Proving a set is closed but not open












0















Suppose we have that set $S = { (x,y) in R^2 : x>0, y = sin(1/x)
}$. Prove that $S$ is closed but not open.




${bf Relevant ; definitions:}$ A set $S$ is open if for every $s in S$, one can find an $r>0$ so that $|x-s| < r$ for all $x in mathbb{R}^n$. $S$ is closed if $S^c$ is open.



Attempt:



Now, the problem here is to make an $r>0$ so that the ball $B(r,x)$ is in the complement of $S$ for all $x in S^c$. (here x =(x,y)) Now, if $x in S^c$, then $x < 0$ and $y neq sin (1/x)$. Here is where I get stuck. I dont know how to construct such an $r$. Any help?






real-analysis calculus







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  • What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
    – Yadati Kiran
    Jan 4 at 8:12












  • no because there can be found some (x',y') where ball intersects sin(1/x)
    – Jimmy Sabater
    Jan 4 at 8:13








  • 1




    I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
    – W. mu
    Jan 4 at 8:13








  • 1




    In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
    – Math_QED
    Jan 4 at 8:51
















0















Suppose we have that set $S = { (x,y) in R^2 : x>0, y = sin(1/x)
}$. Prove that $S$ is closed but not open.




${bf Relevant ; definitions:}$ A set $S$ is open if for every $s in S$, one can find an $r>0$ so that $|x-s| < r$ for all $x in mathbb{R}^n$. $S$ is closed if $S^c$ is open.



Attempt:



Now, the problem here is to make an $r>0$ so that the ball $B(r,x)$ is in the complement of $S$ for all $x in S^c$. (here x =(x,y)) Now, if $x in S^c$, then $x < 0$ and $y neq sin (1/x)$. Here is where I get stuck. I dont know how to construct such an $r$. Any help?






real-analysis calculus







share|cite|improve this question






















  • What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
    – Yadati Kiran
    Jan 4 at 8:12












  • no because there can be found some (x',y') where ball intersects sin(1/x)
    – Jimmy Sabater
    Jan 4 at 8:13








  • 1




    I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
    – W. mu
    Jan 4 at 8:13








  • 1




    In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
    – Math_QED
    Jan 4 at 8:51














0












0








0








Suppose we have that set $S = { (x,y) in R^2 : x>0, y = sin(1/x)
}$. Prove that $S$ is closed but not open.




${bf Relevant ; definitions:}$ A set $S$ is open if for every $s in S$, one can find an $r>0$ so that $|x-s| < r$ for all $x in mathbb{R}^n$. $S$ is closed if $S^c$ is open.



Attempt:



Now, the problem here is to make an $r>0$ so that the ball $B(r,x)$ is in the complement of $S$ for all $x in S^c$. (here x =(x,y)) Now, if $x in S^c$, then $x < 0$ and $y neq sin (1/x)$. Here is where I get stuck. I dont know how to construct such an $r$. Any help?






real-analysis calculus







share|cite|improve this question














Suppose we have that set $S = { (x,y) in R^2 : x>0, y = sin(1/x)
}$. Prove that $S$ is closed but not open.




${bf Relevant ; definitions:}$ A set $S$ is open if for every $s in S$, one can find an $r>0$ so that $|x-s| < r$ for all $x in mathbb{R}^n$. $S$ is closed if $S^c$ is open.



Attempt:



Now, the problem here is to make an $r>0$ so that the ball $B(r,x)$ is in the complement of $S$ for all $x in S^c$. (here x =(x,y)) Now, if $x in S^c$, then $x < 0$ and $y neq sin (1/x)$. Here is where I get stuck. I dont know how to construct such an $r$. Any help?






real-analysis calculus




real-analysis calculus






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 8:01









Jimmy SabaterJimmy Sabater

1,980219




1,980219












  • What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
    – Yadati Kiran
    Jan 4 at 8:12












  • no because there can be found some (x',y') where ball intersects sin(1/x)
    – Jimmy Sabater
    Jan 4 at 8:13








  • 1




    I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
    – W. mu
    Jan 4 at 8:13








  • 1




    In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
    – Math_QED
    Jan 4 at 8:51


















  • What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
    – Yadati Kiran
    Jan 4 at 8:12












  • no because there can be found some (x',y') where ball intersects sin(1/x)
    – Jimmy Sabater
    Jan 4 at 8:13








  • 1




    I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
    – W. mu
    Jan 4 at 8:13








  • 1




    In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
    – Math_QED
    Jan 4 at 8:51
















What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
– Yadati Kiran
Jan 4 at 8:12






What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
– Yadati Kiran
Jan 4 at 8:12














no because there can be found some (x',y') where ball intersects sin(1/x)
– Jimmy Sabater
Jan 4 at 8:13






no because there can be found some (x',y') where ball intersects sin(1/x)
– Jimmy Sabater
Jan 4 at 8:13






1




1




I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
– W. mu
Jan 4 at 8:13






I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
– W. mu
Jan 4 at 8:13






1




1




In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
– Math_QED
Jan 4 at 8:51




In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
– Math_QED
Jan 4 at 8:51










1 Answer
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$(frac 1 {npi},0) in S$ for each $n$ and $(frac 1 {npi},0) to (0,0)$ so the set is not closed. It is not open either. $(1,sin, 1)$ is in $S$ but there is no $r>0$ such that $B((1,sin, 1),r) subset S$?: if there is such an $r$ we get a contradiction as follows: let $0<|y-sin (1)| <r$. Then $(1,y) in B((1,sin, 1),r)$ but $(1,sin (1))$ is not in $S$. (If it is in $S$ then we must have $y =sin(1)$ which is a contradiction.






share|cite|improve this answer























  • Yes, using this definition is easy. The challenge here is to use first principles.
    – Jimmy Sabater
    Jan 4 at 8:23










  • @JimmySabater In your attempt you are trying to prove that $S^{c}$ is open. This is not true. First principles or otherwise you cannot prove a wrong statement. I have shown you why $S^{c}$ is not open.
    – Kavi Rama Murthy
    Jan 4 at 8:26










  • Kavi.Kindly elaborate on your last line: B( (1, sin 1), r), r >0, not a subset of S, I do not get it. Thanks.
    – Peter Szilas
    Jan 4 at 9:16










  • @PeterSzilas I have given the details now.
    – Kavi Rama Murthy
    Jan 4 at 9:25










  • Kavi.Thank you.
    – Peter Szilas
    Jan 4 at 9:38











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$(frac 1 {npi},0) in S$ for each $n$ and $(frac 1 {npi},0) to (0,0)$ so the set is not closed. It is not open either. $(1,sin, 1)$ is in $S$ but there is no $r>0$ such that $B((1,sin, 1),r) subset S$?: if there is such an $r$ we get a contradiction as follows: let $0<|y-sin (1)| <r$. Then $(1,y) in B((1,sin, 1),r)$ but $(1,sin (1))$ is not in $S$. (If it is in $S$ then we must have $y =sin(1)$ which is a contradiction.






share|cite|improve this answer























  • Yes, using this definition is easy. The challenge here is to use first principles.
    – Jimmy Sabater
    Jan 4 at 8:23










  • @JimmySabater In your attempt you are trying to prove that $S^{c}$ is open. This is not true. First principles or otherwise you cannot prove a wrong statement. I have shown you why $S^{c}$ is not open.
    – Kavi Rama Murthy
    Jan 4 at 8:26










  • Kavi.Kindly elaborate on your last line: B( (1, sin 1), r), r >0, not a subset of S, I do not get it. Thanks.
    – Peter Szilas
    Jan 4 at 9:16










  • @PeterSzilas I have given the details now.
    – Kavi Rama Murthy
    Jan 4 at 9:25










  • Kavi.Thank you.
    – Peter Szilas
    Jan 4 at 9:38
















1














$(frac 1 {npi},0) in S$ for each $n$ and $(frac 1 {npi},0) to (0,0)$ so the set is not closed. It is not open either. $(1,sin, 1)$ is in $S$ but there is no $r>0$ such that $B((1,sin, 1),r) subset S$?: if there is such an $r$ we get a contradiction as follows: let $0<|y-sin (1)| <r$. Then $(1,y) in B((1,sin, 1),r)$ but $(1,sin (1))$ is not in $S$. (If it is in $S$ then we must have $y =sin(1)$ which is a contradiction.






share|cite|improve this answer























  • Yes, using this definition is easy. The challenge here is to use first principles.
    – Jimmy Sabater
    Jan 4 at 8:23










  • @JimmySabater In your attempt you are trying to prove that $S^{c}$ is open. This is not true. First principles or otherwise you cannot prove a wrong statement. I have shown you why $S^{c}$ is not open.
    – Kavi Rama Murthy
    Jan 4 at 8:26










  • Kavi.Kindly elaborate on your last line: B( (1, sin 1), r), r >0, not a subset of S, I do not get it. Thanks.
    – Peter Szilas
    Jan 4 at 9:16










  • @PeterSzilas I have given the details now.
    – Kavi Rama Murthy
    Jan 4 at 9:25










  • Kavi.Thank you.
    – Peter Szilas
    Jan 4 at 9:38














1












1








1






$(frac 1 {npi},0) in S$ for each $n$ and $(frac 1 {npi},0) to (0,0)$ so the set is not closed. It is not open either. $(1,sin, 1)$ is in $S$ but there is no $r>0$ such that $B((1,sin, 1),r) subset S$?: if there is such an $r$ we get a contradiction as follows: let $0<|y-sin (1)| <r$. Then $(1,y) in B((1,sin, 1),r)$ but $(1,sin (1))$ is not in $S$. (If it is in $S$ then we must have $y =sin(1)$ which is a contradiction.






share|cite|improve this answer














$(frac 1 {npi},0) in S$ for each $n$ and $(frac 1 {npi},0) to (0,0)$ so the set is not closed. It is not open either. $(1,sin, 1)$ is in $S$ but there is no $r>0$ such that $B((1,sin, 1),r) subset S$?: if there is such an $r$ we get a contradiction as follows: let $0<|y-sin (1)| <r$. Then $(1,y) in B((1,sin, 1),r)$ but $(1,sin (1))$ is not in $S$. (If it is in $S$ then we must have $y =sin(1)$ which is a contradiction.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 9:25

























answered Jan 4 at 8:18









Kavi Rama MurthyKavi Rama Murthy

51.7k32055




51.7k32055












  • Yes, using this definition is easy. The challenge here is to use first principles.
    – Jimmy Sabater
    Jan 4 at 8:23










  • @JimmySabater In your attempt you are trying to prove that $S^{c}$ is open. This is not true. First principles or otherwise you cannot prove a wrong statement. I have shown you why $S^{c}$ is not open.
    – Kavi Rama Murthy
    Jan 4 at 8:26










  • Kavi.Kindly elaborate on your last line: B( (1, sin 1), r), r >0, not a subset of S, I do not get it. Thanks.
    – Peter Szilas
    Jan 4 at 9:16










  • @PeterSzilas I have given the details now.
    – Kavi Rama Murthy
    Jan 4 at 9:25










  • Kavi.Thank you.
    – Peter Szilas
    Jan 4 at 9:38


















  • Yes, using this definition is easy. The challenge here is to use first principles.
    – Jimmy Sabater
    Jan 4 at 8:23










  • @JimmySabater In your attempt you are trying to prove that $S^{c}$ is open. This is not true. First principles or otherwise you cannot prove a wrong statement. I have shown you why $S^{c}$ is not open.
    – Kavi Rama Murthy
    Jan 4 at 8:26










  • Kavi.Kindly elaborate on your last line: B( (1, sin 1), r), r >0, not a subset of S, I do not get it. Thanks.
    – Peter Szilas
    Jan 4 at 9:16










  • @PeterSzilas I have given the details now.
    – Kavi Rama Murthy
    Jan 4 at 9:25










  • Kavi.Thank you.
    – Peter Szilas
    Jan 4 at 9:38
















Yes, using this definition is easy. The challenge here is to use first principles.
– Jimmy Sabater
Jan 4 at 8:23




Yes, using this definition is easy. The challenge here is to use first principles.
– Jimmy Sabater
Jan 4 at 8:23












@JimmySabater In your attempt you are trying to prove that $S^{c}$ is open. This is not true. First principles or otherwise you cannot prove a wrong statement. I have shown you why $S^{c}$ is not open.
– Kavi Rama Murthy
Jan 4 at 8:26




@JimmySabater In your attempt you are trying to prove that $S^{c}$ is open. This is not true. First principles or otherwise you cannot prove a wrong statement. I have shown you why $S^{c}$ is not open.
– Kavi Rama Murthy
Jan 4 at 8:26












Kavi.Kindly elaborate on your last line: B( (1, sin 1), r), r >0, not a subset of S, I do not get it. Thanks.
– Peter Szilas
Jan 4 at 9:16




Kavi.Kindly elaborate on your last line: B( (1, sin 1), r), r >0, not a subset of S, I do not get it. Thanks.
– Peter Szilas
Jan 4 at 9:16












@PeterSzilas I have given the details now.
– Kavi Rama Murthy
Jan 4 at 9:25




@PeterSzilas I have given the details now.
– Kavi Rama Murthy
Jan 4 at 9:25












Kavi.Thank you.
– Peter Szilas
Jan 4 at 9:38




Kavi.Thank you.
– Peter Szilas
Jan 4 at 9:38


















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