I want to find a topological embedding $f : X rightarrow Y$ and $g: Y rightarrow X$, yet $X$ is not...
$begingroup$
This is related to a problem of showing that none of the intervals $(0,1], (0,1), [0,1]$ are homeomorphic to another, since it's in the same problem block. I've tried using $f(x) = 1 - x , f(x) = 1/(x-1)$ between varying pairs of those intervals to no avail. Do I need a function as complicated as the topologist's sine to show this?
general-topology connectedness
edited Dec 10 '18 at 3:46
user1101010
7751730
asked Sep 24 '13 at 23:15
Hermit with AdjointHermit with Adjoint
9,11552458
$endgroup$
add a comment |
$begingroup$
This is related to a problem of showing that none of the intervals $(0,1], (0,1), [0,1]$ are homeomorphic to another, since it's in the same problem block. I've tried using $f(x) = 1 - x , f(x) = 1/(x-1)$ between varying pairs of those intervals to no avail. Do I need a function as complicated as the topologist's sine to show this?
general-topology connectedness
edited Dec 10 '18 at 3:46
user1101010
7751730
asked Sep 24 '13 at 23:15
Hermit with AdjointHermit with Adjoint
9,11552458
$endgroup$
2
$begingroup$
You can embedd each interval in the other one by $f(x)=frac13+frac13 x$
$endgroup$
– Stefan Hamcke
Sep 24 '13 at 23:19
$begingroup$
Ah, that makes sense now. I guess I kept looking for a surjection as well.
$endgroup$
– Hermit with Adjoint
Sep 24 '13 at 23:23
1
$begingroup$
With your additional condition (that it is also surjection), it seems to be identical to this question
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:02
$begingroup$
This questin is about the other problem you mentioned: Is there exist a homemoorphism between either pair of $(0,1),(0,1],[0,1]$; math.stackexchange.com/questions/240414/…
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:03
add a comment |
$begingroup$
This is related to a problem of showing that none of the intervals $(0,1], (0,1), [0,1]$ are homeomorphic to another, since it's in the same problem block. I've tried using $f(x) = 1 - x , f(x) = 1/(x-1)$ between varying pairs of those intervals to no avail. Do I need a function as complicated as the topologist's sine to show this?
general-topology connectedness
edited Dec 10 '18 at 3:46
user1101010
7751730
asked Sep 24 '13 at 23:15
Hermit with AdjointHermit with Adjoint
9,11552458
$endgroup$
This is related to a problem of showing that none of the intervals $(0,1], (0,1), [0,1]$ are homeomorphic to another, since it's in the same problem block. I've tried using $f(x) = 1 - x , f(x) = 1/(x-1)$ between varying pairs of those intervals to no avail. Do I need a function as complicated as the topologist's sine to show this?
general-topology connectedness
general-topology connectedness
edited Dec 10 '18 at 3:46
user1101010
7751730
asked Sep 24 '13 at 23:15
Hermit with AdjointHermit with Adjoint
9,11552458
edited Dec 10 '18 at 3:46
user1101010
7751730
asked Sep 24 '13 at 23:15
Hermit with AdjointHermit with Adjoint
9,11552458
edited Dec 10 '18 at 3:46
user1101010
7751730
edited Dec 10 '18 at 3:46
user1101010
7751730
edited Dec 10 '18 at 3:46
user1101010
7751730
7751730
asked Sep 24 '13 at 23:15
Hermit with AdjointHermit with Adjoint
9,11552458
asked Sep 24 '13 at 23:15
Hermit with AdjointHermit with Adjoint
9,11552458
asked Sep 24 '13 at 23:15
Hermit with AdjointHermit with Adjoint
9,11552458
9,11552458
2
$begingroup$
You can embedd each interval in the other one by $f(x)=frac13+frac13 x$
$endgroup$
– Stefan Hamcke
Sep 24 '13 at 23:19
$begingroup$
Ah, that makes sense now. I guess I kept looking for a surjection as well.
$endgroup$
– Hermit with Adjoint
Sep 24 '13 at 23:23
1
$begingroup$
With your additional condition (that it is also surjection), it seems to be identical to this question
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:02
$begingroup$
This questin is about the other problem you mentioned: Is there exist a homemoorphism between either pair of $(0,1),(0,1],[0,1]$; math.stackexchange.com/questions/240414/…
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:03
add a comment |
2
$begingroup$
You can embedd each interval in the other one by $f(x)=frac13+frac13 x$
$endgroup$
– Stefan Hamcke
Sep 24 '13 at 23:19
$begingroup$
Ah, that makes sense now. I guess I kept looking for a surjection as well.
$endgroup$
– Hermit with Adjoint
Sep 24 '13 at 23:23
1
$begingroup$
With your additional condition (that it is also surjection), it seems to be identical to this question
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:02
$begingroup$
This questin is about the other problem you mentioned: Is there exist a homemoorphism between either pair of $(0,1),(0,1],[0,1]$; math.stackexchange.com/questions/240414/…
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:03
2
2
$begingroup$
You can embedd each interval in the other one by $f(x)=frac13+frac13 x$
$endgroup$
– Stefan Hamcke
Sep 24 '13 at 23:19
$begingroup$
You can embedd each interval in the other one by $f(x)=frac13+frac13 x$
$endgroup$
– Stefan Hamcke
Sep 24 '13 at 23:19
$begingroup$
Ah, that makes sense now. I guess I kept looking for a surjection as well.
$endgroup$
– Hermit with Adjoint
Sep 24 '13 at 23:23
$begingroup$
Ah, that makes sense now. I guess I kept looking for a surjection as well.
$endgroup$
– Hermit with Adjoint
Sep 24 '13 at 23:23
1
1
$begingroup$
With your additional condition (that it is also surjection), it seems to be identical to this question
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:02
$begingroup$
With your additional condition (that it is also surjection), it seems to be identical to this question
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:02
$begingroup$
This questin is about the other problem you mentioned: Is there exist a homemoorphism between either pair of $(0,1),(0,1],[0,1]$; math.stackexchange.com/questions/240414/…
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:03
$begingroup$
This questin is about the other problem you mentioned: Is there exist a homemoorphism between either pair of $(0,1),(0,1],[0,1]$; math.stackexchange.com/questions/240414/…
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:03
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
Define $f : (0, 1) to [0, 1)$ to be the inclusion. This is an embedding.
Define $g : [0, 1) to (0, 1)$ by $displaystyle t mapsto frac12t + frac12$. This embeds into $[1/2, 1) subseteq (0, 1)$.
But $(0, 1)$ and $[0, 1)$ are not homeomorphic. If you remove any point from $(0, 1)$ you get two disconnected components. However, if you remove $0$ from $[0, 1)$ you still have only one component.
$endgroup$
add a comment |
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$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
Define $f : (0, 1) to [0, 1)$ to be the inclusion. This is an embedding.
Define $g : [0, 1) to (0, 1)$ by $displaystyle t mapsto frac12t + frac12$. This embeds into $[1/2, 1) subseteq (0, 1)$.
But $(0, 1)$ and $[0, 1)$ are not homeomorphic. If you remove any point from $(0, 1)$ you get two disconnected components. However, if you remove $0$ from $[0, 1)$ you still have only one component.
$endgroup$
add a comment |
$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
Define $f : (0, 1) to [0, 1)$ to be the inclusion. This is an embedding.
Define $g : [0, 1) to (0, 1)$ by $displaystyle t mapsto frac12t + frac12$. This embeds into $[1/2, 1) subseteq (0, 1)$.
But $(0, 1)$ and $[0, 1)$ are not homeomorphic. If you remove any point from $(0, 1)$ you get two disconnected components. However, if you remove $0$ from $[0, 1)$ you still have only one component.
$endgroup$
add a comment |
$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
Define $f : (0, 1) to [0, 1)$ to be the inclusion. This is an embedding.
Define $g : [0, 1) to (0, 1)$ by $displaystyle t mapsto frac12t + frac12$. This embeds into $[1/2, 1) subseteq (0, 1)$.
But $(0, 1)$ and $[0, 1)$ are not homeomorphic. If you remove any point from $(0, 1)$ you get two disconnected components. However, if you remove $0$ from $[0, 1)$ you still have only one component.
$endgroup$
This community wiki solution is intended to clear the question from the unanswered queue.
Define $f : (0, 1) to [0, 1)$ to be the inclusion. This is an embedding.
Define $g : [0, 1) to (0, 1)$ by $displaystyle t mapsto frac12t + frac12$. This embeds into $[1/2, 1) subseteq (0, 1)$.
But $(0, 1)$ and $[0, 1)$ are not homeomorphic. If you remove any point from $(0, 1)$ you get two disconnected components. However, if you remove $0$ from $[0, 1)$ you still have only one component.
answered Dec 10 '18 at 0:40
community wiki
Robert Cardona
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$begingroup$
You can embedd each interval in the other one by $f(x)=frac13+frac13 x$
$endgroup$
– Stefan Hamcke
Sep 24 '13 at 23:19
$begingroup$
Ah, that makes sense now. I guess I kept looking for a surjection as well.
$endgroup$
– Hermit with Adjoint
Sep 24 '13 at 23:23
1
$begingroup$
With your additional condition (that it is also surjection), it seems to be identical to this question
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:02
$begingroup$
This questin is about the other problem you mentioned: Is there exist a homemoorphism between either pair of $(0,1),(0,1],[0,1]$; math.stackexchange.com/questions/240414/…
$endgroup$
– Martin Sleziak
Sep 26 '13 at 9:03