A cover of Locally connected space with certain compactness property












2












$begingroup$


Suppose $X$ is a locally connected Hausdorff space. If $X$ is $sigma$-compact and locally compact, is it always possible to find a countable set of precompact connected open sets ${U_n}$ (which covers $X$), with the following property?




(*) Any $U_i$ interects at most finitely many $U_j$.




I can produce a precompact connected open cover of $X$ by local compactness and local connectedness, and by $sigma$-compactness, I can choose a countable subcover ${V_n}$.



The problem is how can I meet the additional requirement (*)?



I know that such space must be paracompact but I am not sure if this is related. How can I modify the sets $V_n$ so that the connectedness is preserved?






general-topology compactness connectedness paracompactness







share|cite|improve this question











$endgroup$












  • $begingroup$
    By countable do you mean at most countable or countably infinity? An empty set is open, precompact and connected. Do you mean an open cover? And if possible, could you please add more details on how you came up with the problem? Because by a first look it seems like you are just sticking different notions of compactness together.
    $endgroup$
    – William Sun
    Dec 8 '18 at 4:09










  • $begingroup$
    @WilliamSun I meant at most countable. The construction appears here (page 10, Thm 1.6). Maybe, as the author said, it is a well-known construction.
    $endgroup$
    – Eclipse Sun
    Dec 8 '18 at 4:23
















2












$begingroup$


Suppose $X$ is a locally connected Hausdorff space. If $X$ is $sigma$-compact and locally compact, is it always possible to find a countable set of precompact connected open sets ${U_n}$ (which covers $X$), with the following property?




(*) Any $U_i$ interects at most finitely many $U_j$.




I can produce a precompact connected open cover of $X$ by local compactness and local connectedness, and by $sigma$-compactness, I can choose a countable subcover ${V_n}$.



The problem is how can I meet the additional requirement (*)?



I know that such space must be paracompact but I am not sure if this is related. How can I modify the sets $V_n$ so that the connectedness is preserved?






general-topology compactness connectedness paracompactness







share|cite|improve this question











$endgroup$












  • $begingroup$
    By countable do you mean at most countable or countably infinity? An empty set is open, precompact and connected. Do you mean an open cover? And if possible, could you please add more details on how you came up with the problem? Because by a first look it seems like you are just sticking different notions of compactness together.
    $endgroup$
    – William Sun
    Dec 8 '18 at 4:09










  • $begingroup$
    @WilliamSun I meant at most countable. The construction appears here (page 10, Thm 1.6). Maybe, as the author said, it is a well-known construction.
    $endgroup$
    – Eclipse Sun
    Dec 8 '18 at 4:23














2












2








2





$begingroup$


Suppose $X$ is a locally connected Hausdorff space. If $X$ is $sigma$-compact and locally compact, is it always possible to find a countable set of precompact connected open sets ${U_n}$ (which covers $X$), with the following property?




(*) Any $U_i$ interects at most finitely many $U_j$.




I can produce a precompact connected open cover of $X$ by local compactness and local connectedness, and by $sigma$-compactness, I can choose a countable subcover ${V_n}$.



The problem is how can I meet the additional requirement (*)?



I know that such space must be paracompact but I am not sure if this is related. How can I modify the sets $V_n$ so that the connectedness is preserved?






general-topology compactness connectedness paracompactness







share|cite|improve this question











$endgroup$




Suppose $X$ is a locally connected Hausdorff space. If $X$ is $sigma$-compact and locally compact, is it always possible to find a countable set of precompact connected open sets ${U_n}$ (which covers $X$), with the following property?




(*) Any $U_i$ interects at most finitely many $U_j$.




I can produce a precompact connected open cover of $X$ by local compactness and local connectedness, and by $sigma$-compactness, I can choose a countable subcover ${V_n}$.



The problem is how can I meet the additional requirement (*)?



I know that such space must be paracompact but I am not sure if this is related. How can I modify the sets $V_n$ so that the connectedness is preserved?






general-topology compactness connectedness paracompactness




general-topology compactness connectedness paracompactness






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 5:38









Eric Wofsey

182k13209337




182k13209337










asked Dec 8 '18 at 3:25









Eclipse SunEclipse Sun

6,9841437




6,9841437












  • $begingroup$
    By countable do you mean at most countable or countably infinity? An empty set is open, precompact and connected. Do you mean an open cover? And if possible, could you please add more details on how you came up with the problem? Because by a first look it seems like you are just sticking different notions of compactness together.
    $endgroup$
    – William Sun
    Dec 8 '18 at 4:09










  • $begingroup$
    @WilliamSun I meant at most countable. The construction appears here (page 10, Thm 1.6). Maybe, as the author said, it is a well-known construction.
    $endgroup$
    – Eclipse Sun
    Dec 8 '18 at 4:23


















  • $begingroup$
    By countable do you mean at most countable or countably infinity? An empty set is open, precompact and connected. Do you mean an open cover? And if possible, could you please add more details on how you came up with the problem? Because by a first look it seems like you are just sticking different notions of compactness together.
    $endgroup$
    – William Sun
    Dec 8 '18 at 4:09










  • $begingroup$
    @WilliamSun I meant at most countable. The construction appears here (page 10, Thm 1.6). Maybe, as the author said, it is a well-known construction.
    $endgroup$
    – Eclipse Sun
    Dec 8 '18 at 4:23
















$begingroup$
By countable do you mean at most countable or countably infinity? An empty set is open, precompact and connected. Do you mean an open cover? And if possible, could you please add more details on how you came up with the problem? Because by a first look it seems like you are just sticking different notions of compactness together.
$endgroup$
– William Sun
Dec 8 '18 at 4:09




$begingroup$
By countable do you mean at most countable or countably infinity? An empty set is open, precompact and connected. Do you mean an open cover? And if possible, could you please add more details on how you came up with the problem? Because by a first look it seems like you are just sticking different notions of compactness together.
$endgroup$
– William Sun
Dec 8 '18 at 4:09












$begingroup$
@WilliamSun I meant at most countable. The construction appears here (page 10, Thm 1.6). Maybe, as the author said, it is a well-known construction.
$endgroup$
– Eclipse Sun
Dec 8 '18 at 4:23




$begingroup$
@WilliamSun I meant at most countable. The construction appears here (page 10, Thm 1.6). Maybe, as the author said, it is a well-known construction.
$endgroup$
– Eclipse Sun
Dec 8 '18 at 4:23










1 Answer
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$begingroup$

We will say a set is "good" if it is precompact, connected, and open. Fix a sequence of compact sets $(K_n)$ which cover $X$. Pick finitely many good sets $U_0,dots,U_{n_0}$ which cover $K_0$. Note that $C_0=overline{U_0}$ is compact, so we can also pick finitely many good sets $U_{n_0+1},dots,U_{n_1}$ which cover it.



Now we repeat this process, but make sure none of our new good sets intersect $U_0$. Let $V_0=U_0cupdotscup U_{n_1}$ be the union of all the good sets we've chosen so far. We can pick finitely many good sets $U_{n_1+1},dots,U_{n_2}subseteq Xsetminus C_0$ which cover the compact set $K_1setminus V_0$. We can then let $C_1=overline{U_1}$ and also pick finitely many good sets $U_{n_2+1},dots,U_{n_3}subseteq Xsetminus C_0$ which cover $C_1setminus V_0$.



We continue the process, alternating between picking good sets to cover one of the $K_n$'s, and picking good sets which cover the closure of one of the good sets we chose already. At each stage, we choose all of our new good sets to be disjoint from the closures of the good sets we've covered already. In the end, we get a sequence $(U_n)$ of good sets which cover every $K_n$ and such that only finitely many of them intersect (the closure of) each $U_n$.



(Note that all this argument used about the good sets is that they are a precompact and a basis for the topology. More generally, the same argument shows that given any basis for the topology of a locally compact $sigma$-compact Hausdorff space, there is a countable cover by precompact basis sets such that each one intersects only finitely many others. Taking the basis to consist of all open sets which refine a given open cover, this is one way to prove such a space is paracompact.)






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    $begingroup$

    We will say a set is "good" if it is precompact, connected, and open. Fix a sequence of compact sets $(K_n)$ which cover $X$. Pick finitely many good sets $U_0,dots,U_{n_0}$ which cover $K_0$. Note that $C_0=overline{U_0}$ is compact, so we can also pick finitely many good sets $U_{n_0+1},dots,U_{n_1}$ which cover it.



    Now we repeat this process, but make sure none of our new good sets intersect $U_0$. Let $V_0=U_0cupdotscup U_{n_1}$ be the union of all the good sets we've chosen so far. We can pick finitely many good sets $U_{n_1+1},dots,U_{n_2}subseteq Xsetminus C_0$ which cover the compact set $K_1setminus V_0$. We can then let $C_1=overline{U_1}$ and also pick finitely many good sets $U_{n_2+1},dots,U_{n_3}subseteq Xsetminus C_0$ which cover $C_1setminus V_0$.



    We continue the process, alternating between picking good sets to cover one of the $K_n$'s, and picking good sets which cover the closure of one of the good sets we chose already. At each stage, we choose all of our new good sets to be disjoint from the closures of the good sets we've covered already. In the end, we get a sequence $(U_n)$ of good sets which cover every $K_n$ and such that only finitely many of them intersect (the closure of) each $U_n$.



    (Note that all this argument used about the good sets is that they are a precompact and a basis for the topology. More generally, the same argument shows that given any basis for the topology of a locally compact $sigma$-compact Hausdorff space, there is a countable cover by precompact basis sets such that each one intersects only finitely many others. Taking the basis to consist of all open sets which refine a given open cover, this is one way to prove such a space is paracompact.)






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      We will say a set is "good" if it is precompact, connected, and open. Fix a sequence of compact sets $(K_n)$ which cover $X$. Pick finitely many good sets $U_0,dots,U_{n_0}$ which cover $K_0$. Note that $C_0=overline{U_0}$ is compact, so we can also pick finitely many good sets $U_{n_0+1},dots,U_{n_1}$ which cover it.



      Now we repeat this process, but make sure none of our new good sets intersect $U_0$. Let $V_0=U_0cupdotscup U_{n_1}$ be the union of all the good sets we've chosen so far. We can pick finitely many good sets $U_{n_1+1},dots,U_{n_2}subseteq Xsetminus C_0$ which cover the compact set $K_1setminus V_0$. We can then let $C_1=overline{U_1}$ and also pick finitely many good sets $U_{n_2+1},dots,U_{n_3}subseteq Xsetminus C_0$ which cover $C_1setminus V_0$.



      We continue the process, alternating between picking good sets to cover one of the $K_n$'s, and picking good sets which cover the closure of one of the good sets we chose already. At each stage, we choose all of our new good sets to be disjoint from the closures of the good sets we've covered already. In the end, we get a sequence $(U_n)$ of good sets which cover every $K_n$ and such that only finitely many of them intersect (the closure of) each $U_n$.



      (Note that all this argument used about the good sets is that they are a precompact and a basis for the topology. More generally, the same argument shows that given any basis for the topology of a locally compact $sigma$-compact Hausdorff space, there is a countable cover by precompact basis sets such that each one intersects only finitely many others. Taking the basis to consist of all open sets which refine a given open cover, this is one way to prove such a space is paracompact.)






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        We will say a set is "good" if it is precompact, connected, and open. Fix a sequence of compact sets $(K_n)$ which cover $X$. Pick finitely many good sets $U_0,dots,U_{n_0}$ which cover $K_0$. Note that $C_0=overline{U_0}$ is compact, so we can also pick finitely many good sets $U_{n_0+1},dots,U_{n_1}$ which cover it.



        Now we repeat this process, but make sure none of our new good sets intersect $U_0$. Let $V_0=U_0cupdotscup U_{n_1}$ be the union of all the good sets we've chosen so far. We can pick finitely many good sets $U_{n_1+1},dots,U_{n_2}subseteq Xsetminus C_0$ which cover the compact set $K_1setminus V_0$. We can then let $C_1=overline{U_1}$ and also pick finitely many good sets $U_{n_2+1},dots,U_{n_3}subseteq Xsetminus C_0$ which cover $C_1setminus V_0$.



        We continue the process, alternating between picking good sets to cover one of the $K_n$'s, and picking good sets which cover the closure of one of the good sets we chose already. At each stage, we choose all of our new good sets to be disjoint from the closures of the good sets we've covered already. In the end, we get a sequence $(U_n)$ of good sets which cover every $K_n$ and such that only finitely many of them intersect (the closure of) each $U_n$.



        (Note that all this argument used about the good sets is that they are a precompact and a basis for the topology. More generally, the same argument shows that given any basis for the topology of a locally compact $sigma$-compact Hausdorff space, there is a countable cover by precompact basis sets such that each one intersects only finitely many others. Taking the basis to consist of all open sets which refine a given open cover, this is one way to prove such a space is paracompact.)






        share|cite|improve this answer











        $endgroup$



        We will say a set is "good" if it is precompact, connected, and open. Fix a sequence of compact sets $(K_n)$ which cover $X$. Pick finitely many good sets $U_0,dots,U_{n_0}$ which cover $K_0$. Note that $C_0=overline{U_0}$ is compact, so we can also pick finitely many good sets $U_{n_0+1},dots,U_{n_1}$ which cover it.



        Now we repeat this process, but make sure none of our new good sets intersect $U_0$. Let $V_0=U_0cupdotscup U_{n_1}$ be the union of all the good sets we've chosen so far. We can pick finitely many good sets $U_{n_1+1},dots,U_{n_2}subseteq Xsetminus C_0$ which cover the compact set $K_1setminus V_0$. We can then let $C_1=overline{U_1}$ and also pick finitely many good sets $U_{n_2+1},dots,U_{n_3}subseteq Xsetminus C_0$ which cover $C_1setminus V_0$.



        We continue the process, alternating between picking good sets to cover one of the $K_n$'s, and picking good sets which cover the closure of one of the good sets we chose already. At each stage, we choose all of our new good sets to be disjoint from the closures of the good sets we've covered already. In the end, we get a sequence $(U_n)$ of good sets which cover every $K_n$ and such that only finitely many of them intersect (the closure of) each $U_n$.



        (Note that all this argument used about the good sets is that they are a precompact and a basis for the topology. More generally, the same argument shows that given any basis for the topology of a locally compact $sigma$-compact Hausdorff space, there is a countable cover by precompact basis sets such that each one intersects only finitely many others. Taking the basis to consist of all open sets which refine a given open cover, this is one way to prove such a space is paracompact.)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 8 '18 at 5:44

























        answered Dec 8 '18 at 5:38









        Eric WofseyEric Wofsey

        182k13209337




        182k13209337






























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