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Why isn't the quotient space $X=mathbbR^2/(mathbbR times 0)$ first-countable?
The Next CEO of Stack OverflowSequential compactness vs. countable compactnessThe Line with Uncountably Many Origins: Second Countable?Why isn't the half-disk topology separable?Do spaces satisfying the first axiom of countability have monotone decreasing bases for every point?Hausdorff Non-First-Countable Quotient Spaces of Ordered SquareA quotient space that does not satisfy the first countability axiomSecond Countable, First Countable, and Separable SpacesWhat is a first countable, limit compact space that is not sequentially compact?A first countable, countably compact space is sequentially compactA first countable hemicompact space is locally compact
$begingroup$
Consider the quotient space $X=mathbbR^2/(mathbbR times 0)$ I'm supposed to prove it isn't first countable. It seems to me that this space is homeomorphic to two open triangles joined at their apexes with that point added. That point is the problematic one, but why isn't the collection of open balls with descending rational radii centered at this point intersected with $X$ a countable basis?
Thank you for your answers.
Edit: I found this problem on my university's website. We did this same problem at school but first-countability was replaced with local compactness. It could be possible that the space is in fact first-countable, and the problem statement on the website was incorrect.
general-topology
asked yesterday
InsertNameHereInsertNameHere
773
$endgroup$
|
show 4 more comments
$begingroup$
Consider the quotient space $X=mathbbR^2/(mathbbR times 0)$ I'm supposed to prove it isn't first countable. It seems to me that this space is homeomorphic to two open triangles joined at their apexes with that point added. That point is the problematic one, but why isn't the collection of open balls with descending rational radii centered at this point intersected with $X$ a countable basis?
Thank you for your answers.
Edit: I found this problem on my university's website. We did this same problem at school but first-countability was replaced with local compactness. It could be possible that the space is in fact first-countable, and the problem statement on the website was incorrect.
general-topology
asked yesterday
InsertNameHereInsertNameHere
773
$endgroup$
1
$begingroup$
Consider the neighbourhood $(x, y)in Bbb R^2mid -e^-x^2<y<e^-x^2$ of the problematic point. That's not contained in any of your rational-radius neighbourhoods. So that's a counter to your one example. Haven't figured out a general proof, though.
$endgroup$
– Arthur
yesterday
$begingroup$
Maybe we can use diagonalization? Observe your statement is equivalent to: There doesn't exist a countable collection of open sets $U_n$ such that for every open set $U$ containing $mathbb Rtimes 0$, $mathbb Rtimes 0subset U_nsubset U$ for some $n$.
$endgroup$
– YuiTo Cheng
yesterday
$begingroup$
@Arthur, I thought it was the other way around, i. e. that any open set containing the point has to contain a basis element?
$endgroup$
– InsertNameHere
yesterday
1
$begingroup$
@InsertNameHere You're right. I didn't keep my phrasing straight. Still, my point stands: that set is a witness that your proposed basis isn't actually a basis.
$endgroup$
– Arthur
yesterday
$begingroup$
What is the equivalence relation being used to get the quotient space?
$endgroup$
– Nemo
yesterday
|
show 4 more comments
$begingroup$
Consider the quotient space $X=mathbbR^2/(mathbbR times 0)$ I'm supposed to prove it isn't first countable. It seems to me that this space is homeomorphic to two open triangles joined at their apexes with that point added. That point is the problematic one, but why isn't the collection of open balls with descending rational radii centered at this point intersected with $X$ a countable basis?
Thank you for your answers.
Edit: I found this problem on my university's website. We did this same problem at school but first-countability was replaced with local compactness. It could be possible that the space is in fact first-countable, and the problem statement on the website was incorrect.
general-topology
asked yesterday
InsertNameHereInsertNameHere
773
$endgroup$
Consider the quotient space $X=mathbbR^2/(mathbbR times 0)$ I'm supposed to prove it isn't first countable. It seems to me that this space is homeomorphic to two open triangles joined at their apexes with that point added. That point is the problematic one, but why isn't the collection of open balls with descending rational radii centered at this point intersected with $X$ a countable basis?
Thank you for your answers.
Edit: I found this problem on my university's website. We did this same problem at school but first-countability was replaced with local compactness. It could be possible that the space is in fact first-countable, and the problem statement on the website was incorrect.
general-topology
general-topology
asked yesterday
InsertNameHereInsertNameHere
773
asked yesterday
InsertNameHereInsertNameHere
773
edited yesterday
InsertNameHere
asked yesterday
InsertNameHereInsertNameHere
773
asked yesterday
InsertNameHereInsertNameHere
773
asked yesterday
InsertNameHereInsertNameHere
773
773
1
$begingroup$
Consider the neighbourhood $(x, y)in Bbb R^2mid -e^-x^2<y<e^-x^2$ of the problematic point. That's not contained in any of your rational-radius neighbourhoods. So that's a counter to your one example. Haven't figured out a general proof, though.
$endgroup$
– Arthur
yesterday
$begingroup$
Maybe we can use diagonalization? Observe your statement is equivalent to: There doesn't exist a countable collection of open sets $U_n$ such that for every open set $U$ containing $mathbb Rtimes 0$, $mathbb Rtimes 0subset U_nsubset U$ for some $n$.
$endgroup$
– YuiTo Cheng
yesterday
$begingroup$
@Arthur, I thought it was the other way around, i. e. that any open set containing the point has to contain a basis element?
$endgroup$
– InsertNameHere
yesterday
1
$begingroup$
@InsertNameHere You're right. I didn't keep my phrasing straight. Still, my point stands: that set is a witness that your proposed basis isn't actually a basis.
$endgroup$
– Arthur
yesterday
$begingroup$
What is the equivalence relation being used to get the quotient space?
$endgroup$
– Nemo
yesterday
|
show 4 more comments
1
$begingroup$
Consider the neighbourhood $(x, y)in Bbb R^2mid -e^-x^2<y<e^-x^2$ of the problematic point. That's not contained in any of your rational-radius neighbourhoods. So that's a counter to your one example. Haven't figured out a general proof, though.
$endgroup$
– Arthur
yesterday
$begingroup$
Maybe we can use diagonalization? Observe your statement is equivalent to: There doesn't exist a countable collection of open sets $U_n$ such that for every open set $U$ containing $mathbb Rtimes 0$, $mathbb Rtimes 0subset U_nsubset U$ for some $n$.
$endgroup$
– YuiTo Cheng
yesterday
$begingroup$
@Arthur, I thought it was the other way around, i. e. that any open set containing the point has to contain a basis element?
$endgroup$
– InsertNameHere
yesterday
1
$begingroup$
@InsertNameHere You're right. I didn't keep my phrasing straight. Still, my point stands: that set is a witness that your proposed basis isn't actually a basis.
$endgroup$
– Arthur
yesterday
$begingroup$
What is the equivalence relation being used to get the quotient space?
$endgroup$
– Nemo
yesterday
1
1
$begingroup$
Consider the neighbourhood $(x, y)in Bbb R^2mid -e^-x^2<y<e^-x^2$ of the problematic point. That's not contained in any of your rational-radius neighbourhoods. So that's a counter to your one example. Haven't figured out a general proof, though.
$endgroup$
– Arthur
yesterday
$begingroup$
Consider the neighbourhood $(x, y)in Bbb R^2mid -e^-x^2<y<e^-x^2$ of the problematic point. That's not contained in any of your rational-radius neighbourhoods. So that's a counter to your one example. Haven't figured out a general proof, though.
$endgroup$
– Arthur
yesterday
$begingroup$
Maybe we can use diagonalization? Observe your statement is equivalent to: There doesn't exist a countable collection of open sets $U_n$ such that for every open set $U$ containing $mathbb Rtimes 0$, $mathbb Rtimes 0subset U_nsubset U$ for some $n$.
$endgroup$
– YuiTo Cheng
yesterday
$begingroup$
Maybe we can use diagonalization? Observe your statement is equivalent to: There doesn't exist a countable collection of open sets $U_n$ such that for every open set $U$ containing $mathbb Rtimes 0$, $mathbb Rtimes 0subset U_nsubset U$ for some $n$.
$endgroup$
– YuiTo Cheng
yesterday
$begingroup$
@Arthur, I thought it was the other way around, i. e. that any open set containing the point has to contain a basis element?
$endgroup$
– InsertNameHere
yesterday
$begingroup$
@Arthur, I thought it was the other way around, i. e. that any open set containing the point has to contain a basis element?
$endgroup$
– InsertNameHere
yesterday
1
1
$begingroup$
@InsertNameHere You're right. I didn't keep my phrasing straight. Still, my point stands: that set is a witness that your proposed basis isn't actually a basis.
$endgroup$
– Arthur
yesterday
$begingroup$
@InsertNameHere You're right. I didn't keep my phrasing straight. Still, my point stands: that set is a witness that your proposed basis isn't actually a basis.
$endgroup$
– Arthur
yesterday
$begingroup$
What is the equivalence relation being used to get the quotient space?
$endgroup$
– Nemo
yesterday
$begingroup$
What is the equivalence relation being used to get the quotient space?
$endgroup$
– Nemo
yesterday
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$mathbbR times 0$ is collapsed to a point $ast in X=mathbbR^2/(mathbbR times 0)$. This point does not have a countable base of open neighborhoods.
$X$ is endowed with the quotient topology. This means that open neigborhoods of $ast$ are in a $1$-$1$-correspondence with open subsets of $mathbbR^2$ containing $mathbbR times 0$. Let us denote the collection of these sets by $mathfrakU$.
Therefore, as YuiTo Cheng remarked in his comments, we have to show that there does not exist a countable collection of sets $U_ n in mathfrakU$ such that for every $U in mathfrakU$, $U_n subset U$ for some $n$.
Let us consider any countable collection of $U_ n in mathfrakU$. For each $n in mathbbN$ there exists a number $u_n > 0$ such that $(n,u_n) in U_n$. The set $D = (n,u_n) mid n in mathbbN $ is closed in $mathbbR^2$, thus $U = mathbbR^2 setminus D in mathfrakU$. But by construction no $U_n$ can be contained in $U$.
Remark:
This generalizes as follows. Let $X,Y$ be topological spaces such that $X$ contains an infinite subset $A subset X$ without limit points and let $y in Y$ be a point such that $ y $ is not open in $Y$. Then $(X times Y)/(X times y )$ is not first countable.
answered yesterday
Paul FrostPaul Frost
11.7k3935
$endgroup$
add a comment |
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$begingroup$
$mathbbR times 0$ is collapsed to a point $ast in X=mathbbR^2/(mathbbR times 0)$. This point does not have a countable base of open neighborhoods.
$X$ is endowed with the quotient topology. This means that open neigborhoods of $ast$ are in a $1$-$1$-correspondence with open subsets of $mathbbR^2$ containing $mathbbR times 0$. Let us denote the collection of these sets by $mathfrakU$.
Therefore, as YuiTo Cheng remarked in his comments, we have to show that there does not exist a countable collection of sets $U_ n in mathfrakU$ such that for every $U in mathfrakU$, $U_n subset U$ for some $n$.
Let us consider any countable collection of $U_ n in mathfrakU$. For each $n in mathbbN$ there exists a number $u_n > 0$ such that $(n,u_n) in U_n$. The set $D = (n,u_n) mid n in mathbbN $ is closed in $mathbbR^2$, thus $U = mathbbR^2 setminus D in mathfrakU$. But by construction no $U_n$ can be contained in $U$.
Remark:
This generalizes as follows. Let $X,Y$ be topological spaces such that $X$ contains an infinite subset $A subset X$ without limit points and let $y in Y$ be a point such that $ y $ is not open in $Y$. Then $(X times Y)/(X times y )$ is not first countable.
answered yesterday
Paul FrostPaul Frost
11.7k3935
$endgroup$
add a comment |
$begingroup$
$mathbbR times 0$ is collapsed to a point $ast in X=mathbbR^2/(mathbbR times 0)$. This point does not have a countable base of open neighborhoods.
$X$ is endowed with the quotient topology. This means that open neigborhoods of $ast$ are in a $1$-$1$-correspondence with open subsets of $mathbbR^2$ containing $mathbbR times 0$. Let us denote the collection of these sets by $mathfrakU$.
Therefore, as YuiTo Cheng remarked in his comments, we have to show that there does not exist a countable collection of sets $U_ n in mathfrakU$ such that for every $U in mathfrakU$, $U_n subset U$ for some $n$.
Let us consider any countable collection of $U_ n in mathfrakU$. For each $n in mathbbN$ there exists a number $u_n > 0$ such that $(n,u_n) in U_n$. The set $D = (n,u_n) mid n in mathbbN $ is closed in $mathbbR^2$, thus $U = mathbbR^2 setminus D in mathfrakU$. But by construction no $U_n$ can be contained in $U$.
Remark:
This generalizes as follows. Let $X,Y$ be topological spaces such that $X$ contains an infinite subset $A subset X$ without limit points and let $y in Y$ be a point such that $ y $ is not open in $Y$. Then $(X times Y)/(X times y )$ is not first countable.
answered yesterday
Paul FrostPaul Frost
11.7k3935
$endgroup$
add a comment |
$begingroup$
$mathbbR times 0$ is collapsed to a point $ast in X=mathbbR^2/(mathbbR times 0)$. This point does not have a countable base of open neighborhoods.
$X$ is endowed with the quotient topology. This means that open neigborhoods of $ast$ are in a $1$-$1$-correspondence with open subsets of $mathbbR^2$ containing $mathbbR times 0$. Let us denote the collection of these sets by $mathfrakU$.
Therefore, as YuiTo Cheng remarked in his comments, we have to show that there does not exist a countable collection of sets $U_ n in mathfrakU$ such that for every $U in mathfrakU$, $U_n subset U$ for some $n$.
Let us consider any countable collection of $U_ n in mathfrakU$. For each $n in mathbbN$ there exists a number $u_n > 0$ such that $(n,u_n) in U_n$. The set $D = (n,u_n) mid n in mathbbN $ is closed in $mathbbR^2$, thus $U = mathbbR^2 setminus D in mathfrakU$. But by construction no $U_n$ can be contained in $U$.
Remark:
This generalizes as follows. Let $X,Y$ be topological spaces such that $X$ contains an infinite subset $A subset X$ without limit points and let $y in Y$ be a point such that $ y $ is not open in $Y$. Then $(X times Y)/(X times y )$ is not first countable.
answered yesterday
Paul FrostPaul Frost
11.7k3935
$endgroup$
$mathbbR times 0$ is collapsed to a point $ast in X=mathbbR^2/(mathbbR times 0)$. This point does not have a countable base of open neighborhoods.
$X$ is endowed with the quotient topology. This means that open neigborhoods of $ast$ are in a $1$-$1$-correspondence with open subsets of $mathbbR^2$ containing $mathbbR times 0$. Let us denote the collection of these sets by $mathfrakU$.
Therefore, as YuiTo Cheng remarked in his comments, we have to show that there does not exist a countable collection of sets $U_ n in mathfrakU$ such that for every $U in mathfrakU$, $U_n subset U$ for some $n$.
Let us consider any countable collection of $U_ n in mathfrakU$. For each $n in mathbbN$ there exists a number $u_n > 0$ such that $(n,u_n) in U_n$. The set $D = (n,u_n) mid n in mathbbN $ is closed in $mathbbR^2$, thus $U = mathbbR^2 setminus D in mathfrakU$. But by construction no $U_n$ can be contained in $U$.
Remark:
This generalizes as follows. Let $X,Y$ be topological spaces such that $X$ contains an infinite subset $A subset X$ without limit points and let $y in Y$ be a point such that $ y $ is not open in $Y$. Then $(X times Y)/(X times y )$ is not first countable.
answered yesterday
Paul FrostPaul Frost
11.7k3935
edited yesterday
answered yesterday
Paul FrostPaul Frost
11.7k3935
answered yesterday
Paul FrostPaul Frost
11.7k3935
answered yesterday
Paul FrostPaul Frost
11.7k3935
11.7k3935
add a comment |
add a comment |
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$begingroup$
Consider the neighbourhood $(x, y)in Bbb R^2mid -e^-x^2<y<e^-x^2$ of the problematic point. That's not contained in any of your rational-radius neighbourhoods. So that's a counter to your one example. Haven't figured out a general proof, though.
$endgroup$
– Arthur
yesterday
$begingroup$
Maybe we can use diagonalization? Observe your statement is equivalent to: There doesn't exist a countable collection of open sets $U_n$ such that for every open set $U$ containing $mathbb Rtimes 0$, $mathbb Rtimes 0subset U_nsubset U$ for some $n$.
$endgroup$
– YuiTo Cheng
yesterday
$begingroup$
@Arthur, I thought it was the other way around, i. e. that any open set containing the point has to contain a basis element?
$endgroup$
– InsertNameHere
yesterday
1
$begingroup$
@InsertNameHere You're right. I didn't keep my phrasing straight. Still, my point stands: that set is a witness that your proposed basis isn't actually a basis.
$endgroup$
– Arthur
yesterday
$begingroup$
What is the equivalence relation being used to get the quotient space?
$endgroup$
– Nemo
yesterday