Is my understanding of limit point compactness correct with respect to $[0,1]^{omega}$ with the uniform...
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The following is an exercise problem about limit point compactness from the book "Topology" by Munkres (2nd edition).
Exercise 1 in Section 28: Give $[0,1]^{omega}$ the uniform topology. Find an infinite subset of this space that has no limit point.
Here is a solution:
Let $d$ denote the uniform metric. Choose $c in (0,1]$. Let $A = { 0, c }^{omega} subset [0,1]^{omega}$. Note that if $a$ and $b$ are distinct points in $A$ then $d(a,b) = c$. For any $x in X$ the open ball $B_d(x, c/3)$ has diameter less than or equal $2c/3$, hence $B_d(x, c/3)$ cannot contain more than one point of $A$. It follows that $x$ is not a limit point of $A$.
I am confused about the following points:
- The open ball $B_d(x, c/3)$ has diameter less than or equal $2c/3$.` Why is equal here possible?
- Why not consider the points $x in A$ and $x in X land x notin A$ separately? After all, for any point $x in X land x notin A$, we have to show that there exists a neighborhood around $x$ which does not intersect $A$ (instead of just showing that the ball $B_d(x,c/3)$ cannot contain more than one point of $A$).
Is my understanding correct?
general-topology proof-verification compactness
edited Dec 21 '18 at 14:36
Saaqib Mahmood
7,81842480
asked Feb 18 '14 at 14:49
hengxinhengxin
1,5881428
$endgroup$
add a comment |
$begingroup$
The following is an exercise problem about limit point compactness from the book "Topology" by Munkres (2nd edition).
Exercise 1 in Section 28: Give $[0,1]^{omega}$ the uniform topology. Find an infinite subset of this space that has no limit point.
Here is a solution:
Let $d$ denote the uniform metric. Choose $c in (0,1]$. Let $A = { 0, c }^{omega} subset [0,1]^{omega}$. Note that if $a$ and $b$ are distinct points in $A$ then $d(a,b) = c$. For any $x in X$ the open ball $B_d(x, c/3)$ has diameter less than or equal $2c/3$, hence $B_d(x, c/3)$ cannot contain more than one point of $A$. It follows that $x$ is not a limit point of $A$.
I am confused about the following points:
- The open ball $B_d(x, c/3)$ has diameter less than or equal $2c/3$.` Why is equal here possible?
- Why not consider the points $x in A$ and $x in X land x notin A$ separately? After all, for any point $x in X land x notin A$, we have to show that there exists a neighborhood around $x$ which does not intersect $A$ (instead of just showing that the ball $B_d(x,c/3)$ cannot contain more than one point of $A$).
Is my understanding correct?
general-topology proof-verification compactness
edited Dec 21 '18 at 14:36
Saaqib Mahmood
7,81842480
asked Feb 18 '14 at 14:49
hengxinhengxin
1,5881428
$endgroup$
add a comment |
$begingroup$
The following is an exercise problem about limit point compactness from the book "Topology" by Munkres (2nd edition).
Exercise 1 in Section 28: Give $[0,1]^{omega}$ the uniform topology. Find an infinite subset of this space that has no limit point.
Here is a solution:
Let $d$ denote the uniform metric. Choose $c in (0,1]$. Let $A = { 0, c }^{omega} subset [0,1]^{omega}$. Note that if $a$ and $b$ are distinct points in $A$ then $d(a,b) = c$. For any $x in X$ the open ball $B_d(x, c/3)$ has diameter less than or equal $2c/3$, hence $B_d(x, c/3)$ cannot contain more than one point of $A$. It follows that $x$ is not a limit point of $A$.
I am confused about the following points:
- The open ball $B_d(x, c/3)$ has diameter less than or equal $2c/3$.` Why is equal here possible?
- Why not consider the points $x in A$ and $x in X land x notin A$ separately? After all, for any point $x in X land x notin A$, we have to show that there exists a neighborhood around $x$ which does not intersect $A$ (instead of just showing that the ball $B_d(x,c/3)$ cannot contain more than one point of $A$).
Is my understanding correct?
general-topology proof-verification compactness
edited Dec 21 '18 at 14:36
Saaqib Mahmood
7,81842480
asked Feb 18 '14 at 14:49
hengxinhengxin
1,5881428
$endgroup$
The following is an exercise problem about limit point compactness from the book "Topology" by Munkres (2nd edition).
Exercise 1 in Section 28: Give $[0,1]^{omega}$ the uniform topology. Find an infinite subset of this space that has no limit point.
Here is a solution:
Let $d$ denote the uniform metric. Choose $c in (0,1]$. Let $A = { 0, c }^{omega} subset [0,1]^{omega}$. Note that if $a$ and $b$ are distinct points in $A$ then $d(a,b) = c$. For any $x in X$ the open ball $B_d(x, c/3)$ has diameter less than or equal $2c/3$, hence $B_d(x, c/3)$ cannot contain more than one point of $A$. It follows that $x$ is not a limit point of $A$.
I am confused about the following points:
- The open ball $B_d(x, c/3)$ has diameter less than or equal $2c/3$.` Why is equal here possible?
- Why not consider the points $x in A$ and $x in X land x notin A$ separately? After all, for any point $x in X land x notin A$, we have to show that there exists a neighborhood around $x$ which does not intersect $A$ (instead of just showing that the ball $B_d(x,c/3)$ cannot contain more than one point of $A$).
Is my understanding correct?
general-topology proof-verification compactness
general-topology proof-verification compactness
edited Dec 21 '18 at 14:36
Saaqib Mahmood
7,81842480
asked Feb 18 '14 at 14:49
hengxinhengxin
1,5881428
edited Dec 21 '18 at 14:36
Saaqib Mahmood
7,81842480
asked Feb 18 '14 at 14:49
hengxinhengxin
1,5881428
edited Dec 21 '18 at 14:36
Saaqib Mahmood
7,81842480
edited Dec 21 '18 at 14:36
Saaqib Mahmood
7,81842480
edited Dec 21 '18 at 14:36
Saaqib Mahmood
7,81842480
7,81842480
asked Feb 18 '14 at 14:49
hengxinhengxin
1,5881428
asked Feb 18 '14 at 14:49
hengxinhengxin
1,5881428
asked Feb 18 '14 at 14:49
hengxinhengxin
1,5881428
1,5881428
add a comment |
add a comment |
1 Answer
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- An open ball $B_epsilon(x)$ has the diameter $sup{d(y,y')mid d(y,x)<ϵ,d(y',x)<ϵ}$. Since any such pair $y,y'$ has a distance $<2ϵ$ by the triangle equality, the $sup$ can be $2ϵ$
- Note that $[0,1]$ is Hausdorff, so $[0,1]^omega$ is Hausdorff, too. It follows that if some neighborhood of $xin X$ contains only finitely many points of $Asetminus{x}$, then there is some neighborhood of $x$ containing none of them.
answered Feb 18 '14 at 15:14
Stefan HamckeStefan Hamcke
21.8k42879
$endgroup$
add a comment |
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$begingroup$
- An open ball $B_epsilon(x)$ has the diameter $sup{d(y,y')mid d(y,x)<ϵ,d(y',x)<ϵ}$. Since any such pair $y,y'$ has a distance $<2ϵ$ by the triangle equality, the $sup$ can be $2ϵ$
- Note that $[0,1]$ is Hausdorff, so $[0,1]^omega$ is Hausdorff, too. It follows that if some neighborhood of $xin X$ contains only finitely many points of $Asetminus{x}$, then there is some neighborhood of $x$ containing none of them.
answered Feb 18 '14 at 15:14
Stefan HamckeStefan Hamcke
21.8k42879
$endgroup$
add a comment |
$begingroup$
- An open ball $B_epsilon(x)$ has the diameter $sup{d(y,y')mid d(y,x)<ϵ,d(y',x)<ϵ}$. Since any such pair $y,y'$ has a distance $<2ϵ$ by the triangle equality, the $sup$ can be $2ϵ$
- Note that $[0,1]$ is Hausdorff, so $[0,1]^omega$ is Hausdorff, too. It follows that if some neighborhood of $xin X$ contains only finitely many points of $Asetminus{x}$, then there is some neighborhood of $x$ containing none of them.
answered Feb 18 '14 at 15:14
Stefan HamckeStefan Hamcke
21.8k42879
$endgroup$
add a comment |
$begingroup$
- An open ball $B_epsilon(x)$ has the diameter $sup{d(y,y')mid d(y,x)<ϵ,d(y',x)<ϵ}$. Since any such pair $y,y'$ has a distance $<2ϵ$ by the triangle equality, the $sup$ can be $2ϵ$
- Note that $[0,1]$ is Hausdorff, so $[0,1]^omega$ is Hausdorff, too. It follows that if some neighborhood of $xin X$ contains only finitely many points of $Asetminus{x}$, then there is some neighborhood of $x$ containing none of them.
answered Feb 18 '14 at 15:14
Stefan HamckeStefan Hamcke
21.8k42879
$endgroup$
- An open ball $B_epsilon(x)$ has the diameter $sup{d(y,y')mid d(y,x)<ϵ,d(y',x)<ϵ}$. Since any such pair $y,y'$ has a distance $<2ϵ$ by the triangle equality, the $sup$ can be $2ϵ$
- Note that $[0,1]$ is Hausdorff, so $[0,1]^omega$ is Hausdorff, too. It follows that if some neighborhood of $xin X$ contains only finitely many points of $Asetminus{x}$, then there is some neighborhood of $x$ containing none of them.
answered Feb 18 '14 at 15:14
Stefan HamckeStefan Hamcke
21.8k42879
answered Feb 18 '14 at 15:14
Stefan HamckeStefan Hamcke
21.8k42879
answered Feb 18 '14 at 15:14
Stefan HamckeStefan Hamcke
21.8k42879
answered Feb 18 '14 at 15:14
Stefan HamckeStefan Hamcke
21.8k42879
21.8k42879
add a comment |
add a comment |
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