A space is Hausdorff iff…












5












$begingroup$


Here's an exercise from Marco Manetti's "Topologia" book (ex. 3.59 in the italian version) that i'm stuck on:




Prove that a topological space $X$ is $mathrm{T2}iff{x}=bigcaplimits_{Uin,mathcal{I}(x)}kern{-7pt}overline{U};;;forall xin X$




Partial proof:
$(Rightarrow)$ Let $xin X$, suppose that $yin X$ and $xneq y$. They have disjoint neighborhoods because $X$ is $mathrm{T2}$. So $exists A,Bsubset X$ open and disjoint, such that $xin A$ and $yin B$, and as they're disjoint $Asubset B^{kern{1pt}mathrm{c}}$ and $B^{kern{1pt}mathrm{c}}$ is a closed neighborhood of $x$, choosing such a $B_y$ for each $yneq x$ yields a family of closed neighborhoods of $x$, and $bigcaplimits_{yneq x}kern{-3pt}overline{B^{kern{1pt}mathrm{c}}_y}=bigcaplimits_{yneq x}kern{-3pt}B^{kern{1pt}mathrm{c}}_y={x}$.



$(Leftarrow)$ My initial idea was to prove that if in a space $X$ with the property on the right we assume that given a pair of points $x,yin X$ they have no $Ainmathcal{I}(x),Binmathcal{I}(y)$ open and disjoint we have a contradiction, but all my ideas seemed ineffective, how should i proceed?






general-topology separation-axioms







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    Here's an exercise from Marco Manetti's "Topologia" book (ex. 3.59 in the italian version) that i'm stuck on:




    Prove that a topological space $X$ is $mathrm{T2}iff{x}=bigcaplimits_{Uin,mathcal{I}(x)}kern{-7pt}overline{U};;;forall xin X$




    Partial proof:
    $(Rightarrow)$ Let $xin X$, suppose that $yin X$ and $xneq y$. They have disjoint neighborhoods because $X$ is $mathrm{T2}$. So $exists A,Bsubset X$ open and disjoint, such that $xin A$ and $yin B$, and as they're disjoint $Asubset B^{kern{1pt}mathrm{c}}$ and $B^{kern{1pt}mathrm{c}}$ is a closed neighborhood of $x$, choosing such a $B_y$ for each $yneq x$ yields a family of closed neighborhoods of $x$, and $bigcaplimits_{yneq x}kern{-3pt}overline{B^{kern{1pt}mathrm{c}}_y}=bigcaplimits_{yneq x}kern{-3pt}B^{kern{1pt}mathrm{c}}_y={x}$.



    $(Leftarrow)$ My initial idea was to prove that if in a space $X$ with the property on the right we assume that given a pair of points $x,yin X$ they have no $Ainmathcal{I}(x),Binmathcal{I}(y)$ open and disjoint we have a contradiction, but all my ideas seemed ineffective, how should i proceed?






    general-topology separation-axioms







    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      Here's an exercise from Marco Manetti's "Topologia" book (ex. 3.59 in the italian version) that i'm stuck on:




      Prove that a topological space $X$ is $mathrm{T2}iff{x}=bigcaplimits_{Uin,mathcal{I}(x)}kern{-7pt}overline{U};;;forall xin X$




      Partial proof:
      $(Rightarrow)$ Let $xin X$, suppose that $yin X$ and $xneq y$. They have disjoint neighborhoods because $X$ is $mathrm{T2}$. So $exists A,Bsubset X$ open and disjoint, such that $xin A$ and $yin B$, and as they're disjoint $Asubset B^{kern{1pt}mathrm{c}}$ and $B^{kern{1pt}mathrm{c}}$ is a closed neighborhood of $x$, choosing such a $B_y$ for each $yneq x$ yields a family of closed neighborhoods of $x$, and $bigcaplimits_{yneq x}kern{-3pt}overline{B^{kern{1pt}mathrm{c}}_y}=bigcaplimits_{yneq x}kern{-3pt}B^{kern{1pt}mathrm{c}}_y={x}$.



      $(Leftarrow)$ My initial idea was to prove that if in a space $X$ with the property on the right we assume that given a pair of points $x,yin X$ they have no $Ainmathcal{I}(x),Binmathcal{I}(y)$ open and disjoint we have a contradiction, but all my ideas seemed ineffective, how should i proceed?






      general-topology separation-axioms







      share|cite|improve this question











      $endgroup$




      Here's an exercise from Marco Manetti's "Topologia" book (ex. 3.59 in the italian version) that i'm stuck on:




      Prove that a topological space $X$ is $mathrm{T2}iff{x}=bigcaplimits_{Uin,mathcal{I}(x)}kern{-7pt}overline{U};;;forall xin X$




      Partial proof:
      $(Rightarrow)$ Let $xin X$, suppose that $yin X$ and $xneq y$. They have disjoint neighborhoods because $X$ is $mathrm{T2}$. So $exists A,Bsubset X$ open and disjoint, such that $xin A$ and $yin B$, and as they're disjoint $Asubset B^{kern{1pt}mathrm{c}}$ and $B^{kern{1pt}mathrm{c}}$ is a closed neighborhood of $x$, choosing such a $B_y$ for each $yneq x$ yields a family of closed neighborhoods of $x$, and $bigcaplimits_{yneq x}kern{-3pt}overline{B^{kern{1pt}mathrm{c}}_y}=bigcaplimits_{yneq x}kern{-3pt}B^{kern{1pt}mathrm{c}}_y={x}$.



      $(Leftarrow)$ My initial idea was to prove that if in a space $X$ with the property on the right we assume that given a pair of points $x,yin X$ they have no $Ainmathcal{I}(x),Binmathcal{I}(y)$ open and disjoint we have a contradiction, but all my ideas seemed ineffective, how should i proceed?






      general-topology separation-axioms




      general-topology separation-axioms






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 27 '18 at 21:39







      Ladooscuro

















      asked Dec 27 '18 at 20:12









      LadooscuroLadooscuro

      284




      284






















          2 Answers
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          active

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          3












          $begingroup$

          Suppose that $X$ is $T_2$. If $x,yin X$, there are $Uinmathcal{I}(x)$ and $Vinmathcal{I}(Y)$ such that $Ucap V=emptyset$. Let $O$ be an open set such that $yin O$ and $Osubset V$. Then $O^complement$ is a closed set that containes $U$. Therefore, $overline Usubset O^complement$ and therefore $ynotinoverline U$. Sinse this occurs for every $yin X$, $bigcap_{Uinmathcal{I}(x)}overline{U}={x}$.



          Now, suppose that $X$ is not $T_2$. Take $x,yin X$ such that, for any $Uinmathcal{I}(x)$ and any $Vinmathcal{I}(y)$, $Ucap Vneqemptyset$. It's not hard to prove that $yinbigcap_{Uinmathcal{I}(x)}overline{U}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the suggestion!
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 20:35










          • $begingroup$
            Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 20:45










          • $begingroup$
            Indeed it does.
            $endgroup$
            – José Carlos Santos
            Dec 27 '18 at 20:45



















          3












          $begingroup$

          Starting the proof with “let $x,yin X$” is not the right way.



          Suppose $X$ is Hausdorff and let $xin X$. Suppose $yin X$ and $yne x$. Then there exist $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$ with $Ucap V=emptyset$; in particular $ynotinbar{U}$. Thus $ynotinbigcap_{Uinmathcal{I}(x)}bar{U}$.



          Suppose $X$ is not Hausdorff and let $x,yin X$, with $xne y$, such that $Ucap Vneemptyset$, for every $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$. In particular, $yinbar{U}$, for every $Uinmathcal{I}(x)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Care to explain why that's not the right way? except for not assuming x neq y
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 21:24






          • 1




            $begingroup$
            @Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
            $endgroup$
            – egreg
            Dec 27 '18 at 21:36










          • $begingroup$
            Thanks, i'll keep it in mind
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 21:37











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          $begingroup$

          Suppose that $X$ is $T_2$. If $x,yin X$, there are $Uinmathcal{I}(x)$ and $Vinmathcal{I}(Y)$ such that $Ucap V=emptyset$. Let $O$ be an open set such that $yin O$ and $Osubset V$. Then $O^complement$ is a closed set that containes $U$. Therefore, $overline Usubset O^complement$ and therefore $ynotinoverline U$. Sinse this occurs for every $yin X$, $bigcap_{Uinmathcal{I}(x)}overline{U}={x}$.



          Now, suppose that $X$ is not $T_2$. Take $x,yin X$ such that, for any $Uinmathcal{I}(x)$ and any $Vinmathcal{I}(y)$, $Ucap Vneqemptyset$. It's not hard to prove that $yinbigcap_{Uinmathcal{I}(x)}overline{U}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the suggestion!
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 20:35










          • $begingroup$
            Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 20:45










          • $begingroup$
            Indeed it does.
            $endgroup$
            – José Carlos Santos
            Dec 27 '18 at 20:45
















          3












          $begingroup$

          Suppose that $X$ is $T_2$. If $x,yin X$, there are $Uinmathcal{I}(x)$ and $Vinmathcal{I}(Y)$ such that $Ucap V=emptyset$. Let $O$ be an open set such that $yin O$ and $Osubset V$. Then $O^complement$ is a closed set that containes $U$. Therefore, $overline Usubset O^complement$ and therefore $ynotinoverline U$. Sinse this occurs for every $yin X$, $bigcap_{Uinmathcal{I}(x)}overline{U}={x}$.



          Now, suppose that $X$ is not $T_2$. Take $x,yin X$ such that, for any $Uinmathcal{I}(x)$ and any $Vinmathcal{I}(y)$, $Ucap Vneqemptyset$. It's not hard to prove that $yinbigcap_{Uinmathcal{I}(x)}overline{U}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the suggestion!
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 20:35










          • $begingroup$
            Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 20:45










          • $begingroup$
            Indeed it does.
            $endgroup$
            – José Carlos Santos
            Dec 27 '18 at 20:45














          3












          3








          3





          $begingroup$

          Suppose that $X$ is $T_2$. If $x,yin X$, there are $Uinmathcal{I}(x)$ and $Vinmathcal{I}(Y)$ such that $Ucap V=emptyset$. Let $O$ be an open set such that $yin O$ and $Osubset V$. Then $O^complement$ is a closed set that containes $U$. Therefore, $overline Usubset O^complement$ and therefore $ynotinoverline U$. Sinse this occurs for every $yin X$, $bigcap_{Uinmathcal{I}(x)}overline{U}={x}$.



          Now, suppose that $X$ is not $T_2$. Take $x,yin X$ such that, for any $Uinmathcal{I}(x)$ and any $Vinmathcal{I}(y)$, $Ucap Vneqemptyset$. It's not hard to prove that $yinbigcap_{Uinmathcal{I}(x)}overline{U}$.






          share|cite|improve this answer









          $endgroup$



          Suppose that $X$ is $T_2$. If $x,yin X$, there are $Uinmathcal{I}(x)$ and $Vinmathcal{I}(Y)$ such that $Ucap V=emptyset$. Let $O$ be an open set such that $yin O$ and $Osubset V$. Then $O^complement$ is a closed set that containes $U$. Therefore, $overline Usubset O^complement$ and therefore $ynotinoverline U$. Sinse this occurs for every $yin X$, $bigcap_{Uinmathcal{I}(x)}overline{U}={x}$.



          Now, suppose that $X$ is not $T_2$. Take $x,yin X$ such that, for any $Uinmathcal{I}(x)$ and any $Vinmathcal{I}(y)$, $Ucap Vneqemptyset$. It's not hard to prove that $yinbigcap_{Uinmathcal{I}(x)}overline{U}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 20:22









          José Carlos SantosJosé Carlos Santos

          168k22132236




          168k22132236












          • $begingroup$
            Thanks for the suggestion!
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 20:35










          • $begingroup$
            Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 20:45










          • $begingroup$
            Indeed it does.
            $endgroup$
            – José Carlos Santos
            Dec 27 '18 at 20:45


















          • $begingroup$
            Thanks for the suggestion!
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 20:35










          • $begingroup$
            Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 20:45










          • $begingroup$
            Indeed it does.
            $endgroup$
            – José Carlos Santos
            Dec 27 '18 at 20:45
















          $begingroup$
          Thanks for the suggestion!
          $endgroup$
          – Ladooscuro
          Dec 27 '18 at 20:35




          $begingroup$
          Thanks for the suggestion!
          $endgroup$
          – Ladooscuro
          Dec 27 '18 at 20:35












          $begingroup$
          Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
          $endgroup$
          – Ladooscuro
          Dec 27 '18 at 20:45




          $begingroup$
          Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
          $endgroup$
          – Ladooscuro
          Dec 27 '18 at 20:45












          $begingroup$
          Indeed it does.
          $endgroup$
          – José Carlos Santos
          Dec 27 '18 at 20:45




          $begingroup$
          Indeed it does.
          $endgroup$
          – José Carlos Santos
          Dec 27 '18 at 20:45











          3












          $begingroup$

          Starting the proof with “let $x,yin X$” is not the right way.



          Suppose $X$ is Hausdorff and let $xin X$. Suppose $yin X$ and $yne x$. Then there exist $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$ with $Ucap V=emptyset$; in particular $ynotinbar{U}$. Thus $ynotinbigcap_{Uinmathcal{I}(x)}bar{U}$.



          Suppose $X$ is not Hausdorff and let $x,yin X$, with $xne y$, such that $Ucap Vneemptyset$, for every $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$. In particular, $yinbar{U}$, for every $Uinmathcal{I}(x)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Care to explain why that's not the right way? except for not assuming x neq y
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 21:24






          • 1




            $begingroup$
            @Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
            $endgroup$
            – egreg
            Dec 27 '18 at 21:36










          • $begingroup$
            Thanks, i'll keep it in mind
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 21:37
















          3












          $begingroup$

          Starting the proof with “let $x,yin X$” is not the right way.



          Suppose $X$ is Hausdorff and let $xin X$. Suppose $yin X$ and $yne x$. Then there exist $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$ with $Ucap V=emptyset$; in particular $ynotinbar{U}$. Thus $ynotinbigcap_{Uinmathcal{I}(x)}bar{U}$.



          Suppose $X$ is not Hausdorff and let $x,yin X$, with $xne y$, such that $Ucap Vneemptyset$, for every $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$. In particular, $yinbar{U}$, for every $Uinmathcal{I}(x)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Care to explain why that's not the right way? except for not assuming x neq y
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 21:24






          • 1




            $begingroup$
            @Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
            $endgroup$
            – egreg
            Dec 27 '18 at 21:36










          • $begingroup$
            Thanks, i'll keep it in mind
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 21:37














          3












          3








          3





          $begingroup$

          Starting the proof with “let $x,yin X$” is not the right way.



          Suppose $X$ is Hausdorff and let $xin X$. Suppose $yin X$ and $yne x$. Then there exist $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$ with $Ucap V=emptyset$; in particular $ynotinbar{U}$. Thus $ynotinbigcap_{Uinmathcal{I}(x)}bar{U}$.



          Suppose $X$ is not Hausdorff and let $x,yin X$, with $xne y$, such that $Ucap Vneemptyset$, for every $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$. In particular, $yinbar{U}$, for every $Uinmathcal{I}(x)$.






          share|cite|improve this answer









          $endgroup$



          Starting the proof with “let $x,yin X$” is not the right way.



          Suppose $X$ is Hausdorff and let $xin X$. Suppose $yin X$ and $yne x$. Then there exist $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$ with $Ucap V=emptyset$; in particular $ynotinbar{U}$. Thus $ynotinbigcap_{Uinmathcal{I}(x)}bar{U}$.



          Suppose $X$ is not Hausdorff and let $x,yin X$, with $xne y$, such that $Ucap Vneemptyset$, for every $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$. In particular, $yinbar{U}$, for every $Uinmathcal{I}(x)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 21:16









          egregegreg

          184k1486205




          184k1486205












          • $begingroup$
            Care to explain why that's not the right way? except for not assuming x neq y
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 21:24






          • 1




            $begingroup$
            @Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
            $endgroup$
            – egreg
            Dec 27 '18 at 21:36










          • $begingroup$
            Thanks, i'll keep it in mind
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 21:37


















          • $begingroup$
            Care to explain why that's not the right way? except for not assuming x neq y
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 21:24






          • 1




            $begingroup$
            @Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
            $endgroup$
            – egreg
            Dec 27 '18 at 21:36










          • $begingroup$
            Thanks, i'll keep it in mind
            $endgroup$
            – Ladooscuro
            Dec 27 '18 at 21:37
















          $begingroup$
          Care to explain why that's not the right way? except for not assuming x neq y
          $endgroup$
          – Ladooscuro
          Dec 27 '18 at 21:24




          $begingroup$
          Care to explain why that's not the right way? except for not assuming x neq y
          $endgroup$
          – Ladooscuro
          Dec 27 '18 at 21:24




          1




          1




          $begingroup$
          @Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
          $endgroup$
          – egreg
          Dec 27 '18 at 21:36




          $begingroup$
          @Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
          $endgroup$
          – egreg
          Dec 27 '18 at 21:36












          $begingroup$
          Thanks, i'll keep it in mind
          $endgroup$
          – Ladooscuro
          Dec 27 '18 at 21:37




          $begingroup$
          Thanks, i'll keep it in mind
          $endgroup$
          – Ladooscuro
          Dec 27 '18 at 21:37


















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          Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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          Dieringhausen

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          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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