Is A = ${(0,0)}cup {(x,sin(1/x))|0 < x le 1}$ $subseteq mathbb{R}^2$ under the usual metric on $mathbb{R}$...
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Is A = ${(0,0)}cup {(x,sin(1/x))|0 < x le 1}$ $subseteq mathbb{R}^2$ under the usual metric on $mathbb{R}$ is compact ?
I thinks yes , because A is closed and bounded
Is its True??
general-topology
asked Nov 21 at 11:53
Messi fifa
50111
add a comment |
up vote
0
down vote
favorite
Is A = ${(0,0)}cup {(x,sin(1/x))|0 < x le 1}$ $subseteq mathbb{R}^2$ under the usual metric on $mathbb{R}$ is compact ?
I thinks yes , because A is closed and bounded
Is its True??
general-topology
asked Nov 21 at 11:53
Messi fifa
50111
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is A = ${(0,0)}cup {(x,sin(1/x))|0 < x le 1}$ $subseteq mathbb{R}^2$ under the usual metric on $mathbb{R}$ is compact ?
I thinks yes , because A is closed and bounded
Is its True??
general-topology
asked Nov 21 at 11:53
Messi fifa
50111
Is A = ${(0,0)}cup {(x,sin(1/x))|0 < x le 1}$ $subseteq mathbb{R}^2$ under the usual metric on $mathbb{R}$ is compact ?
I thinks yes , because A is closed and bounded
Is its True??
general-topology
general-topology
asked Nov 21 at 11:53
Messi fifa
50111
asked Nov 21 at 11:53
Messi fifa
50111
asked Nov 21 at 11:53
Messi fifa
50111
asked Nov 21 at 11:53
Messi fifa
50111
asked Nov 21 at 11:53
Messi fifa
50111
50111
add a comment |
add a comment |
1 Answer
1
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5
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accepted
The set is not closed. $(0,1)$ is a point in the closure but it is not in this set. [Note that $(frac 1 {(4n+1)pi /2}, sin ((4n+1)pi /2)) to (0,1)$].
answered Nov 21 at 11:55
Kavi Rama Murthy
45.9k31854
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The set is not closed. $(0,1)$ is a point in the closure but it is not in this set. [Note that $(frac 1 {(4n+1)pi /2}, sin ((4n+1)pi /2)) to (0,1)$].
answered Nov 21 at 11:55
Kavi Rama Murthy
45.9k31854
add a comment |
up vote
5
down vote
accepted
The set is not closed. $(0,1)$ is a point in the closure but it is not in this set. [Note that $(frac 1 {(4n+1)pi /2}, sin ((4n+1)pi /2)) to (0,1)$].
answered Nov 21 at 11:55
Kavi Rama Murthy
45.9k31854
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The set is not closed. $(0,1)$ is a point in the closure but it is not in this set. [Note that $(frac 1 {(4n+1)pi /2}, sin ((4n+1)pi /2)) to (0,1)$].
answered Nov 21 at 11:55
Kavi Rama Murthy
45.9k31854
The set is not closed. $(0,1)$ is a point in the closure but it is not in this set. [Note that $(frac 1 {(4n+1)pi /2}, sin ((4n+1)pi /2)) to (0,1)$].
answered Nov 21 at 11:55
Kavi Rama Murthy
45.9k31854
answered Nov 21 at 11:55
Kavi Rama Murthy
45.9k31854
answered Nov 21 at 11:55
Kavi Rama Murthy
45.9k31854
answered Nov 21 at 11:55
Kavi Rama Murthy
45.9k31854
45.9k31854
add a comment |
add a comment |
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