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
    18 hours ago










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










  • $begingroup$
    Actually, this one came to me also. I'm not sure
    $endgroup$
    – MathCosmo
    18 hours 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
    17 hours 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
    17 hours 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 17 hours ago







MathCosmo

















asked 18 hours 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
    18 hours ago










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










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












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




$begingroup$
Actually, this one came to me also. I'm not sure
$endgroup$
– MathCosmo
18 hours 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
17 hours 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
17 hours 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
17 hours 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
17 hours 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
      17 hours 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
      17 hours 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
      15 hours ago






    • 1




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










    • $begingroup$
      @AndreaMori Thank you very much.
      $endgroup$
      – Bertrand Wittgenstein's Ghost
      8 hours 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 16 hours ago

























          answered 17 hours 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
                17 hours 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
                17 hours 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
                15 hours ago






              • 1




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










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                8 hours 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
                17 hours 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
                17 hours 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
                15 hours ago






              • 1




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










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                8 hours 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 16 hours ago

























              answered 17 hours 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
                17 hours 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
                17 hours 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
                15 hours ago






              • 1




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










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                8 hours 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
                17 hours 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
                17 hours 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
                15 hours ago






              • 1




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










              • $begingroup$
                @AndreaMori Thank you very much.
                $endgroup$
                – Bertrand Wittgenstein's Ghost
                8 hours 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
              17 hours 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
              17 hours 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
              17 hours 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
              17 hours 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
              15 hours 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
              15 hours 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
              15 hours ago




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












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




              $begingroup$
              @AndreaMori Thank you very much.
              $endgroup$
              – Bertrand Wittgenstein's Ghost
              8 hours 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 10 hours ago









                  NealNeal

                  24k24087




                  24k24087






























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