Union of lines $ { y = x/n : n in mathbb N+ }$ not homeomorphic to infinite wedge sum of lines?
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As is described in the title, I believe $ { y = x/n : n in mathbb N+ }$ is homeomorphic to the infinite wedge sum $bigvee _infty mathbb R $, since the natural bijection is continuous at the crossing point in both direction. But a friend of mine told me it was wrong.
Another related question which appears on Hatcher's text is the union of circles centered $(n,0)$ with radius $n$. Again, it is claimed that it is not homeomorphic to the infinite wedge $bigvee _infty S^1 $, and I can't figure out the reason.
Could anybody explain the two baffling questions please? Thanks!!
general-topology algebraic-topology homotopy-theory
asked Nov 19 at 8:52
Dromeda
303
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up vote
1
down vote
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As is described in the title, I believe $ { y = x/n : n in mathbb N+ }$ is homeomorphic to the infinite wedge sum $bigvee _infty mathbb R $, since the natural bijection is continuous at the crossing point in both direction. But a friend of mine told me it was wrong.
Another related question which appears on Hatcher's text is the union of circles centered $(n,0)$ with radius $n$. Again, it is claimed that it is not homeomorphic to the infinite wedge $bigvee _infty S^1 $, and I can't figure out the reason.
Could anybody explain the two baffling questions please? Thanks!!
general-topology algebraic-topology homotopy-theory
asked Nov 19 at 8:52
Dromeda
303
In short, the neighborhoods of the origin for both subspaces of $mathbb{R}^2$ are "uniform"-they contain a uniformly positive length from every component of the union. This doesn't hold for the wedge sums.
– Kevin Carlson
Nov 19 at 17:42
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
As is described in the title, I believe $ { y = x/n : n in mathbb N+ }$ is homeomorphic to the infinite wedge sum $bigvee _infty mathbb R $, since the natural bijection is continuous at the crossing point in both direction. But a friend of mine told me it was wrong.
Another related question which appears on Hatcher's text is the union of circles centered $(n,0)$ with radius $n$. Again, it is claimed that it is not homeomorphic to the infinite wedge $bigvee _infty S^1 $, and I can't figure out the reason.
Could anybody explain the two baffling questions please? Thanks!!
general-topology algebraic-topology homotopy-theory
asked Nov 19 at 8:52
Dromeda
303
As is described in the title, I believe $ { y = x/n : n in mathbb N+ }$ is homeomorphic to the infinite wedge sum $bigvee _infty mathbb R $, since the natural bijection is continuous at the crossing point in both direction. But a friend of mine told me it was wrong.
Another related question which appears on Hatcher's text is the union of circles centered $(n,0)$ with radius $n$. Again, it is claimed that it is not homeomorphic to the infinite wedge $bigvee _infty S^1 $, and I can't figure out the reason.
Could anybody explain the two baffling questions please? Thanks!!
general-topology algebraic-topology homotopy-theory
general-topology algebraic-topology homotopy-theory
asked Nov 19 at 8:52
Dromeda
303
asked Nov 19 at 8:52
Dromeda
303
asked Nov 19 at 8:52
Dromeda
303
asked Nov 19 at 8:52
Dromeda
303
asked Nov 19 at 8:52
Dromeda
303
303
In short, the neighborhoods of the origin for both subspaces of $mathbb{R}^2$ are "uniform"-they contain a uniformly positive length from every component of the union. This doesn't hold for the wedge sums.
– Kevin Carlson
Nov 19 at 17:42
add a comment |
In short, the neighborhoods of the origin for both subspaces of $mathbb{R}^2$ are "uniform"-they contain a uniformly positive length from every component of the union. This doesn't hold for the wedge sums.
– Kevin Carlson
Nov 19 at 17:42
In short, the neighborhoods of the origin for both subspaces of $mathbb{R}^2$ are "uniform"-they contain a uniformly positive length from every component of the union. This doesn't hold for the wedge sums.
– Kevin Carlson
Nov 19 at 17:42
In short, the neighborhoods of the origin for both subspaces of $mathbb{R}^2$ are "uniform"-they contain a uniformly positive length from every component of the union. This doesn't hold for the wedge sums.
– Kevin Carlson
Nov 19 at 17:42
add a comment |
1 Answer
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In both cases the subspace topology that your union inherits from the plane is metrizable.
The corresponding wedge, a quotient of the union, is not. At the vertex the quotient does not have a countable neighbourhood base: given a countable sequence $langle U_n:ninmathbb{N}rangle$ of neighbourhoods take in the $k$th space a neighbourhood $O_k$ of the point corresponding to the vertex that is a proper subset of $bigcap_{nle k}U_n$. Then the $O_k$ determine a neighbourhood $O$ of the vertex that contains none of the $U_n$.
answered Nov 19 at 9:23
hartkp
1,27965
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
In both cases the subspace topology that your union inherits from the plane is metrizable.
The corresponding wedge, a quotient of the union, is not. At the vertex the quotient does not have a countable neighbourhood base: given a countable sequence $langle U_n:ninmathbb{N}rangle$ of neighbourhoods take in the $k$th space a neighbourhood $O_k$ of the point corresponding to the vertex that is a proper subset of $bigcap_{nle k}U_n$. Then the $O_k$ determine a neighbourhood $O$ of the vertex that contains none of the $U_n$.
answered Nov 19 at 9:23
hartkp
1,27965
add a comment |
up vote
2
down vote
accepted
In both cases the subspace topology that your union inherits from the plane is metrizable.
The corresponding wedge, a quotient of the union, is not. At the vertex the quotient does not have a countable neighbourhood base: given a countable sequence $langle U_n:ninmathbb{N}rangle$ of neighbourhoods take in the $k$th space a neighbourhood $O_k$ of the point corresponding to the vertex that is a proper subset of $bigcap_{nle k}U_n$. Then the $O_k$ determine a neighbourhood $O$ of the vertex that contains none of the $U_n$.
answered Nov 19 at 9:23
hartkp
1,27965
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
In both cases the subspace topology that your union inherits from the plane is metrizable.
The corresponding wedge, a quotient of the union, is not. At the vertex the quotient does not have a countable neighbourhood base: given a countable sequence $langle U_n:ninmathbb{N}rangle$ of neighbourhoods take in the $k$th space a neighbourhood $O_k$ of the point corresponding to the vertex that is a proper subset of $bigcap_{nle k}U_n$. Then the $O_k$ determine a neighbourhood $O$ of the vertex that contains none of the $U_n$.
answered Nov 19 at 9:23
hartkp
1,27965
In both cases the subspace topology that your union inherits from the plane is metrizable.
The corresponding wedge, a quotient of the union, is not. At the vertex the quotient does not have a countable neighbourhood base: given a countable sequence $langle U_n:ninmathbb{N}rangle$ of neighbourhoods take in the $k$th space a neighbourhood $O_k$ of the point corresponding to the vertex that is a proper subset of $bigcap_{nle k}U_n$. Then the $O_k$ determine a neighbourhood $O$ of the vertex that contains none of the $U_n$.
answered Nov 19 at 9:23
hartkp
1,27965
answered Nov 19 at 9:23
hartkp
1,27965
answered Nov 19 at 9:23
hartkp
1,27965
answered Nov 19 at 9:23
hartkp
1,27965
1,27965
add a comment |
add a comment |
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In short, the neighborhoods of the origin for both subspaces of $mathbb{R}^2$ are "uniform"-they contain a uniformly positive length from every component of the union. This doesn't hold for the wedge sums.
– Kevin Carlson
Nov 19 at 17:42