For the existence of one-point compactification, do we need locally compactness?












0














In the book Topology by Munkres, at page 184, it is given the existence and uniqueness of one point compactification of a locally compact Hausdorff space; however, in the existence part, I can't see where we needed the locally compactness of that space, and this raised the question:



Does one point compactification of a Hausdorff space always exist (even though it is not unique) ?



See the proof in the book;



(sorry for the images; they are just for reference for those that doesn't have the book with them)



enter image description hereenter image description hereenter image description here










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  • 5




    If you remove a point from a compact Hausdorff space, you get a locally compact space.
    – Lord Shark the Unknown
    Dec 26 at 8:53






  • 1




    If $X$ is Hausdorff but not locally compact its one point compactification won't be Hausdorff, but it can be constructed in the same way
    – Alessandro Codenotti
    Dec 26 at 9:02






  • 1




    For every space there is the one-point-Alexandrov compactification.
    – drhab
    Dec 26 at 9:03






  • 1




    "we can choose a compact set in $X$..." last paragraph
    – Alessandro Codenotti
    Dec 26 at 9:07






  • 1




    In topology a property P is called hereditary when ($Xsubset Y$ and $Y$ has property P$)implies (X $ has property P).
    – DanielWainfleet
    2 days ago


















0














In the book Topology by Munkres, at page 184, it is given the existence and uniqueness of one point compactification of a locally compact Hausdorff space; however, in the existence part, I can't see where we needed the locally compactness of that space, and this raised the question:



Does one point compactification of a Hausdorff space always exist (even though it is not unique) ?



See the proof in the book;



(sorry for the images; they are just for reference for those that doesn't have the book with them)



enter image description hereenter image description hereenter image description here










share|cite|improve this question




















  • 5




    If you remove a point from a compact Hausdorff space, you get a locally compact space.
    – Lord Shark the Unknown
    Dec 26 at 8:53






  • 1




    If $X$ is Hausdorff but not locally compact its one point compactification won't be Hausdorff, but it can be constructed in the same way
    – Alessandro Codenotti
    Dec 26 at 9:02






  • 1




    For every space there is the one-point-Alexandrov compactification.
    – drhab
    Dec 26 at 9:03






  • 1




    "we can choose a compact set in $X$..." last paragraph
    – Alessandro Codenotti
    Dec 26 at 9:07






  • 1




    In topology a property P is called hereditary when ($Xsubset Y$ and $Y$ has property P$)implies (X $ has property P).
    – DanielWainfleet
    2 days ago
















0












0








0







In the book Topology by Munkres, at page 184, it is given the existence and uniqueness of one point compactification of a locally compact Hausdorff space; however, in the existence part, I can't see where we needed the locally compactness of that space, and this raised the question:



Does one point compactification of a Hausdorff space always exist (even though it is not unique) ?



See the proof in the book;



(sorry for the images; they are just for reference for those that doesn't have the book with them)



enter image description hereenter image description hereenter image description here










share|cite|improve this question















In the book Topology by Munkres, at page 184, it is given the existence and uniqueness of one point compactification of a locally compact Hausdorff space; however, in the existence part, I can't see where we needed the locally compactness of that space, and this raised the question:



Does one point compactification of a Hausdorff space always exist (even though it is not unique) ?



See the proof in the book;



(sorry for the images; they are just for reference for those that doesn't have the book with them)



enter image description hereenter image description hereenter image description here







general-topology compactness separation-axioms compactification






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share|cite|improve this question













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edited Dec 26 at 9:20









Martin Sleziak

44.7k7115270




44.7k7115270










asked Dec 26 at 8:52









onurcanbektas

3,3251936




3,3251936








  • 5




    If you remove a point from a compact Hausdorff space, you get a locally compact space.
    – Lord Shark the Unknown
    Dec 26 at 8:53






  • 1




    If $X$ is Hausdorff but not locally compact its one point compactification won't be Hausdorff, but it can be constructed in the same way
    – Alessandro Codenotti
    Dec 26 at 9:02






  • 1




    For every space there is the one-point-Alexandrov compactification.
    – drhab
    Dec 26 at 9:03






  • 1




    "we can choose a compact set in $X$..." last paragraph
    – Alessandro Codenotti
    Dec 26 at 9:07






  • 1




    In topology a property P is called hereditary when ($Xsubset Y$ and $Y$ has property P$)implies (X $ has property P).
    – DanielWainfleet
    2 days ago
















  • 5




    If you remove a point from a compact Hausdorff space, you get a locally compact space.
    – Lord Shark the Unknown
    Dec 26 at 8:53






  • 1




    If $X$ is Hausdorff but not locally compact its one point compactification won't be Hausdorff, but it can be constructed in the same way
    – Alessandro Codenotti
    Dec 26 at 9:02






  • 1




    For every space there is the one-point-Alexandrov compactification.
    – drhab
    Dec 26 at 9:03






  • 1




    "we can choose a compact set in $X$..." last paragraph
    – Alessandro Codenotti
    Dec 26 at 9:07






  • 1




    In topology a property P is called hereditary when ($Xsubset Y$ and $Y$ has property P$)implies (X $ has property P).
    – DanielWainfleet
    2 days ago










5




5




If you remove a point from a compact Hausdorff space, you get a locally compact space.
– Lord Shark the Unknown
Dec 26 at 8:53




If you remove a point from a compact Hausdorff space, you get a locally compact space.
– Lord Shark the Unknown
Dec 26 at 8:53




1




1




If $X$ is Hausdorff but not locally compact its one point compactification won't be Hausdorff, but it can be constructed in the same way
– Alessandro Codenotti
Dec 26 at 9:02




If $X$ is Hausdorff but not locally compact its one point compactification won't be Hausdorff, but it can be constructed in the same way
– Alessandro Codenotti
Dec 26 at 9:02




1




1




For every space there is the one-point-Alexandrov compactification.
– drhab
Dec 26 at 9:03




For every space there is the one-point-Alexandrov compactification.
– drhab
Dec 26 at 9:03




1




1




"we can choose a compact set in $X$..." last paragraph
– Alessandro Codenotti
Dec 26 at 9:07




"we can choose a compact set in $X$..." last paragraph
– Alessandro Codenotti
Dec 26 at 9:07




1




1




In topology a property P is called hereditary when ($Xsubset Y$ and $Y$ has property P$)implies (X $ has property P).
– DanielWainfleet
2 days ago






In topology a property P is called hereditary when ($Xsubset Y$ and $Y$ has property P$)implies (X $ has property P).
– DanielWainfleet
2 days ago












1 Answer
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oldest

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For any space $X$ we can construct a space $alpha(X)$, the Aleksandrov extension of $X$ by defining a space $Y$ as Munkres does with the extra provision that we take all complements of closed compact subsets of $X$ as the extra neighbourhoods for $infty$. One can easily check that $alpha(X)$ is then compact.



The "closed" is needed in general because if e.g. $X$ is not Hausdorff it could have some compact subset $K$ which is not closed, and then (if we were to omit the closed condition) $(Xsetminus K) cup {infty}$ would be open while its intersection with $X$ would be $Xsetminus K$, which was not open, so if we left out the closed condition $X$ would not have the same topology as a subspace of $alpha(X)$ as originally, going against the idea of an extension/compactification: we want to embed $X$ in a larger space with better properties, so in the larger space it should be a subspace with the same topology that it had originally.



If we want $Y = alpha(X)$ to be Hausdorff, (so in particular $X$ should then be Hausdorff, as a subspace of $Y$) we need to be able to separate $infty$ from every point $x$ in $X$. As a neighbourhood of $infty$ is of the form ${infty} cup X setminus C$, with $C$ compact and closed, every point $x$ should then have a neighbourhood that sits inside a compact closed set, i.e. $X$ must be locally compact.



So $alpha(X)$ can always be defined such that $alpha(X)setminus X$ is a point and $X$ is a subspace of $alpha(X)$ and it is always compact (regardless of $X$) but $alpha(X)$ is Hausdorff iff $X$ is locally compact and Hausdorff. A special case is when $X$ is already Hausdorff and compact, in which case we add an isolated point $infty$ (as $X$ can be taken as $C$, a compact closed subset) and we get that $X$ is not dense in $alpha(X)$.



Normally we only consider Hausdorff compactifications and in that case the local compactness is needed for the Hausdorffness of the construction $alpha(X)$. And also because then $X$ is an open subset of a compact Hausdorff space and thus locally compact for that reason.






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    For any space $X$ we can construct a space $alpha(X)$, the Aleksandrov extension of $X$ by defining a space $Y$ as Munkres does with the extra provision that we take all complements of closed compact subsets of $X$ as the extra neighbourhoods for $infty$. One can easily check that $alpha(X)$ is then compact.



    The "closed" is needed in general because if e.g. $X$ is not Hausdorff it could have some compact subset $K$ which is not closed, and then (if we were to omit the closed condition) $(Xsetminus K) cup {infty}$ would be open while its intersection with $X$ would be $Xsetminus K$, which was not open, so if we left out the closed condition $X$ would not have the same topology as a subspace of $alpha(X)$ as originally, going against the idea of an extension/compactification: we want to embed $X$ in a larger space with better properties, so in the larger space it should be a subspace with the same topology that it had originally.



    If we want $Y = alpha(X)$ to be Hausdorff, (so in particular $X$ should then be Hausdorff, as a subspace of $Y$) we need to be able to separate $infty$ from every point $x$ in $X$. As a neighbourhood of $infty$ is of the form ${infty} cup X setminus C$, with $C$ compact and closed, every point $x$ should then have a neighbourhood that sits inside a compact closed set, i.e. $X$ must be locally compact.



    So $alpha(X)$ can always be defined such that $alpha(X)setminus X$ is a point and $X$ is a subspace of $alpha(X)$ and it is always compact (regardless of $X$) but $alpha(X)$ is Hausdorff iff $X$ is locally compact and Hausdorff. A special case is when $X$ is already Hausdorff and compact, in which case we add an isolated point $infty$ (as $X$ can be taken as $C$, a compact closed subset) and we get that $X$ is not dense in $alpha(X)$.



    Normally we only consider Hausdorff compactifications and in that case the local compactness is needed for the Hausdorffness of the construction $alpha(X)$. And also because then $X$ is an open subset of a compact Hausdorff space and thus locally compact for that reason.






    share|cite|improve this answer




























      4














      For any space $X$ we can construct a space $alpha(X)$, the Aleksandrov extension of $X$ by defining a space $Y$ as Munkres does with the extra provision that we take all complements of closed compact subsets of $X$ as the extra neighbourhoods for $infty$. One can easily check that $alpha(X)$ is then compact.



      The "closed" is needed in general because if e.g. $X$ is not Hausdorff it could have some compact subset $K$ which is not closed, and then (if we were to omit the closed condition) $(Xsetminus K) cup {infty}$ would be open while its intersection with $X$ would be $Xsetminus K$, which was not open, so if we left out the closed condition $X$ would not have the same topology as a subspace of $alpha(X)$ as originally, going against the idea of an extension/compactification: we want to embed $X$ in a larger space with better properties, so in the larger space it should be a subspace with the same topology that it had originally.



      If we want $Y = alpha(X)$ to be Hausdorff, (so in particular $X$ should then be Hausdorff, as a subspace of $Y$) we need to be able to separate $infty$ from every point $x$ in $X$. As a neighbourhood of $infty$ is of the form ${infty} cup X setminus C$, with $C$ compact and closed, every point $x$ should then have a neighbourhood that sits inside a compact closed set, i.e. $X$ must be locally compact.



      So $alpha(X)$ can always be defined such that $alpha(X)setminus X$ is a point and $X$ is a subspace of $alpha(X)$ and it is always compact (regardless of $X$) but $alpha(X)$ is Hausdorff iff $X$ is locally compact and Hausdorff. A special case is when $X$ is already Hausdorff and compact, in which case we add an isolated point $infty$ (as $X$ can be taken as $C$, a compact closed subset) and we get that $X$ is not dense in $alpha(X)$.



      Normally we only consider Hausdorff compactifications and in that case the local compactness is needed for the Hausdorffness of the construction $alpha(X)$. And also because then $X$ is an open subset of a compact Hausdorff space and thus locally compact for that reason.






      share|cite|improve this answer


























        4












        4








        4






        For any space $X$ we can construct a space $alpha(X)$, the Aleksandrov extension of $X$ by defining a space $Y$ as Munkres does with the extra provision that we take all complements of closed compact subsets of $X$ as the extra neighbourhoods for $infty$. One can easily check that $alpha(X)$ is then compact.



        The "closed" is needed in general because if e.g. $X$ is not Hausdorff it could have some compact subset $K$ which is not closed, and then (if we were to omit the closed condition) $(Xsetminus K) cup {infty}$ would be open while its intersection with $X$ would be $Xsetminus K$, which was not open, so if we left out the closed condition $X$ would not have the same topology as a subspace of $alpha(X)$ as originally, going against the idea of an extension/compactification: we want to embed $X$ in a larger space with better properties, so in the larger space it should be a subspace with the same topology that it had originally.



        If we want $Y = alpha(X)$ to be Hausdorff, (so in particular $X$ should then be Hausdorff, as a subspace of $Y$) we need to be able to separate $infty$ from every point $x$ in $X$. As a neighbourhood of $infty$ is of the form ${infty} cup X setminus C$, with $C$ compact and closed, every point $x$ should then have a neighbourhood that sits inside a compact closed set, i.e. $X$ must be locally compact.



        So $alpha(X)$ can always be defined such that $alpha(X)setminus X$ is a point and $X$ is a subspace of $alpha(X)$ and it is always compact (regardless of $X$) but $alpha(X)$ is Hausdorff iff $X$ is locally compact and Hausdorff. A special case is when $X$ is already Hausdorff and compact, in which case we add an isolated point $infty$ (as $X$ can be taken as $C$, a compact closed subset) and we get that $X$ is not dense in $alpha(X)$.



        Normally we only consider Hausdorff compactifications and in that case the local compactness is needed for the Hausdorffness of the construction $alpha(X)$. And also because then $X$ is an open subset of a compact Hausdorff space and thus locally compact for that reason.






        share|cite|improve this answer














        For any space $X$ we can construct a space $alpha(X)$, the Aleksandrov extension of $X$ by defining a space $Y$ as Munkres does with the extra provision that we take all complements of closed compact subsets of $X$ as the extra neighbourhoods for $infty$. One can easily check that $alpha(X)$ is then compact.



        The "closed" is needed in general because if e.g. $X$ is not Hausdorff it could have some compact subset $K$ which is not closed, and then (if we were to omit the closed condition) $(Xsetminus K) cup {infty}$ would be open while its intersection with $X$ would be $Xsetminus K$, which was not open, so if we left out the closed condition $X$ would not have the same topology as a subspace of $alpha(X)$ as originally, going against the idea of an extension/compactification: we want to embed $X$ in a larger space with better properties, so in the larger space it should be a subspace with the same topology that it had originally.



        If we want $Y = alpha(X)$ to be Hausdorff, (so in particular $X$ should then be Hausdorff, as a subspace of $Y$) we need to be able to separate $infty$ from every point $x$ in $X$. As a neighbourhood of $infty$ is of the form ${infty} cup X setminus C$, with $C$ compact and closed, every point $x$ should then have a neighbourhood that sits inside a compact closed set, i.e. $X$ must be locally compact.



        So $alpha(X)$ can always be defined such that $alpha(X)setminus X$ is a point and $X$ is a subspace of $alpha(X)$ and it is always compact (regardless of $X$) but $alpha(X)$ is Hausdorff iff $X$ is locally compact and Hausdorff. A special case is when $X$ is already Hausdorff and compact, in which case we add an isolated point $infty$ (as $X$ can be taken as $C$, a compact closed subset) and we get that $X$ is not dense in $alpha(X)$.



        Normally we only consider Hausdorff compactifications and in that case the local compactness is needed for the Hausdorffness of the construction $alpha(X)$. And also because then $X$ is an open subset of a compact Hausdorff space and thus locally compact for that reason.







        share|cite|improve this answer














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        edited 2 days ago

























        answered Dec 26 at 12:28









        Henno Brandsma

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