Is $left(x, frac{1}{x}cdot cosleft( frac{1}{x^2}right)right)cup{(0,0)}$ a connected set?
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Is $$ left{left(x, frac{1}{x}cdot cosleft( frac{1}{x^2}right)right) mid x in mathbb{R} setminus {0}right} cup { (0,0)} subset mathbb{R^2}$$ a connected set?
I tried proving by contradiction that it is connected , but it didn't seem to lead anywhere. It would be easy to prove that it is not connected if it did not contain $(0,0)$.
connectedness
edited Nov 17 at 10:29
b00n heT
10.2k12134
asked Nov 17 at 10:17
user560461
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up vote
1
down vote
favorite
Is $$ left{left(x, frac{1}{x}cdot cosleft( frac{1}{x^2}right)right) mid x in mathbb{R} setminus {0}right} cup { (0,0)} subset mathbb{R^2}$$ a connected set?
I tried proving by contradiction that it is connected , but it didn't seem to lead anywhere. It would be easy to prove that it is not connected if it did not contain $(0,0)$.
connectedness
edited Nov 17 at 10:29
b00n heT
10.2k12134
asked Nov 17 at 10:17
user560461
525
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is $$ left{left(x, frac{1}{x}cdot cosleft( frac{1}{x^2}right)right) mid x in mathbb{R} setminus {0}right} cup { (0,0)} subset mathbb{R^2}$$ a connected set?
I tried proving by contradiction that it is connected , but it didn't seem to lead anywhere. It would be easy to prove that it is not connected if it did not contain $(0,0)$.
connectedness
edited Nov 17 at 10:29
b00n heT
10.2k12134
asked Nov 17 at 10:17
user560461
525
Is $$ left{left(x, frac{1}{x}cdot cosleft( frac{1}{x^2}right)right) mid x in mathbb{R} setminus {0}right} cup { (0,0)} subset mathbb{R^2}$$ a connected set?
I tried proving by contradiction that it is connected , but it didn't seem to lead anywhere. It would be easy to prove that it is not connected if it did not contain $(0,0)$.
connectedness
connectedness
edited Nov 17 at 10:29
b00n heT
10.2k12134
asked Nov 17 at 10:17
user560461
525
edited Nov 17 at 10:29
b00n heT
10.2k12134
asked Nov 17 at 10:17
user560461
525
edited Nov 17 at 10:29
b00n heT
10.2k12134
edited Nov 17 at 10:29
b00n heT
10.2k12134
edited Nov 17 at 10:29
b00n heT
10.2k12134
10.2k12134
asked Nov 17 at 10:17
user560461
525
asked Nov 17 at 10:17
user560461
525
asked Nov 17 at 10:17
user560461
525
525
add a comment |
add a comment |
1 Answer
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${(x,frac 1 x cos(frac 1 {x^{2}})): x>0}$ is connected because it is a continuous image of a connected set. Now $(0,0)$ belongs to the closure of this set because $(0,0)=lim (x_n,frac 1 {x_n} cos(frac 1 {x_n^{2}}))$ where $x_n=((n+frac 1 2)pi)^{-1/2}$. Hence ${(x,frac 1 x cos(frac 1 {x^{2}})): x geq0}$ is connected and ${(x,frac 1 x cos(frac 1 {x^{2}})): xleq 0}$ is connected is connected by a similar argument. The union of these two is also connected because they have a point in common.
answered Nov 17 at 11:52
Kavi Rama Murthy
43.3k31751
add a comment |
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
${(x,frac 1 x cos(frac 1 {x^{2}})): x>0}$ is connected because it is a continuous image of a connected set. Now $(0,0)$ belongs to the closure of this set because $(0,0)=lim (x_n,frac 1 {x_n} cos(frac 1 {x_n^{2}}))$ where $x_n=((n+frac 1 2)pi)^{-1/2}$. Hence ${(x,frac 1 x cos(frac 1 {x^{2}})): x geq0}$ is connected and ${(x,frac 1 x cos(frac 1 {x^{2}})): xleq 0}$ is connected is connected by a similar argument. The union of these two is also connected because they have a point in common.
answered Nov 17 at 11:52
Kavi Rama Murthy
43.3k31751
add a comment |
up vote
0
down vote
${(x,frac 1 x cos(frac 1 {x^{2}})): x>0}$ is connected because it is a continuous image of a connected set. Now $(0,0)$ belongs to the closure of this set because $(0,0)=lim (x_n,frac 1 {x_n} cos(frac 1 {x_n^{2}}))$ where $x_n=((n+frac 1 2)pi)^{-1/2}$. Hence ${(x,frac 1 x cos(frac 1 {x^{2}})): x geq0}$ is connected and ${(x,frac 1 x cos(frac 1 {x^{2}})): xleq 0}$ is connected is connected by a similar argument. The union of these two is also connected because they have a point in common.
answered Nov 17 at 11:52
Kavi Rama Murthy
43.3k31751
add a comment |
up vote
0
down vote
up vote
0
down vote
${(x,frac 1 x cos(frac 1 {x^{2}})): x>0}$ is connected because it is a continuous image of a connected set. Now $(0,0)$ belongs to the closure of this set because $(0,0)=lim (x_n,frac 1 {x_n} cos(frac 1 {x_n^{2}}))$ where $x_n=((n+frac 1 2)pi)^{-1/2}$. Hence ${(x,frac 1 x cos(frac 1 {x^{2}})): x geq0}$ is connected and ${(x,frac 1 x cos(frac 1 {x^{2}})): xleq 0}$ is connected is connected by a similar argument. The union of these two is also connected because they have a point in common.
answered Nov 17 at 11:52
Kavi Rama Murthy
43.3k31751
${(x,frac 1 x cos(frac 1 {x^{2}})): x>0}$ is connected because it is a continuous image of a connected set. Now $(0,0)$ belongs to the closure of this set because $(0,0)=lim (x_n,frac 1 {x_n} cos(frac 1 {x_n^{2}}))$ where $x_n=((n+frac 1 2)pi)^{-1/2}$. Hence ${(x,frac 1 x cos(frac 1 {x^{2}})): x geq0}$ is connected and ${(x,frac 1 x cos(frac 1 {x^{2}})): xleq 0}$ is connected is connected by a similar argument. The union of these two is also connected because they have a point in common.
answered Nov 17 at 11:52
Kavi Rama Murthy
43.3k31751
answered Nov 17 at 11:52
Kavi Rama Murthy
43.3k31751
answered Nov 17 at 11:52
Kavi Rama Murthy
43.3k31751
answered Nov 17 at 11:52
Kavi Rama Murthy
43.3k31751
43.3k31751
add a comment |
add a comment |
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