Give an example of a space which is locally compact at all but one point.Two non-homeomorphic connected, hausdorff, locally compact spaces whose one-point compactifications are homeomorphicone-point compactification of a compact spaceIs there an open subspace of a locally compact space that is not locally compact?Does there exist a Hausdorff group which is not locally compact?Is there a locally compact, locally connected, Hausdorff and second countable space that is “nowhere locally Euclidean”?Closed subset of locally compact Hausdorff spaceAny example of a connected space that is not locally connected?$(0,2)$ is a locally compact Hausdorff space. Then what will be that one point which will make $(0,2)$ a compact Hausdorff space?Is there a locally compact space which is not a k-spaceCan anyone give me a example of space which is limit point compact but not compact?(Except the popular sets $S_Omega$ and $Z times Y$ )
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Give an example of a space which is locally compact at all but one point.
Two non-homeomorphic connected, hausdorff, locally compact spaces whose one-point compactifications are homeomorphicone-point compactification of a compact spaceIs there an open subspace of a locally compact space that is not locally compact?Does there exist a Hausdorff group which is not locally compact?Is there a locally compact, locally connected, Hausdorff and second countable space that is “nowhere locally Euclidean”?Closed subset of locally compact Hausdorff spaceAny example of a connected space that is not locally connected?$(0,2)$ is a locally compact Hausdorff space. Then what will be that one point which will make $(0,2)$ a compact Hausdorff space?Is there a locally compact space which is not a k-spaceCan anyone give me a example of space which is limit point compact but not compact?(Except the popular sets $S_Omega$ and $Z times Y$ )
$begingroup$
Give an example of a space which is locally compact at all but one point.
My professor told me that
Wedge product of infinitely many circles
Will do the job. But I'm not getting his answer. Other examples will be appreciated too.
general-topology algebraic-topology
$endgroup$
add a comment |
$begingroup$
Give an example of a space which is locally compact at all but one point.
My professor told me that
Wedge product of infinitely many circles
Will do the job. But I'm not getting his answer. Other examples will be appreciated too.
general-topology algebraic-topology
$endgroup$
1
$begingroup$
How about the real line plus the point at infinity with a neighborhood around infinity is a set that contains all numbers with sufficiently high magnitude? Then I don't think the point infinity has a compact neighborhood
$endgroup$
– Mark
yesterday
$begingroup$
Actually, this one came to me also. I'm not sure
$endgroup$
– MathCosmo
yesterday
1
$begingroup$
@Mark : since $Bbb R$ is homeomorphic to the open interval $(0,1)$ adding the point $infty$ with the right half-lines as neighborhoods of it will make the space homeomorphic to $(0,1]$ which is certainly locally compact at every point.
$endgroup$
– Andrea Mori
yesterday
$begingroup$
I was suggesting that each open set of infinity has right and left half lines. But I see that would give homeomorphism to a circle
$endgroup$
– Mark
yesterday
add a comment |
$begingroup$
Give an example of a space which is locally compact at all but one point.
My professor told me that
Wedge product of infinitely many circles
Will do the job. But I'm not getting his answer. Other examples will be appreciated too.
general-topology algebraic-topology
$endgroup$
Give an example of a space which is locally compact at all but one point.
My professor told me that
Wedge product of infinitely many circles
Will do the job. But I'm not getting his answer. Other examples will be appreciated too.
general-topology algebraic-topology
general-topology algebraic-topology
edited yesterday
MathCosmo
asked yesterday
MathCosmoMathCosmo
322314
322314
1
$begingroup$
How about the real line plus the point at infinity with a neighborhood around infinity is a set that contains all numbers with sufficiently high magnitude? Then I don't think the point infinity has a compact neighborhood
$endgroup$
– Mark
yesterday
$begingroup$
Actually, this one came to me also. I'm not sure
$endgroup$
– MathCosmo
yesterday
1
$begingroup$
@Mark : since $Bbb R$ is homeomorphic to the open interval $(0,1)$ adding the point $infty$ with the right half-lines as neighborhoods of it will make the space homeomorphic to $(0,1]$ which is certainly locally compact at every point.
$endgroup$
– Andrea Mori
yesterday
$begingroup$
I was suggesting that each open set of infinity has right and left half lines. But I see that would give homeomorphism to a circle
$endgroup$
– Mark
yesterday
add a comment |
1
$begingroup$
How about the real line plus the point at infinity with a neighborhood around infinity is a set that contains all numbers with sufficiently high magnitude? Then I don't think the point infinity has a compact neighborhood
$endgroup$
– Mark
yesterday
$begingroup$
Actually, this one came to me also. I'm not sure
$endgroup$
– MathCosmo
yesterday
1
$begingroup$
@Mark : since $Bbb R$ is homeomorphic to the open interval $(0,1)$ adding the point $infty$ with the right half-lines as neighborhoods of it will make the space homeomorphic to $(0,1]$ which is certainly locally compact at every point.
$endgroup$
– Andrea Mori
yesterday
$begingroup$
I was suggesting that each open set of infinity has right and left half lines. But I see that would give homeomorphism to a circle
$endgroup$
– Mark
yesterday
1
1
$begingroup$
How about the real line plus the point at infinity with a neighborhood around infinity is a set that contains all numbers with sufficiently high magnitude? Then I don't think the point infinity has a compact neighborhood
$endgroup$
– Mark
yesterday
$begingroup$
How about the real line plus the point at infinity with a neighborhood around infinity is a set that contains all numbers with sufficiently high magnitude? Then I don't think the point infinity has a compact neighborhood
$endgroup$
– Mark
yesterday
$begingroup$
Actually, this one came to me also. I'm not sure
$endgroup$
– MathCosmo
yesterday
$begingroup$
Actually, this one came to me also. I'm not sure
$endgroup$
– MathCosmo
yesterday
1
1
$begingroup$
@Mark : since $Bbb R$ is homeomorphic to the open interval $(0,1)$ adding the point $infty$ with the right half-lines as neighborhoods of it will make the space homeomorphic to $(0,1]$ which is certainly locally compact at every point.
$endgroup$
– Andrea Mori
yesterday
$begingroup$
@Mark : since $Bbb R$ is homeomorphic to the open interval $(0,1)$ adding the point $infty$ with the right half-lines as neighborhoods of it will make the space homeomorphic to $(0,1]$ which is certainly locally compact at every point.
$endgroup$
– Andrea Mori
yesterday
$begingroup$
I was suggesting that each open set of infinity has right and left half lines. But I see that would give homeomorphism to a circle
$endgroup$
– Mark
yesterday
$begingroup$
I was suggesting that each open set of infinity has right and left half lines. But I see that would give homeomorphism to a circle
$endgroup$
– Mark
yesterday
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A variation of the wedge space that your professor suggested: the hedgehog of infinite "spininess":
Let $X = [0,1] times mathbbR$, where we identify all points $(0,t)$ to a single point (equivalence class, really, we coudl call it plain $0$, the origin) with the metric:
$$dleft((x,t), (x',t')right) =
begincases
|x-x'| & text if t=t' \
|x| + |x'| & text if t neq t' \
endcases$$
so the space looks like $mathbbR$ many copies of $[0,1]$ glued together at their $0$, metrically. If does not matter what representation $(0,t)$ we choose for $0$ in this computation (so the metric is well-defined).
At all points $(x,t)$ different from $0$ the space behave locally just like $(0,1]$ so is locally compact, but no neighbourhood of $0$ is compact: for some ball neighbourhood $B(0,r)$ for some $r>0$ pick $0 < s < r$ and note that $(s,t): t in mathbbR$ is an infinite closed and discrete subset of $B(0,r)$ showing its non-compactness. So I could have sufficed with index set $mathbbN$ in the second coordinate for this...)
For a nice overview at an elementary level of such "hedgehog spaces", see this paper, e.g. They can be quite useful.
$endgroup$
add a comment |
$begingroup$
HINT: In slightly more elementary terms what your professor is suggesting is the following.
Consider infinitely many (a numerable infinity will suffice) copies $C_1$, $C_2$, $C_3$, ... of the circle $S^1$ with its natural topology and on each copy fix a point
$$
P_iin C_i.
$$
Now consider the quotient
$$
X=left.bigcup_iC_iright/sim
$$
where the only non trivial equivalence is $P_1sim P_2sim P_3simcdots$. Let $Pin X$ be the class of the points $P_i$. What is a neighborhood of $P$?
(Added later)
More generally you can consider a numerable family of pointed locally compact spaces $T_i, P_iin T_i_iinBbb N$ and form the quotient space
$$
X=left.bigcup_iinBbb NT_iright/sim
$$
where the $P_i$ are declared equivalent to each other, and $Pin X$ their common class. Then $X$ is locally compact at each of its points $Qneq P$ but $P$ doesn't have a compact neighborhood since every open set $Pin Usubset X$ is of the form
$$
U=left.bigcup_iinBbb NA_iright/sim
$$
with $A_i$ open in $T_i$ and $A_icap A_j=P$ for every $ineq j$.
$endgroup$
$begingroup$
This not my question, but I am curious. What is neighborhood of $P$? Can you explain, I am having a hard time understanding. Thanks.
$endgroup$
– Bertrand Wittgenstein's Ghost
yesterday
$begingroup$
We need not take circles, just identifying the $0$'s in countably many copies of $[0,1]$ will do. This is not the same as my example in another answer, but this has a finer, non-metric topology, if we use the quotient topology.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
@HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
$endgroup$
– Andrea Mori
yesterday
1
$begingroup$
@BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
$endgroup$
– Andrea Mori
yesterday
$begingroup$
@AndreaMori Thank you very much.
$endgroup$
– Bertrand Wittgenstein's Ghost
yesterday
add a comment |
$begingroup$
Here is a different example. Take the universal cover of a disk in $mathbbR^2$: $widetildeB(0,1)−0$. Endow it with the Euclidean metric it inherits from the disk, and then take its completion with respect to that metric.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A variation of the wedge space that your professor suggested: the hedgehog of infinite "spininess":
Let $X = [0,1] times mathbbR$, where we identify all points $(0,t)$ to a single point (equivalence class, really, we coudl call it plain $0$, the origin) with the metric:
$$dleft((x,t), (x',t')right) =
begincases
|x-x'| & text if t=t' \
|x| + |x'| & text if t neq t' \
endcases$$
so the space looks like $mathbbR$ many copies of $[0,1]$ glued together at their $0$, metrically. If does not matter what representation $(0,t)$ we choose for $0$ in this computation (so the metric is well-defined).
At all points $(x,t)$ different from $0$ the space behave locally just like $(0,1]$ so is locally compact, but no neighbourhood of $0$ is compact: for some ball neighbourhood $B(0,r)$ for some $r>0$ pick $0 < s < r$ and note that $(s,t): t in mathbbR$ is an infinite closed and discrete subset of $B(0,r)$ showing its non-compactness. So I could have sufficed with index set $mathbbN$ in the second coordinate for this...)
For a nice overview at an elementary level of such "hedgehog spaces", see this paper, e.g. They can be quite useful.
$endgroup$
add a comment |
$begingroup$
A variation of the wedge space that your professor suggested: the hedgehog of infinite "spininess":
Let $X = [0,1] times mathbbR$, where we identify all points $(0,t)$ to a single point (equivalence class, really, we coudl call it plain $0$, the origin) with the metric:
$$dleft((x,t), (x',t')right) =
begincases
|x-x'| & text if t=t' \
|x| + |x'| & text if t neq t' \
endcases$$
so the space looks like $mathbbR$ many copies of $[0,1]$ glued together at their $0$, metrically. If does not matter what representation $(0,t)$ we choose for $0$ in this computation (so the metric is well-defined).
At all points $(x,t)$ different from $0$ the space behave locally just like $(0,1]$ so is locally compact, but no neighbourhood of $0$ is compact: for some ball neighbourhood $B(0,r)$ for some $r>0$ pick $0 < s < r$ and note that $(s,t): t in mathbbR$ is an infinite closed and discrete subset of $B(0,r)$ showing its non-compactness. So I could have sufficed with index set $mathbbN$ in the second coordinate for this...)
For a nice overview at an elementary level of such "hedgehog spaces", see this paper, e.g. They can be quite useful.
$endgroup$
add a comment |
$begingroup$
A variation of the wedge space that your professor suggested: the hedgehog of infinite "spininess":
Let $X = [0,1] times mathbbR$, where we identify all points $(0,t)$ to a single point (equivalence class, really, we coudl call it plain $0$, the origin) with the metric:
$$dleft((x,t), (x',t')right) =
begincases
|x-x'| & text if t=t' \
|x| + |x'| & text if t neq t' \
endcases$$
so the space looks like $mathbbR$ many copies of $[0,1]$ glued together at their $0$, metrically. If does not matter what representation $(0,t)$ we choose for $0$ in this computation (so the metric is well-defined).
At all points $(x,t)$ different from $0$ the space behave locally just like $(0,1]$ so is locally compact, but no neighbourhood of $0$ is compact: for some ball neighbourhood $B(0,r)$ for some $r>0$ pick $0 < s < r$ and note that $(s,t): t in mathbbR$ is an infinite closed and discrete subset of $B(0,r)$ showing its non-compactness. So I could have sufficed with index set $mathbbN$ in the second coordinate for this...)
For a nice overview at an elementary level of such "hedgehog spaces", see this paper, e.g. They can be quite useful.
$endgroup$
A variation of the wedge space that your professor suggested: the hedgehog of infinite "spininess":
Let $X = [0,1] times mathbbR$, where we identify all points $(0,t)$ to a single point (equivalence class, really, we coudl call it plain $0$, the origin) with the metric:
$$dleft((x,t), (x',t')right) =
begincases
|x-x'| & text if t=t' \
|x| + |x'| & text if t neq t' \
endcases$$
so the space looks like $mathbbR$ many copies of $[0,1]$ glued together at their $0$, metrically. If does not matter what representation $(0,t)$ we choose for $0$ in this computation (so the metric is well-defined).
At all points $(x,t)$ different from $0$ the space behave locally just like $(0,1]$ so is locally compact, but no neighbourhood of $0$ is compact: for some ball neighbourhood $B(0,r)$ for some $r>0$ pick $0 < s < r$ and note that $(s,t): t in mathbbR$ is an infinite closed and discrete subset of $B(0,r)$ showing its non-compactness. So I could have sufficed with index set $mathbbN$ in the second coordinate for this...)
For a nice overview at an elementary level of such "hedgehog spaces", see this paper, e.g. They can be quite useful.
edited yesterday
answered yesterday
Henno BrandsmaHenno Brandsma
114k348123
114k348123
add a comment |
add a comment |
$begingroup$
HINT: In slightly more elementary terms what your professor is suggesting is the following.
Consider infinitely many (a numerable infinity will suffice) copies $C_1$, $C_2$, $C_3$, ... of the circle $S^1$ with its natural topology and on each copy fix a point
$$
P_iin C_i.
$$
Now consider the quotient
$$
X=left.bigcup_iC_iright/sim
$$
where the only non trivial equivalence is $P_1sim P_2sim P_3simcdots$. Let $Pin X$ be the class of the points $P_i$. What is a neighborhood of $P$?
(Added later)
More generally you can consider a numerable family of pointed locally compact spaces $T_i, P_iin T_i_iinBbb N$ and form the quotient space
$$
X=left.bigcup_iinBbb NT_iright/sim
$$
where the $P_i$ are declared equivalent to each other, and $Pin X$ their common class. Then $X$ is locally compact at each of its points $Qneq P$ but $P$ doesn't have a compact neighborhood since every open set $Pin Usubset X$ is of the form
$$
U=left.bigcup_iinBbb NA_iright/sim
$$
with $A_i$ open in $T_i$ and $A_icap A_j=P$ for every $ineq j$.
$endgroup$
$begingroup$
This not my question, but I am curious. What is neighborhood of $P$? Can you explain, I am having a hard time understanding. Thanks.
$endgroup$
– Bertrand Wittgenstein's Ghost
yesterday
$begingroup$
We need not take circles, just identifying the $0$'s in countably many copies of $[0,1]$ will do. This is not the same as my example in another answer, but this has a finer, non-metric topology, if we use the quotient topology.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
@HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
$endgroup$
– Andrea Mori
yesterday
1
$begingroup$
@BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
$endgroup$
– Andrea Mori
yesterday
$begingroup$
@AndreaMori Thank you very much.
$endgroup$
– Bertrand Wittgenstein's Ghost
yesterday
add a comment |
$begingroup$
HINT: In slightly more elementary terms what your professor is suggesting is the following.
Consider infinitely many (a numerable infinity will suffice) copies $C_1$, $C_2$, $C_3$, ... of the circle $S^1$ with its natural topology and on each copy fix a point
$$
P_iin C_i.
$$
Now consider the quotient
$$
X=left.bigcup_iC_iright/sim
$$
where the only non trivial equivalence is $P_1sim P_2sim P_3simcdots$. Let $Pin X$ be the class of the points $P_i$. What is a neighborhood of $P$?
(Added later)
More generally you can consider a numerable family of pointed locally compact spaces $T_i, P_iin T_i_iinBbb N$ and form the quotient space
$$
X=left.bigcup_iinBbb NT_iright/sim
$$
where the $P_i$ are declared equivalent to each other, and $Pin X$ their common class. Then $X$ is locally compact at each of its points $Qneq P$ but $P$ doesn't have a compact neighborhood since every open set $Pin Usubset X$ is of the form
$$
U=left.bigcup_iinBbb NA_iright/sim
$$
with $A_i$ open in $T_i$ and $A_icap A_j=P$ for every $ineq j$.
$endgroup$
$begingroup$
This not my question, but I am curious. What is neighborhood of $P$? Can you explain, I am having a hard time understanding. Thanks.
$endgroup$
– Bertrand Wittgenstein's Ghost
yesterday
$begingroup$
We need not take circles, just identifying the $0$'s in countably many copies of $[0,1]$ will do. This is not the same as my example in another answer, but this has a finer, non-metric topology, if we use the quotient topology.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
@HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
$endgroup$
– Andrea Mori
yesterday
1
$begingroup$
@BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
$endgroup$
– Andrea Mori
yesterday
$begingroup$
@AndreaMori Thank you very much.
$endgroup$
– Bertrand Wittgenstein's Ghost
yesterday
add a comment |
$begingroup$
HINT: In slightly more elementary terms what your professor is suggesting is the following.
Consider infinitely many (a numerable infinity will suffice) copies $C_1$, $C_2$, $C_3$, ... of the circle $S^1$ with its natural topology and on each copy fix a point
$$
P_iin C_i.
$$
Now consider the quotient
$$
X=left.bigcup_iC_iright/sim
$$
where the only non trivial equivalence is $P_1sim P_2sim P_3simcdots$. Let $Pin X$ be the class of the points $P_i$. What is a neighborhood of $P$?
(Added later)
More generally you can consider a numerable family of pointed locally compact spaces $T_i, P_iin T_i_iinBbb N$ and form the quotient space
$$
X=left.bigcup_iinBbb NT_iright/sim
$$
where the $P_i$ are declared equivalent to each other, and $Pin X$ their common class. Then $X$ is locally compact at each of its points $Qneq P$ but $P$ doesn't have a compact neighborhood since every open set $Pin Usubset X$ is of the form
$$
U=left.bigcup_iinBbb NA_iright/sim
$$
with $A_i$ open in $T_i$ and $A_icap A_j=P$ for every $ineq j$.
$endgroup$
HINT: In slightly more elementary terms what your professor is suggesting is the following.
Consider infinitely many (a numerable infinity will suffice) copies $C_1$, $C_2$, $C_3$, ... of the circle $S^1$ with its natural topology and on each copy fix a point
$$
P_iin C_i.
$$
Now consider the quotient
$$
X=left.bigcup_iC_iright/sim
$$
where the only non trivial equivalence is $P_1sim P_2sim P_3simcdots$. Let $Pin X$ be the class of the points $P_i$. What is a neighborhood of $P$?
(Added later)
More generally you can consider a numerable family of pointed locally compact spaces $T_i, P_iin T_i_iinBbb N$ and form the quotient space
$$
X=left.bigcup_iinBbb NT_iright/sim
$$
where the $P_i$ are declared equivalent to each other, and $Pin X$ their common class. Then $X$ is locally compact at each of its points $Qneq P$ but $P$ doesn't have a compact neighborhood since every open set $Pin Usubset X$ is of the form
$$
U=left.bigcup_iinBbb NA_iright/sim
$$
with $A_i$ open in $T_i$ and $A_icap A_j=P$ for every $ineq j$.
edited yesterday
answered yesterday
Andrea MoriAndrea Mori
20.1k13466
20.1k13466
$begingroup$
This not my question, but I am curious. What is neighborhood of $P$? Can you explain, I am having a hard time understanding. Thanks.
$endgroup$
– Bertrand Wittgenstein's Ghost
yesterday
$begingroup$
We need not take circles, just identifying the $0$'s in countably many copies of $[0,1]$ will do. This is not the same as my example in another answer, but this has a finer, non-metric topology, if we use the quotient topology.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
@HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
$endgroup$
– Andrea Mori
yesterday
1
$begingroup$
@BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
$endgroup$
– Andrea Mori
yesterday
$begingroup$
@AndreaMori Thank you very much.
$endgroup$
– Bertrand Wittgenstein's Ghost
yesterday
add a comment |
$begingroup$
This not my question, but I am curious. What is neighborhood of $P$? Can you explain, I am having a hard time understanding. Thanks.
$endgroup$
– Bertrand Wittgenstein's Ghost
yesterday
$begingroup$
We need not take circles, just identifying the $0$'s in countably many copies of $[0,1]$ will do. This is not the same as my example in another answer, but this has a finer, non-metric topology, if we use the quotient topology.
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– Henno Brandsma
yesterday
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@HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
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– Andrea Mori
yesterday
1
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@BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
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– Andrea Mori
yesterday
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@AndreaMori Thank you very much.
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– Bertrand Wittgenstein's Ghost
yesterday
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This not my question, but I am curious. What is neighborhood of $P$? Can you explain, I am having a hard time understanding. Thanks.
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– Bertrand Wittgenstein's Ghost
yesterday
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This not my question, but I am curious. What is neighborhood of $P$? Can you explain, I am having a hard time understanding. Thanks.
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– Bertrand Wittgenstein's Ghost
yesterday
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We need not take circles, just identifying the $0$'s in countably many copies of $[0,1]$ will do. This is not the same as my example in another answer, but this has a finer, non-metric topology, if we use the quotient topology.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
We need not take circles, just identifying the $0$'s in countably many copies of $[0,1]$ will do. This is not the same as my example in another answer, but this has a finer, non-metric topology, if we use the quotient topology.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
@HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
$endgroup$
– Andrea Mori
yesterday
$begingroup$
@HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
$endgroup$
– Andrea Mori
yesterday
1
1
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@BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
$endgroup$
– Andrea Mori
yesterday
$begingroup$
@BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
$endgroup$
– Andrea Mori
yesterday
$begingroup$
@AndreaMori Thank you very much.
$endgroup$
– Bertrand Wittgenstein's Ghost
yesterday
$begingroup$
@AndreaMori Thank you very much.
$endgroup$
– Bertrand Wittgenstein's Ghost
yesterday
add a comment |
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Here is a different example. Take the universal cover of a disk in $mathbbR^2$: $widetildeB(0,1)−0$. Endow it with the Euclidean metric it inherits from the disk, and then take its completion with respect to that metric.
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add a comment |
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Here is a different example. Take the universal cover of a disk in $mathbbR^2$: $widetildeB(0,1)−0$. Endow it with the Euclidean metric it inherits from the disk, and then take its completion with respect to that metric.
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add a comment |
$begingroup$
Here is a different example. Take the universal cover of a disk in $mathbbR^2$: $widetildeB(0,1)−0$. Endow it with the Euclidean metric it inherits from the disk, and then take its completion with respect to that metric.
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Here is a different example. Take the universal cover of a disk in $mathbbR^2$: $widetildeB(0,1)−0$. Endow it with the Euclidean metric it inherits from the disk, and then take its completion with respect to that metric.
answered yesterday
NealNeal
24k24087
24k24087
add a comment |
add a comment |
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How about the real line plus the point at infinity with a neighborhood around infinity is a set that contains all numbers with sufficiently high magnitude? Then I don't think the point infinity has a compact neighborhood
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– Mark
yesterday
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Actually, this one came to me also. I'm not sure
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– MathCosmo
yesterday
1
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@Mark : since $Bbb R$ is homeomorphic to the open interval $(0,1)$ adding the point $infty$ with the right half-lines as neighborhoods of it will make the space homeomorphic to $(0,1]$ which is certainly locally compact at every point.
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– Andrea Mori
yesterday
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I was suggesting that each open set of infinity has right and left half lines. But I see that would give homeomorphism to a circle
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– Mark
yesterday