Let $X = { (x, sin(1/x)) : 0 <x leq 1 } cup { (0,y) : -1leq y leq 1}$ and $Y = [0,1)$. Are $X$ and $X...
$begingroup$
Consider $X = { (x, sin(1/x)) : 0 <x leq 1 } cup { (0,y) : -1leq y leq 1}$ as a subspace of $mathbb{R}^2$ and $Y = [0,1)$ as a subspace of $mathbb{R}$. Then which of these options are correct?
$X$ is connected.
$X$ is compact.
$X times Y$ (in product topology) is connected.
$X times Y$ (in product topology) is compact.
My answer is : option $1)$ and option $3)$ are true: by the graph I can show it's connected.
Option $2)$ and option $4)$ are false because the graph is unbounded so it is not compact.
Is this correct? Any hints/solution will be appreciated.
Thank you.
general-topology
edited Dec 14 '18 at 11:02
Brahadeesh
6,46942363
asked Dec 14 '18 at 10:29
jasminejasmine
1,809418
$endgroup$
add a comment |
$begingroup$
Consider $X = { (x, sin(1/x)) : 0 <x leq 1 } cup { (0,y) : -1leq y leq 1}$ as a subspace of $mathbb{R}^2$ and $Y = [0,1)$ as a subspace of $mathbb{R}$. Then which of these options are correct?
$X$ is connected.
$X$ is compact.
$X times Y$ (in product topology) is connected.
$X times Y$ (in product topology) is compact.
My answer is : option $1)$ and option $3)$ are true: by the graph I can show it's connected.
Option $2)$ and option $4)$ are false because the graph is unbounded so it is not compact.
Is this correct? Any hints/solution will be appreciated.
Thank you.
general-topology
edited Dec 14 '18 at 11:02
Brahadeesh
6,46942363
asked Dec 14 '18 at 10:29
jasminejasmine
1,809418
$endgroup$
$begingroup$
The space $X$ is known as the (closed) topologist's sine curve. There are quite a number of posts in this forum.
$endgroup$
– Paul Frost
Dec 14 '18 at 11:59
$begingroup$
Personally, I don't know what graph you're talking about, so I'd be skeptical.
$endgroup$
– Saucy O'Path
Dec 15 '18 at 5:45
add a comment |
$begingroup$
Consider $X = { (x, sin(1/x)) : 0 <x leq 1 } cup { (0,y) : -1leq y leq 1}$ as a subspace of $mathbb{R}^2$ and $Y = [0,1)$ as a subspace of $mathbb{R}$. Then which of these options are correct?
$X$ is connected.
$X$ is compact.
$X times Y$ (in product topology) is connected.
$X times Y$ (in product topology) is compact.
My answer is : option $1)$ and option $3)$ are true: by the graph I can show it's connected.
Option $2)$ and option $4)$ are false because the graph is unbounded so it is not compact.
Is this correct? Any hints/solution will be appreciated.
Thank you.
general-topology
edited Dec 14 '18 at 11:02
Brahadeesh
6,46942363
asked Dec 14 '18 at 10:29
jasminejasmine
1,809418
$endgroup$
Consider $X = { (x, sin(1/x)) : 0 <x leq 1 } cup { (0,y) : -1leq y leq 1}$ as a subspace of $mathbb{R}^2$ and $Y = [0,1)$ as a subspace of $mathbb{R}$. Then which of these options are correct?
$X$ is connected.
$X$ is compact.
$X times Y$ (in product topology) is connected.
$X times Y$ (in product topology) is compact.
My answer is : option $1)$ and option $3)$ are true: by the graph I can show it's connected.
Option $2)$ and option $4)$ are false because the graph is unbounded so it is not compact.
Is this correct? Any hints/solution will be appreciated.
Thank you.
general-topology
general-topology
edited Dec 14 '18 at 11:02
Brahadeesh
6,46942363
asked Dec 14 '18 at 10:29
jasminejasmine
1,809418
edited Dec 14 '18 at 11:02
Brahadeesh
6,46942363
asked Dec 14 '18 at 10:29
jasminejasmine
1,809418
edited Dec 14 '18 at 11:02
Brahadeesh
6,46942363
edited Dec 14 '18 at 11:02
Brahadeesh
6,46942363
edited Dec 14 '18 at 11:02
Brahadeesh
6,46942363
6,46942363
asked Dec 14 '18 at 10:29
jasminejasmine
1,809418
asked Dec 14 '18 at 10:29
jasminejasmine
1,809418
asked Dec 14 '18 at 10:29
jasminejasmine
1,809418
1,809418
$begingroup$
The space $X$ is known as the (closed) topologist's sine curve. There are quite a number of posts in this forum.
$endgroup$
– Paul Frost
Dec 14 '18 at 11:59
$begingroup$
Personally, I don't know what graph you're talking about, so I'd be skeptical.
$endgroup$
– Saucy O'Path
Dec 15 '18 at 5:45
add a comment |
$begingroup$
The space $X$ is known as the (closed) topologist's sine curve. There are quite a number of posts in this forum.
$endgroup$
– Paul Frost
Dec 14 '18 at 11:59
$begingroup$
Personally, I don't know what graph you're talking about, so I'd be skeptical.
$endgroup$
– Saucy O'Path
Dec 15 '18 at 5:45
$begingroup$
The space $X$ is known as the (closed) topologist's sine curve. There are quite a number of posts in this forum.
$endgroup$
– Paul Frost
Dec 14 '18 at 11:59
$begingroup$
The space $X$ is known as the (closed) topologist's sine curve. There are quite a number of posts in this forum.
$endgroup$
– Paul Frost
Dec 14 '18 at 11:59
$begingroup$
Personally, I don't know what graph you're talking about, so I'd be skeptical.
$endgroup$
– Saucy O'Path
Dec 15 '18 at 5:45
$begingroup$
Personally, I don't know what graph you're talking about, so I'd be skeptical.
$endgroup$
– Saucy O'Path
Dec 15 '18 at 5:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No! $X$ is compact, The minimum bounded is $1$. For closedness, the second set is the set of limit points of the first one.
For connectedness of $X$, note that the first set, call $A$, in $X$ is the continous image of $(0,1]$ under $x mapsto sin 1/x$ . Now $(0,1]$ is connected , so is its continuous image $A$. Also $A$ is connected implies $overline{A}$ is conneceted. But $overline{A}=X$ and so $X$ is connected !
The latter two is easy to determine and it is left as an exercise to you :-)
answered Dec 14 '18 at 10:32
Chinnapparaj RChinnapparaj R
5,5182928
$endgroup$
$begingroup$
thanks u @Chinnaapparaj,,,got its now in detail
$endgroup$
– jasmine
Dec 14 '18 at 11:07
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Dec 14 '18 at 14:09
add a comment |
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1 Answer
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1 Answer
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$begingroup$
No! $X$ is compact, The minimum bounded is $1$. For closedness, the second set is the set of limit points of the first one.
For connectedness of $X$, note that the first set, call $A$, in $X$ is the continous image of $(0,1]$ under $x mapsto sin 1/x$ . Now $(0,1]$ is connected , so is its continuous image $A$. Also $A$ is connected implies $overline{A}$ is conneceted. But $overline{A}=X$ and so $X$ is connected !
The latter two is easy to determine and it is left as an exercise to you :-)
answered Dec 14 '18 at 10:32
Chinnapparaj RChinnapparaj R
5,5182928
$endgroup$
$begingroup$
thanks u @Chinnaapparaj,,,got its now in detail
$endgroup$
– jasmine
Dec 14 '18 at 11:07
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Dec 14 '18 at 14:09
add a comment |
$begingroup$
No! $X$ is compact, The minimum bounded is $1$. For closedness, the second set is the set of limit points of the first one.
For connectedness of $X$, note that the first set, call $A$, in $X$ is the continous image of $(0,1]$ under $x mapsto sin 1/x$ . Now $(0,1]$ is connected , so is its continuous image $A$. Also $A$ is connected implies $overline{A}$ is conneceted. But $overline{A}=X$ and so $X$ is connected !
The latter two is easy to determine and it is left as an exercise to you :-)
answered Dec 14 '18 at 10:32
Chinnapparaj RChinnapparaj R
5,5182928
$endgroup$
$begingroup$
thanks u @Chinnaapparaj,,,got its now in detail
$endgroup$
– jasmine
Dec 14 '18 at 11:07
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Dec 14 '18 at 14:09
add a comment |
$begingroup$
No! $X$ is compact, The minimum bounded is $1$. For closedness, the second set is the set of limit points of the first one.
For connectedness of $X$, note that the first set, call $A$, in $X$ is the continous image of $(0,1]$ under $x mapsto sin 1/x$ . Now $(0,1]$ is connected , so is its continuous image $A$. Also $A$ is connected implies $overline{A}$ is conneceted. But $overline{A}=X$ and so $X$ is connected !
The latter two is easy to determine and it is left as an exercise to you :-)
answered Dec 14 '18 at 10:32
Chinnapparaj RChinnapparaj R
5,5182928
$endgroup$
No! $X$ is compact, The minimum bounded is $1$. For closedness, the second set is the set of limit points of the first one.
For connectedness of $X$, note that the first set, call $A$, in $X$ is the continous image of $(0,1]$ under $x mapsto sin 1/x$ . Now $(0,1]$ is connected , so is its continuous image $A$. Also $A$ is connected implies $overline{A}$ is conneceted. But $overline{A}=X$ and so $X$ is connected !
The latter two is easy to determine and it is left as an exercise to you :-)
answered Dec 14 '18 at 10:32
Chinnapparaj RChinnapparaj R
5,5182928
edited Dec 14 '18 at 10:58
answered Dec 14 '18 at 10:32
Chinnapparaj RChinnapparaj R
5,5182928
answered Dec 14 '18 at 10:32
Chinnapparaj RChinnapparaj R
5,5182928
answered Dec 14 '18 at 10:32
Chinnapparaj RChinnapparaj R
5,5182928
5,5182928
$begingroup$
thanks u @Chinnaapparaj,,,got its now in detail
$endgroup$
– jasmine
Dec 14 '18 at 11:07
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Dec 14 '18 at 14:09
add a comment |
$begingroup$
thanks u @Chinnaapparaj,,,got its now in detail
$endgroup$
– jasmine
Dec 14 '18 at 11:07
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Dec 14 '18 at 14:09
$begingroup$
thanks u @Chinnaapparaj,,,got its now in detail
$endgroup$
– jasmine
Dec 14 '18 at 11:07
$begingroup$
thanks u @Chinnaapparaj,,,got its now in detail
$endgroup$
– jasmine
Dec 14 '18 at 11:07
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Dec 14 '18 at 14:09
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Dec 14 '18 at 14:09
add a comment |
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$begingroup$
The space $X$ is known as the (closed) topologist's sine curve. There are quite a number of posts in this forum.
$endgroup$
– Paul Frost
Dec 14 '18 at 11:59
$begingroup$
Personally, I don't know what graph you're talking about, so I'd be skeptical.
$endgroup$
– Saucy O'Path
Dec 15 '18 at 5:45