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 spaceLocally metrizable but not metrizable space$(0,2)$ 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 spaceLocally metrizable but not metrizable space$(0,2)$ 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 $Z times Y$ )






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


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.










share|cite|improve this question











$endgroup$







  • 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
    Mar 26 at 8:10










  • $begingroup$
    Actually, this one came to me also. I'm not sure
    $endgroup$
    – MathCosmo
    Mar 26 at 8:11






  • 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
    Mar 26 at 8:16










  • $begingroup$
    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
    Mar 26 at 8:37

















2












$begingroup$


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.










share|cite|improve this question











$endgroup$







  • 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
    Mar 26 at 8:10










  • $begingroup$
    Actually, this one came to me also. I'm not sure
    $endgroup$
    – MathCosmo
    Mar 26 at 8:11






  • 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
    Mar 26 at 8:16










  • $begingroup$
    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
    Mar 26 at 8:37













2












2








2


1



$begingroup$


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.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 26 at 9:09







MathCosmo

















asked Mar 26 at 8:05









MathCosmoMathCosmo

3503 silver badges14 bronze badges




3503 silver badges14 bronze badges







  • 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
    Mar 26 at 8:10










  • $begingroup$
    Actually, this one came to me also. I'm not sure
    $endgroup$
    – MathCosmo
    Mar 26 at 8:11






  • 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
    Mar 26 at 8:16










  • $begingroup$
    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
    Mar 26 at 8:37












  • 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
    Mar 26 at 8:10










  • $begingroup$
    Actually, this one came to me also. I'm not sure
    $endgroup$
    – MathCosmo
    Mar 26 at 8:11






  • 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
    Mar 26 at 8:16










  • $begingroup$
    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
    Mar 26 at 8:37







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
Mar 26 at 8:10




$begingroup$
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
Mar 26 at 8:10












$begingroup$
Actually, this one came to me also. I'm not sure
$endgroup$
– MathCosmo
Mar 26 at 8:11




$begingroup$
Actually, this one came to me also. I'm not sure
$endgroup$
– MathCosmo
Mar 26 at 8:11




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
Mar 26 at 8:16




$begingroup$
@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
Mar 26 at 8:16












$begingroup$
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
Mar 26 at 8:37




$begingroup$
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
Mar 26 at 8:37










3 Answers
3






active

oldest

votes


















5












$begingroup$

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.






share|cite|improve this answer











$endgroup$




















    6












    $begingroup$

    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 $ineq j$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      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
      Mar 26 at 8:54










    • $begingroup$
      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
      Mar 26 at 9:02










    • $begingroup$
      @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
      $endgroup$
      – Andrea Mori
      Mar 26 at 10:13






    • 1




      $begingroup$
      @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
      $endgroup$
      – Andrea Mori
      Mar 26 at 10:15










    • $begingroup$
      @AndreaMori Thank you very much.
      $endgroup$
      – Bertrand Wittgenstein's Ghost
      Mar 26 at 17:37


















    0












    $begingroup$

    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.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$

















        5












        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$















          5












          5








          5





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 26 at 10:02

























          answered Mar 26 at 9:00









          Henno BrandsmaHenno Brandsma

          126k4 gold badges52 silver badges136 bronze badges




          126k4 gold badges52 silver badges136 bronze badges























              6












              $begingroup$

              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 $ineq j$.






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                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
                Mar 26 at 8:54










              • $begingroup$
                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
                Mar 26 at 9:02










              • $begingroup$
                @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
                $endgroup$
                – Andrea Mori
                Mar 26 at 10:13






              • 1




                $begingroup$
                @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
                $endgroup$
                – Andrea Mori
                Mar 26 at 10:15










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                Mar 26 at 17:37















              6












              $begingroup$

              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 $ineq j$.






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                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
                Mar 26 at 8:54










              • $begingroup$
                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
                Mar 26 at 9:02










              • $begingroup$
                @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
                $endgroup$
                – Andrea Mori
                Mar 26 at 10:13






              • 1




                $begingroup$
                @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
                $endgroup$
                – Andrea Mori
                Mar 26 at 10:15










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                Mar 26 at 17:37













              6












              6








              6





              $begingroup$

              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 $ineq j$.






              share|cite|improve this answer











              $endgroup$



              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 $ineq j$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Mar 26 at 10:12

























              answered Mar 26 at 8:24









              Andrea MoriAndrea Mori

              20.7k1 gold badge35 silver badges68 bronze badges




              20.7k1 gold badge35 silver badges68 bronze badges











              • $begingroup$
                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
                Mar 26 at 8:54










              • $begingroup$
                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
                Mar 26 at 9:02










              • $begingroup$
                @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
                $endgroup$
                – Andrea Mori
                Mar 26 at 10:13






              • 1




                $begingroup$
                @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
                $endgroup$
                – Andrea Mori
                Mar 26 at 10:15










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                Mar 26 at 17:37
















              • $begingroup$
                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
                Mar 26 at 8:54










              • $begingroup$
                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
                Mar 26 at 9:02










              • $begingroup$
                @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
                $endgroup$
                – Andrea Mori
                Mar 26 at 10:13






              • 1




                $begingroup$
                @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
                $endgroup$
                – Andrea Mori
                Mar 26 at 10:15










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                Mar 26 at 17:37















              $begingroup$
              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
              Mar 26 at 8:54




              $begingroup$
              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
              Mar 26 at 8:54












              $begingroup$
              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
              Mar 26 at 9:02




              $begingroup$
              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
              Mar 26 at 9:02












              $begingroup$
              @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
              $endgroup$
              – Andrea Mori
              Mar 26 at 10:13




              $begingroup$
              @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
              $endgroup$
              – Andrea Mori
              Mar 26 at 10:13




              1




              1




              $begingroup$
              @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
              $endgroup$
              – Andrea Mori
              Mar 26 at 10:15




              $begingroup$
              @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
              $endgroup$
              – Andrea Mori
              Mar 26 at 10:15












              $begingroup$
              @AndreaMori Thank you very much.
              $endgroup$
              – Bertrand Wittgenstein's Ghost
              Mar 26 at 17:37




              $begingroup$
              @AndreaMori Thank you very much.
              $endgroup$
              – Bertrand Wittgenstein's Ghost
              Mar 26 at 17:37











              0












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 26 at 15:58









                  NealNeal

                  24.6k2 gold badges41 silver badges89 bronze badges




                  24.6k2 gold badges41 silver badges89 bronze badges



























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