Problem with homeomorphic subspaces
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Are spaces $ A = left{ frac{1}{n} ,middle|, n in mathbb{N} right} $ and $ B = left{ frac{n+1}{n} ,middle|, n in mathbb{N} right} $ as subspaces of $ ( 0 , + infty)$ homeomorphic? Are $operatorname{Cl}(A)$ and $operatorname{Cl}(B)$ homeomorphic?
I would say that $A$ and $B$ aren't homeomorphic because $A$ is compact and $B$ is not compact (continuous function invariants). But when it comes to $operatorname{Cl}(A)$ and $operatorname{Cl}(B)$ they are both compact so it would seem that they might be homeomorphic, but how do I construct the homeomorphism? The problem is the element $1$ of $B$, I can't "send" any element of A to it.
general-topology
edited Nov 18 at 13:25
José Carlos Santos
143k20112211
asked Nov 18 at 13:17
user15269
1608
add a comment |
up vote
0
down vote
favorite
Are spaces $ A = left{ frac{1}{n} ,middle|, n in mathbb{N} right} $ and $ B = left{ frac{n+1}{n} ,middle|, n in mathbb{N} right} $ as subspaces of $ ( 0 , + infty)$ homeomorphic? Are $operatorname{Cl}(A)$ and $operatorname{Cl}(B)$ homeomorphic?
I would say that $A$ and $B$ aren't homeomorphic because $A$ is compact and $B$ is not compact (continuous function invariants). But when it comes to $operatorname{Cl}(A)$ and $operatorname{Cl}(B)$ they are both compact so it would seem that they might be homeomorphic, but how do I construct the homeomorphism? The problem is the element $1$ of $B$, I can't "send" any element of A to it.
general-topology
edited Nov 18 at 13:25
José Carlos Santos
143k20112211
asked Nov 18 at 13:17
user15269
1608
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Are spaces $ A = left{ frac{1}{n} ,middle|, n in mathbb{N} right} $ and $ B = left{ frac{n+1}{n} ,middle|, n in mathbb{N} right} $ as subspaces of $ ( 0 , + infty)$ homeomorphic? Are $operatorname{Cl}(A)$ and $operatorname{Cl}(B)$ homeomorphic?
I would say that $A$ and $B$ aren't homeomorphic because $A$ is compact and $B$ is not compact (continuous function invariants). But when it comes to $operatorname{Cl}(A)$ and $operatorname{Cl}(B)$ they are both compact so it would seem that they might be homeomorphic, but how do I construct the homeomorphism? The problem is the element $1$ of $B$, I can't "send" any element of A to it.
general-topology
edited Nov 18 at 13:25
José Carlos Santos
143k20112211
asked Nov 18 at 13:17
user15269
1608
Are spaces $ A = left{ frac{1}{n} ,middle|, n in mathbb{N} right} $ and $ B = left{ frac{n+1}{n} ,middle|, n in mathbb{N} right} $ as subspaces of $ ( 0 , + infty)$ homeomorphic? Are $operatorname{Cl}(A)$ and $operatorname{Cl}(B)$ homeomorphic?
I would say that $A$ and $B$ aren't homeomorphic because $A$ is compact and $B$ is not compact (continuous function invariants). But when it comes to $operatorname{Cl}(A)$ and $operatorname{Cl}(B)$ they are both compact so it would seem that they might be homeomorphic, but how do I construct the homeomorphism? The problem is the element $1$ of $B$, I can't "send" any element of A to it.
general-topology
general-topology
edited Nov 18 at 13:25
José Carlos Santos
143k20112211
asked Nov 18 at 13:17
user15269
1608
edited Nov 18 at 13:25
José Carlos Santos
143k20112211
asked Nov 18 at 13:17
user15269
1608
edited Nov 18 at 13:25
José Carlos Santos
143k20112211
edited Nov 18 at 13:25
José Carlos Santos
143k20112211
edited Nov 18 at 13:25
José Carlos Santos
143k20112211
143k20112211
asked Nov 18 at 13:17
user15269
1608
asked Nov 18 at 13:17
user15269
1608
asked Nov 18 at 13:17
user15269
1608
1608
add a comment |
add a comment |
1 Answer
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Yes, they are homeomorphic. Just consider$$begin{array}{rccc}fcolon&A&longrightarrow&B\&x&mapsto&x+1;end{array}$$it's a homeomorphism.
However, $overline A$ and $overline B$ are not homeomorphic, since $overline A$ isn't compact, whereas $overline B$ is.
answered Nov 18 at 13:22
José Carlos Santos
143k20112211
But as a subspace of $<0, +infty> $ why isn't $ ClA$ compact? Correct me if I'm wrong but isn't A closed in said subspace (because its complement is open in the subspace) and then $ A = ClA $? It's also bounded so it should be compact?
– user15269
Nov 18 at 13:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes, they are homeomorphic. Just consider$$begin{array}{rccc}fcolon&A&longrightarrow&B\&x&mapsto&x+1;end{array}$$it's a homeomorphism.
However, $overline A$ and $overline B$ are not homeomorphic, since $overline A$ isn't compact, whereas $overline B$ is.
answered Nov 18 at 13:22
José Carlos Santos
143k20112211
But as a subspace of $<0, +infty> $ why isn't $ ClA$ compact? Correct me if I'm wrong but isn't A closed in said subspace (because its complement is open in the subspace) and then $ A = ClA $? It's also bounded so it should be compact?
– user15269
Nov 18 at 13:35
add a comment |
up vote
0
down vote
Yes, they are homeomorphic. Just consider$$begin{array}{rccc}fcolon&A&longrightarrow&B\&x&mapsto&x+1;end{array}$$it's a homeomorphism.
However, $overline A$ and $overline B$ are not homeomorphic, since $overline A$ isn't compact, whereas $overline B$ is.
answered Nov 18 at 13:22
José Carlos Santos
143k20112211
But as a subspace of $<0, +infty> $ why isn't $ ClA$ compact? Correct me if I'm wrong but isn't A closed in said subspace (because its complement is open in the subspace) and then $ A = ClA $? It's also bounded so it should be compact?
– user15269
Nov 18 at 13:35
add a comment |
up vote
0
down vote
up vote
0
down vote
Yes, they are homeomorphic. Just consider$$begin{array}{rccc}fcolon&A&longrightarrow&B\&x&mapsto&x+1;end{array}$$it's a homeomorphism.
However, $overline A$ and $overline B$ are not homeomorphic, since $overline A$ isn't compact, whereas $overline B$ is.
answered Nov 18 at 13:22
José Carlos Santos
143k20112211
Yes, they are homeomorphic. Just consider$$begin{array}{rccc}fcolon&A&longrightarrow&B\&x&mapsto&x+1;end{array}$$it's a homeomorphism.
However, $overline A$ and $overline B$ are not homeomorphic, since $overline A$ isn't compact, whereas $overline B$ is.
answered Nov 18 at 13:22
José Carlos Santos
143k20112211
answered Nov 18 at 13:22
José Carlos Santos
143k20112211
answered Nov 18 at 13:22
José Carlos Santos
143k20112211
answered Nov 18 at 13:22
José Carlos Santos
143k20112211
143k20112211
But as a subspace of $<0, +infty> $ why isn't $ ClA$ compact? Correct me if I'm wrong but isn't A closed in said subspace (because its complement is open in the subspace) and then $ A = ClA $? It's also bounded so it should be compact?
– user15269
Nov 18 at 13:35
add a comment |
But as a subspace of $<0, +infty> $ why isn't $ ClA$ compact? Correct me if I'm wrong but isn't A closed in said subspace (because its complement is open in the subspace) and then $ A = ClA $? It's also bounded so it should be compact?
– user15269
Nov 18 at 13:35
But as a subspace of $<0, +infty> $ why isn't $ ClA$ compact? Correct me if I'm wrong but isn't A closed in said subspace (because its complement is open in the subspace) and then $ A = ClA $? It's also bounded so it should be compact?
– user15269
Nov 18 at 13:35
But as a subspace of $<0, +infty> $ why isn't $ ClA$ compact? Correct me if I'm wrong but isn't A closed in said subspace (because its complement is open in the subspace) and then $ A = ClA $? It's also bounded so it should be compact?
– user15269
Nov 18 at 13:35
add a comment |
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