On the definition of locally connectedness.












1














Why don’t we just define the locally connectedness the same way we define locally compactness, that is, every point has a connected neighborhood?



On the wiki page the weak locally connectedness and connectedness are proved to be “almost identical”, but it does not mention the motivation of the definition.






general-topology







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  • No answer to your question, but relevant for the concept of local (path) connectedness: math.stackexchange.com/q/2999685
    – Paul Frost
    Dec 4 '18 at 22:44
















1














Why don’t we just define the locally connectedness the same way we define locally compactness, that is, every point has a connected neighborhood?



On the wiki page the weak locally connectedness and connectedness are proved to be “almost identical”, but it does not mention the motivation of the definition.






general-topology







share|cite|improve this question






















  • No answer to your question, but relevant for the concept of local (path) connectedness: math.stackexchange.com/q/2999685
    – Paul Frost
    Dec 4 '18 at 22:44














1












1








1


0





Why don’t we just define the locally connectedness the same way we define locally compactness, that is, every point has a connected neighborhood?



On the wiki page the weak locally connectedness and connectedness are proved to be “almost identical”, but it does not mention the motivation of the definition.






general-topology







share|cite|improve this question













Why don’t we just define the locally connectedness the same way we define locally compactness, that is, every point has a connected neighborhood?



On the wiki page the weak locally connectedness and connectedness are proved to be “almost identical”, but it does not mention the motivation of the definition.






general-topology




general-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 21:18









William SunWilliam Sun

471111




471111












  • No answer to your question, but relevant for the concept of local (path) connectedness: math.stackexchange.com/q/2999685
    – Paul Frost
    Dec 4 '18 at 22:44


















  • No answer to your question, but relevant for the concept of local (path) connectedness: math.stackexchange.com/q/2999685
    – Paul Frost
    Dec 4 '18 at 22:44
















No answer to your question, but relevant for the concept of local (path) connectedness: math.stackexchange.com/q/2999685
– Paul Frost
Dec 4 '18 at 22:44




No answer to your question, but relevant for the concept of local (path) connectedness: math.stackexchange.com/q/2999685
– Paul Frost
Dec 4 '18 at 22:44










3 Answers
3






active

oldest

votes


















2














Local compactness is an easier case: we mostly consider that property in the context of Hausdorff spaces, and for Hausdorff spaces it's equivalent to demand one compact neighbourhood or a local base of compact neighbourhoods in the definition.



The latter defintition of localisation of a property is more intuitive: however "close" to a point we are, we "see" compactness. In proofs we often need a compact neighbourhood inside some open set we are working in, e.g. and the local base variant gives us lots of options. One defines certain notions to prove results, and we get better results with a local base variant.



One one the reasons to study local (path-)connectedness is to make a distinction between different kinds of connected spaces. The "one connected neighbourhood" definition would kill that idea as then a connected space is locally connected automatically, chosing $X$ as the connected neighbourhood. There even are more refined notions of local connectedness like "connected im Kleinen" (somewhat old-fashioned nowadays) to distinguish among locally connected spaces in the strong sense.






share|cite|improve this answer































    1














    I think the crucial fact is that (reference)




    A space is locally connected if and only if for every open set U, the connected components of U (in the subspace topology) are open.




    With the other definition (the weak one), this would be false.






    share|cite|improve this answer





























      1














      Consider a space like the closed topologist's sine curve, i.e., the closure $mbox{cl}(G)$ where $G$ is the graph of $sin(1/x)$ for $0 < x le pi$. This space is connected, so every point has a connected neighbourhood, namely the whole space, but has points (in this case the points that lie on the $y$-axis) for which all sufficiently small neighbourhoods are disconnected.






      share|cite|improve this answer























      • I don’t quite get the reason why containing a smaller disconnected neighborhood should be avoided. Consider the case where we choose a point $xinmathbb{R}$, then trivially any neighborhood of $x$ contains a disconnected neighborhood.
        – William Sun
        Dec 4 '18 at 21:39










      • Sorry. I said that wrong. I've fixed the answer.
        – Rob Arthan
        Dec 4 '18 at 21:40













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      Local compactness is an easier case: we mostly consider that property in the context of Hausdorff spaces, and for Hausdorff spaces it's equivalent to demand one compact neighbourhood or a local base of compact neighbourhoods in the definition.



      The latter defintition of localisation of a property is more intuitive: however "close" to a point we are, we "see" compactness. In proofs we often need a compact neighbourhood inside some open set we are working in, e.g. and the local base variant gives us lots of options. One defines certain notions to prove results, and we get better results with a local base variant.



      One one the reasons to study local (path-)connectedness is to make a distinction between different kinds of connected spaces. The "one connected neighbourhood" definition would kill that idea as then a connected space is locally connected automatically, chosing $X$ as the connected neighbourhood. There even are more refined notions of local connectedness like "connected im Kleinen" (somewhat old-fashioned nowadays) to distinguish among locally connected spaces in the strong sense.






      share|cite|improve this answer




























        2














        Local compactness is an easier case: we mostly consider that property in the context of Hausdorff spaces, and for Hausdorff spaces it's equivalent to demand one compact neighbourhood or a local base of compact neighbourhoods in the definition.



        The latter defintition of localisation of a property is more intuitive: however "close" to a point we are, we "see" compactness. In proofs we often need a compact neighbourhood inside some open set we are working in, e.g. and the local base variant gives us lots of options. One defines certain notions to prove results, and we get better results with a local base variant.



        One one the reasons to study local (path-)connectedness is to make a distinction between different kinds of connected spaces. The "one connected neighbourhood" definition would kill that idea as then a connected space is locally connected automatically, chosing $X$ as the connected neighbourhood. There even are more refined notions of local connectedness like "connected im Kleinen" (somewhat old-fashioned nowadays) to distinguish among locally connected spaces in the strong sense.






        share|cite|improve this answer


























          2












          2








          2






          Local compactness is an easier case: we mostly consider that property in the context of Hausdorff spaces, and for Hausdorff spaces it's equivalent to demand one compact neighbourhood or a local base of compact neighbourhoods in the definition.



          The latter defintition of localisation of a property is more intuitive: however "close" to a point we are, we "see" compactness. In proofs we often need a compact neighbourhood inside some open set we are working in, e.g. and the local base variant gives us lots of options. One defines certain notions to prove results, and we get better results with a local base variant.



          One one the reasons to study local (path-)connectedness is to make a distinction between different kinds of connected spaces. The "one connected neighbourhood" definition would kill that idea as then a connected space is locally connected automatically, chosing $X$ as the connected neighbourhood. There even are more refined notions of local connectedness like "connected im Kleinen" (somewhat old-fashioned nowadays) to distinguish among locally connected spaces in the strong sense.






          share|cite|improve this answer














          Local compactness is an easier case: we mostly consider that property in the context of Hausdorff spaces, and for Hausdorff spaces it's equivalent to demand one compact neighbourhood or a local base of compact neighbourhoods in the definition.



          The latter defintition of localisation of a property is more intuitive: however "close" to a point we are, we "see" compactness. In proofs we often need a compact neighbourhood inside some open set we are working in, e.g. and the local base variant gives us lots of options. One defines certain notions to prove results, and we get better results with a local base variant.



          One one the reasons to study local (path-)connectedness is to make a distinction between different kinds of connected spaces. The "one connected neighbourhood" definition would kill that idea as then a connected space is locally connected automatically, chosing $X$ as the connected neighbourhood. There even are more refined notions of local connectedness like "connected im Kleinen" (somewhat old-fashioned nowadays) to distinguish among locally connected spaces in the strong sense.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '18 at 2:55

























          answered Dec 4 '18 at 21:38









          Henno BrandsmaHenno Brandsma

          105k347114




          105k347114























              1














              I think the crucial fact is that (reference)




              A space is locally connected if and only if for every open set U, the connected components of U (in the subspace topology) are open.




              With the other definition (the weak one), this would be false.






              share|cite|improve this answer


























                1














                I think the crucial fact is that (reference)




                A space is locally connected if and only if for every open set U, the connected components of U (in the subspace topology) are open.




                With the other definition (the weak one), this would be false.






                share|cite|improve this answer
























                  1












                  1








                  1






                  I think the crucial fact is that (reference)




                  A space is locally connected if and only if for every open set U, the connected components of U (in the subspace topology) are open.




                  With the other definition (the weak one), this would be false.






                  share|cite|improve this answer












                  I think the crucial fact is that (reference)




                  A space is locally connected if and only if for every open set U, the connected components of U (in the subspace topology) are open.




                  With the other definition (the weak one), this would be false.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 21:26









                  FedericoFederico

                  4,829514




                  4,829514























                      1














                      Consider a space like the closed topologist's sine curve, i.e., the closure $mbox{cl}(G)$ where $G$ is the graph of $sin(1/x)$ for $0 < x le pi$. This space is connected, so every point has a connected neighbourhood, namely the whole space, but has points (in this case the points that lie on the $y$-axis) for which all sufficiently small neighbourhoods are disconnected.






                      share|cite|improve this answer























                      • I don’t quite get the reason why containing a smaller disconnected neighborhood should be avoided. Consider the case where we choose a point $xinmathbb{R}$, then trivially any neighborhood of $x$ contains a disconnected neighborhood.
                        – William Sun
                        Dec 4 '18 at 21:39










                      • Sorry. I said that wrong. I've fixed the answer.
                        – Rob Arthan
                        Dec 4 '18 at 21:40


















                      1














                      Consider a space like the closed topologist's sine curve, i.e., the closure $mbox{cl}(G)$ where $G$ is the graph of $sin(1/x)$ for $0 < x le pi$. This space is connected, so every point has a connected neighbourhood, namely the whole space, but has points (in this case the points that lie on the $y$-axis) for which all sufficiently small neighbourhoods are disconnected.






                      share|cite|improve this answer























                      • I don’t quite get the reason why containing a smaller disconnected neighborhood should be avoided. Consider the case where we choose a point $xinmathbb{R}$, then trivially any neighborhood of $x$ contains a disconnected neighborhood.
                        – William Sun
                        Dec 4 '18 at 21:39










                      • Sorry. I said that wrong. I've fixed the answer.
                        – Rob Arthan
                        Dec 4 '18 at 21:40
















                      1












                      1








                      1






                      Consider a space like the closed topologist's sine curve, i.e., the closure $mbox{cl}(G)$ where $G$ is the graph of $sin(1/x)$ for $0 < x le pi$. This space is connected, so every point has a connected neighbourhood, namely the whole space, but has points (in this case the points that lie on the $y$-axis) for which all sufficiently small neighbourhoods are disconnected.






                      share|cite|improve this answer














                      Consider a space like the closed topologist's sine curve, i.e., the closure $mbox{cl}(G)$ where $G$ is the graph of $sin(1/x)$ for $0 < x le pi$. This space is connected, so every point has a connected neighbourhood, namely the whole space, but has points (in this case the points that lie on the $y$-axis) for which all sufficiently small neighbourhoods are disconnected.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 4 '18 at 21:42

























                      answered Dec 4 '18 at 21:30









                      Rob ArthanRob Arthan

                      29.1k42866




                      29.1k42866












                      • I don’t quite get the reason why containing a smaller disconnected neighborhood should be avoided. Consider the case where we choose a point $xinmathbb{R}$, then trivially any neighborhood of $x$ contains a disconnected neighborhood.
                        – William Sun
                        Dec 4 '18 at 21:39










                      • Sorry. I said that wrong. I've fixed the answer.
                        – Rob Arthan
                        Dec 4 '18 at 21:40




















                      • I don’t quite get the reason why containing a smaller disconnected neighborhood should be avoided. Consider the case where we choose a point $xinmathbb{R}$, then trivially any neighborhood of $x$ contains a disconnected neighborhood.
                        – William Sun
                        Dec 4 '18 at 21:39










                      • Sorry. I said that wrong. I've fixed the answer.
                        – Rob Arthan
                        Dec 4 '18 at 21:40


















                      I don’t quite get the reason why containing a smaller disconnected neighborhood should be avoided. Consider the case where we choose a point $xinmathbb{R}$, then trivially any neighborhood of $x$ contains a disconnected neighborhood.
                      – William Sun
                      Dec 4 '18 at 21:39




                      I don’t quite get the reason why containing a smaller disconnected neighborhood should be avoided. Consider the case where we choose a point $xinmathbb{R}$, then trivially any neighborhood of $x$ contains a disconnected neighborhood.
                      – William Sun
                      Dec 4 '18 at 21:39












                      Sorry. I said that wrong. I've fixed the answer.
                      – Rob Arthan
                      Dec 4 '18 at 21:40






                      Sorry. I said that wrong. I've fixed the answer.
                      – Rob Arthan
                      Dec 4 '18 at 21:40




















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