Give an example of a space which is locally compact at all but one point.












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
    2 days ago










  • $begingroup$
    Actually, this one came to me also. I'm not sure
    $endgroup$
    – MathCosmo
    2 days ago






  • 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
    2 days ago










  • $begingroup$
    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
    2 days ago
















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
    2 days ago










  • $begingroup$
    Actually, this one came to me also. I'm not sure
    $endgroup$
    – MathCosmo
    2 days ago






  • 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
    2 days ago










  • $begingroup$
    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
    2 days ago














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 2 days ago







MathCosmo

















asked 2 days ago









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
    2 days ago










  • $begingroup$
    Actually, this one came to me also. I'm not sure
    $endgroup$
    – MathCosmo
    2 days ago






  • 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
    2 days ago










  • $begingroup$
    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
    2 days ago














  • 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
    2 days ago










  • $begingroup$
    Actually, this one came to me also. I'm not sure
    $endgroup$
    – MathCosmo
    2 days ago






  • 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
    2 days ago










  • $begingroup$
    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
    2 days ago








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
2 days ago




$begingroup$
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
2 days ago












$begingroup$
Actually, this one came to me also. I'm not sure
$endgroup$
– MathCosmo
2 days ago




$begingroup$
Actually, this one came to me also. I'm not sure
$endgroup$
– MathCosmo
2 days ago




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
2 days ago




$begingroup$
@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
2 days ago












$begingroup$
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
2 days ago




$begingroup$
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
2 days ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

A variation of the wedge space that your professor suggested: the hedgehog of infinite "spininess":



Let $X = [0,1] times mathbb{R}$, 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:



$$dleft((x,t), (x',t')right) =
begin{cases}
|x-x'| & text{ if } t=t' \
|x| + |x'| & text{ if } t neq t' \
end{cases}$$



so the space looks like $mathbb{R}$ 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 ${(s,t): t in mathbb{R}}$ is an infinite closed and discrete subset of $B(0,r)$ showing its non-compactness. So I could have sufficed with index set $mathbb{N}$ 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
    $$
    X=left.bigcup_{i}C_iright/sim
    $$

    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
    $$
    X=left.bigcup_{iinBbb N}T_iright/sim
    $$

    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
    $$
    U=left.bigcup_{iinBbb N}A_iright/sim
    $$

    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
      2 days ago










    • $begingroup$
      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
      2 days ago










    • $begingroup$
      @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
      $endgroup$
      – Andrea Mori
      2 days ago






    • 1




      $begingroup$
      @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
      $endgroup$
      – Andrea Mori
      2 days ago










    • $begingroup$
      @AndreaMori Thank you very much.
      $endgroup$
      – Bertrand Wittgenstein's Ghost
      2 days ago



















    0












    $begingroup$

    Here is a different example. Take the universal cover of a disk in $mathbb{R}^2$: $widetilde{B(0,1)−{0}}$. 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 $X = [0,1] times mathbb{R}$, 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:



      $$dleft((x,t), (x',t')right) =
      begin{cases}
      |x-x'| & text{ if } t=t' \
      |x| + |x'| & text{ if } t neq t' \
      end{cases}$$



      so the space looks like $mathbb{R}$ 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 ${(s,t): t in mathbb{R}}$ is an infinite closed and discrete subset of $B(0,r)$ showing its non-compactness. So I could have sufficed with index set $mathbb{N}$ 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 $X = [0,1] times mathbb{R}$, 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:



        $$dleft((x,t), (x',t')right) =
        begin{cases}
        |x-x'| & text{ if } t=t' \
        |x| + |x'| & text{ if } t neq t' \
        end{cases}$$



        so the space looks like $mathbb{R}$ 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 ${(s,t): t in mathbb{R}}$ is an infinite closed and discrete subset of $B(0,r)$ showing its non-compactness. So I could have sufficed with index set $mathbb{N}$ 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 $X = [0,1] times mathbb{R}$, 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:



          $$dleft((x,t), (x',t')right) =
          begin{cases}
          |x-x'| & text{ if } t=t' \
          |x| + |x'| & text{ if } t neq t' \
          end{cases}$$



          so the space looks like $mathbb{R}$ 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 ${(s,t): t in mathbb{R}}$ is an infinite closed and discrete subset of $B(0,r)$ showing its non-compactness. So I could have sufficed with index set $mathbb{N}$ 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 $X = [0,1] times mathbb{R}$, 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:



          $$dleft((x,t), (x',t')right) =
          begin{cases}
          |x-x'| & text{ if } t=t' \
          |x| + |x'| & text{ if } t neq t' \
          end{cases}$$



          so the space looks like $mathbb{R}$ 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 ${(s,t): t in mathbb{R}}$ is an infinite closed and discrete subset of $B(0,r)$ showing its non-compactness. So I could have sufficed with index set $mathbb{N}$ 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 2 days ago

























          answered 2 days ago









          Henno BrandsmaHenno Brandsma

          114k348123




          114k348123























              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
              $$
              X=left.bigcup_{i}C_iright/sim
              $$

              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
              $$
              X=left.bigcup_{iinBbb N}T_iright/sim
              $$

              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
              $$
              U=left.bigcup_{iinBbb N}A_iright/sim
              $$

              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
                2 days ago










              • $begingroup$
                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
                2 days ago










              • $begingroup$
                @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
                $endgroup$
                – Andrea Mori
                2 days ago






              • 1




                $begingroup$
                @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
                $endgroup$
                – Andrea Mori
                2 days ago










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                2 days ago
















              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
              $$
              X=left.bigcup_{i}C_iright/sim
              $$

              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
              $$
              X=left.bigcup_{iinBbb N}T_iright/sim
              $$

              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
              $$
              U=left.bigcup_{iinBbb N}A_iright/sim
              $$

              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
                2 days ago










              • $begingroup$
                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
                2 days ago










              • $begingroup$
                @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
                $endgroup$
                – Andrea Mori
                2 days ago






              • 1




                $begingroup$
                @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
                $endgroup$
                – Andrea Mori
                2 days ago










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                2 days ago














              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
              $$
              X=left.bigcup_{i}C_iright/sim
              $$

              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
              $$
              X=left.bigcup_{iinBbb N}T_iright/sim
              $$

              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
              $$
              U=left.bigcup_{iinBbb N}A_iright/sim
              $$

              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
              $$
              X=left.bigcup_{i}C_iright/sim
              $$

              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
              $$
              X=left.bigcup_{iinBbb N}T_iright/sim
              $$

              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
              $$
              U=left.bigcup_{iinBbb N}A_iright/sim
              $$

              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 2 days ago

























              answered 2 days ago









              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
                2 days ago










              • $begingroup$
                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
                2 days ago










              • $begingroup$
                @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
                $endgroup$
                – Andrea Mori
                2 days ago






              • 1




                $begingroup$
                @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
                $endgroup$
                – Andrea Mori
                2 days ago










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                2 days ago


















              • $begingroup$
                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
                2 days ago










              • $begingroup$
                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
                2 days ago










              • $begingroup$
                @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
                $endgroup$
                – Andrea Mori
                2 days ago






              • 1




                $begingroup$
                @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
                $endgroup$
                – Andrea Mori
                2 days ago










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                2 days ago
















              $begingroup$
              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
              2 days ago




              $begingroup$
              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
              2 days ago












              $begingroup$
              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
              2 days ago




              $begingroup$
              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
              2 days ago












              $begingroup$
              @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
              $endgroup$
              – Andrea Mori
              2 days ago




              $begingroup$
              @HennoBrandsma, of course we not need circles. In fact we can take basically anything. I added a note to my answer.
              $endgroup$
              – Andrea Mori
              2 days ago




              1




              1




              $begingroup$
              @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
              $endgroup$
              – Andrea Mori
              2 days ago




              $begingroup$
              @BertrandWittgenstein'sGhost, I edited my answer including a general construction which should make the basic example more clear
              $endgroup$
              – Andrea Mori
              2 days ago












              $begingroup$
              @AndreaMori Thank you very much.
              $endgroup$
              – Bertrand Wittgenstein's Ghost
              2 days ago




              $begingroup$
              @AndreaMori Thank you very much.
              $endgroup$
              – Bertrand Wittgenstein's Ghost
              2 days ago











              0












              $begingroup$

              Here is a different example. Take the universal cover of a disk in $mathbb{R}^2$: $widetilde{B(0,1)−{0}}$. 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 $mathbb{R}^2$: $widetilde{B(0,1)−{0}}$. 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 $mathbb{R}^2$: $widetilde{B(0,1)−{0}}$. 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 $mathbb{R}^2$: $widetilde{B(0,1)−{0}}$. 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 2 days ago









                  NealNeal

                  24k24087




                  24k24087






























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