Proving a set is closed but not open
$begingroup$
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
asked Jan 4 at 8:01
Jimmy SabaterJimmy Sabater
2,887324
$endgroup$
add a comment |
$begingroup$
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
asked Jan 4 at 8:01
Jimmy SabaterJimmy Sabater
2,887324
$endgroup$
$begingroup$
What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
$endgroup$
– Yadati Kiran
Jan 4 at 8:12
$begingroup$
no because there can be found some (x',y') where ball intersects sin(1/x)
$endgroup$
– Jimmy Sabater
Jan 4 at 8:13
1
$begingroup$
I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
$endgroup$
– W. mu
Jan 4 at 8:13
1
$begingroup$
In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
$endgroup$
– Math_QED
Jan 4 at 8:51
add a comment |
$begingroup$
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
asked Jan 4 at 8:01
Jimmy SabaterJimmy Sabater
2,887324
$endgroup$
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
asked Jan 4 at 8:01
Jimmy SabaterJimmy Sabater
2,887324
asked Jan 4 at 8:01
Jimmy SabaterJimmy Sabater
2,887324
asked Jan 4 at 8:01
Jimmy SabaterJimmy Sabater
2,887324
asked Jan 4 at 8:01
Jimmy SabaterJimmy Sabater
2,887324
asked Jan 4 at 8:01
Jimmy SabaterJimmy Sabater
2,887324
2,887324
$begingroup$
What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
$endgroup$
– Yadati Kiran
Jan 4 at 8:12
$begingroup$
no because there can be found some (x',y') where ball intersects sin(1/x)
$endgroup$
– Jimmy Sabater
Jan 4 at 8:13
1
$begingroup$
I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
$endgroup$
– W. mu
Jan 4 at 8:13
1
$begingroup$
In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
$endgroup$
– Math_QED
Jan 4 at 8:51
add a comment |
$begingroup$
What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
$endgroup$
– Yadati Kiran
Jan 4 at 8:12
$begingroup$
no because there can be found some (x',y') where ball intersects sin(1/x)
$endgroup$
– Jimmy Sabater
Jan 4 at 8:13
1
$begingroup$
I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
$endgroup$
– W. mu
Jan 4 at 8:13
1
$begingroup$
In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
$endgroup$
– Math_QED
Jan 4 at 8:51
$begingroup$
What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
$endgroup$
– Yadati Kiran
Jan 4 at 8:12
$begingroup$
What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
$endgroup$
– Yadati Kiran
Jan 4 at 8:12
$begingroup$
no because there can be found some (x',y') where ball intersects sin(1/x)
$endgroup$
– Jimmy Sabater
Jan 4 at 8:13
$begingroup$
no because there can be found some (x',y') where ball intersects sin(1/x)
$endgroup$
– Jimmy Sabater
Jan 4 at 8:13
1
1
$begingroup$
I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
$endgroup$
– W. mu
Jan 4 at 8:13
$begingroup$
I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
$endgroup$
– W. mu
Jan 4 at 8:13
1
1
$begingroup$
In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
$endgroup$
– Math_QED
Jan 4 at 8:51
$begingroup$
In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
$endgroup$
– Math_QED
Jan 4 at 8:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$(frac 1 {npi},0) in S$ 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.
answered Jan 4 at 8:18
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
$begingroup$
Yes, using this definition is easy. The challenge here is to use first principles.
$endgroup$
– Jimmy Sabater
Jan 4 at 8:23
$begingroup$
@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.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 8:26
$begingroup$
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.
$endgroup$
– Peter Szilas
Jan 4 at 9:16
$begingroup$
@PeterSzilas I have given the details now.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 9:25
$begingroup$
Kavi.Thank you.
$endgroup$
– Peter Szilas
Jan 4 at 9:38
|
show 1 more comment
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$begingroup$
$(frac 1 {npi},0) in S$ 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.
answered Jan 4 at 8:18
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
$begingroup$
Yes, using this definition is easy. The challenge here is to use first principles.
$endgroup$
– Jimmy Sabater
Jan 4 at 8:23
$begingroup$
@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.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 8:26
$begingroup$
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.
$endgroup$
– Peter Szilas
Jan 4 at 9:16
$begingroup$
@PeterSzilas I have given the details now.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 9:25
$begingroup$
Kavi.Thank you.
$endgroup$
– Peter Szilas
Jan 4 at 9:38
|
show 1 more comment
$begingroup$
$(frac 1 {npi},0) in S$ 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.
answered Jan 4 at 8:18
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
$begingroup$
Yes, using this definition is easy. The challenge here is to use first principles.
$endgroup$
– Jimmy Sabater
Jan 4 at 8:23
$begingroup$
@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.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 8:26
$begingroup$
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.
$endgroup$
– Peter Szilas
Jan 4 at 9:16
$begingroup$
@PeterSzilas I have given the details now.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 9:25
$begingroup$
Kavi.Thank you.
$endgroup$
– Peter Szilas
Jan 4 at 9:38
|
show 1 more comment
$begingroup$
$(frac 1 {npi},0) in S$ 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.
answered Jan 4 at 8:18
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
$endgroup$
$(frac 1 {npi},0) in S$ 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.
answered Jan 4 at 8:18
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
edited Jan 4 at 9:25
answered Jan 4 at 8:18
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
answered Jan 4 at 8:18
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
answered Jan 4 at 8:18
Kavi Rama MurthyKavi Rama Murthy
64.5k42665
64.5k42665
$begingroup$
Yes, using this definition is easy. The challenge here is to use first principles.
$endgroup$
– Jimmy Sabater
Jan 4 at 8:23
$begingroup$
@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.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 8:26
$begingroup$
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.
$endgroup$
– Peter Szilas
Jan 4 at 9:16
$begingroup$
@PeterSzilas I have given the details now.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 9:25
$begingroup$
Kavi.Thank you.
$endgroup$
– Peter Szilas
Jan 4 at 9:38
|
show 1 more comment
$begingroup$
Yes, using this definition is easy. The challenge here is to use first principles.
$endgroup$
– Jimmy Sabater
Jan 4 at 8:23
$begingroup$
@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.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 8:26
$begingroup$
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.
$endgroup$
– Peter Szilas
Jan 4 at 9:16
$begingroup$
@PeterSzilas I have given the details now.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 9:25
$begingroup$
Kavi.Thank you.
$endgroup$
– Peter Szilas
Jan 4 at 9:38
$begingroup$
Yes, using this definition is easy. The challenge here is to use first principles.
$endgroup$
– Jimmy Sabater
Jan 4 at 8:23
$begingroup$
Yes, using this definition is easy. The challenge here is to use first principles.
$endgroup$
– Jimmy Sabater
Jan 4 at 8:23
$begingroup$
@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.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 8:26
$begingroup$
@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.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 8:26
$begingroup$
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.
$endgroup$
– Peter Szilas
Jan 4 at 9:16
$begingroup$
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.
$endgroup$
– Peter Szilas
Jan 4 at 9:16
$begingroup$
@PeterSzilas I have given the details now.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 9:25
$begingroup$
@PeterSzilas I have given the details now.
$endgroup$
– Kavi Rama Murthy
Jan 4 at 9:25
$begingroup$
Kavi.Thank you.
$endgroup$
– Peter Szilas
Jan 4 at 9:38
$begingroup$
Kavi.Thank you.
$endgroup$
– Peter Szilas
Jan 4 at 9:38
|
show 1 more comment
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$begingroup$
What about at $(x=0,y)inmathbb{R}^2$? Can you construct such a ball?
$endgroup$
– Yadati Kiran
Jan 4 at 8:12
$begingroup$
no because there can be found some (x',y') where ball intersects sin(1/x)
$endgroup$
– Jimmy Sabater
Jan 4 at 8:13
1
$begingroup$
I think this is wrong. $(0,0)$ is a counterexample. Any ball contains a point in $S$ in $x$-axis at least.
$endgroup$
– W. mu
Jan 4 at 8:13
1
$begingroup$
In general, once you see strict inequality in a set "closed" is probably not true (unless one considers unbounded constructs or something similar).
$endgroup$
– Math_QED
Jan 4 at 8:51