is right-continuous [or you could just patch this into the main proof if you prefer], so $f$ is both left- and right-continuous, so is continuous.






        share|cite|improve this answer









        $endgroup$



        No. Mostly, your third sentence is mangled. Here's a corrected version:



        If $f$ is not left continuous at $c$, then there is some $varepsilon > 0$ such that for any $delta > 0$, there is some $x in (c-delta,c)$ such that $|f(x)-f(c)| geq varepsilon$. Choosing, in particular, for each $n in mathbb{N}$, $delta = frac{1}{n}$, and choosing some $x_n in (c-delta,c)$, we obtain a sequence $(x_n)$ such that $(x_n)to c$ but $|f(x_n)-f(c)| geq varepsilon$ for all $n$. Now, every sequence has a monotone subsequence, so in particular $(x_n)$ has a monotone subsequence $(x_{n_k})$, and $(x_{n_k})to c$, since it's a subsequence of $(x_n)$, and $|f(x_{n_k})-f(c)|geqvarepsilon$ for all $k$. This contradicts our hypothesis about $f$, so $f$ is left continuous.



        An identical proof (with $(c-delta,c)$ replaced by $(c,c+delta)$) shows that $f$ is right-continuous [or you could just patch this into the main proof if you prefer], so $f$ is both left- and right-continuous, so is continuous.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 '18 at 13:27









        user3482749user3482749

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