A space is Hausdorff iff…












4














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









New contributor




Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

























    4














    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









    New contributor




    Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      4












      4








      4







      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









      New contributor




      Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      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









      New contributor




      Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago





















      New contributor




      Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 4 hours ago









      Ladooscuro

      234




      234




      New contributor




      Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes


















          3














          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





















          • Thanks for the suggestion!
            – Ladooscuro
            3 hours ago










          • 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"
            – Ladooscuro
            3 hours ago










          • Indeed it does.
            – José Carlos Santos
            3 hours ago



















          1














          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





















          • Care to explain why that's not the right way? except for not assuming x neq y
            – Ladooscuro
            3 hours ago






          • 1




            @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.
            – egreg
            2 hours ago










          • Thanks, i'll keep it in mind
            – Ladooscuro
            2 hours ago











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });






          Ladooscuro is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054317%2fa-space-is-hausdorff-iff%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          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





















          • Thanks for the suggestion!
            – Ladooscuro
            3 hours ago










          • 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"
            – Ladooscuro
            3 hours ago










          • Indeed it does.
            – José Carlos Santos
            3 hours ago
















          3














          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





















          • Thanks for the suggestion!
            – Ladooscuro
            3 hours ago










          • 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"
            – Ladooscuro
            3 hours ago










          • Indeed it does.
            – José Carlos Santos
            3 hours ago














          3












          3








          3






          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












          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 4 hours ago









          José Carlos Santos

          149k22117219




          149k22117219












          • Thanks for the suggestion!
            – Ladooscuro
            3 hours ago










          • 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"
            – Ladooscuro
            3 hours ago










          • Indeed it does.
            – José Carlos Santos
            3 hours ago


















          • Thanks for the suggestion!
            – Ladooscuro
            3 hours ago










          • 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"
            – Ladooscuro
            3 hours ago










          • Indeed it does.
            – José Carlos Santos
            3 hours ago
















          Thanks for the suggestion!
          – Ladooscuro
          3 hours ago




          Thanks for the suggestion!
          – Ladooscuro
          3 hours ago












          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"
          – Ladooscuro
          3 hours ago




          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"
          – Ladooscuro
          3 hours ago












          Indeed it does.
          – José Carlos Santos
          3 hours ago




          Indeed it does.
          – José Carlos Santos
          3 hours ago











          1














          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





















          • Care to explain why that's not the right way? except for not assuming x neq y
            – Ladooscuro
            3 hours ago






          • 1




            @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.
            – egreg
            2 hours ago










          • Thanks, i'll keep it in mind
            – Ladooscuro
            2 hours ago
















          1














          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





















          • Care to explain why that's not the right way? except for not assuming x neq y
            – Ladooscuro
            3 hours ago






          • 1




            @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.
            – egreg
            2 hours ago










          • Thanks, i'll keep it in mind
            – Ladooscuro
            2 hours ago














          1












          1








          1






          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












          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 3 hours ago









          egreg

          177k1484200




          177k1484200












          • Care to explain why that's not the right way? except for not assuming x neq y
            – Ladooscuro
            3 hours ago






          • 1




            @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.
            – egreg
            2 hours ago










          • Thanks, i'll keep it in mind
            – Ladooscuro
            2 hours ago


















          • Care to explain why that's not the right way? except for not assuming x neq y
            – Ladooscuro
            3 hours ago






          • 1




            @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.
            – egreg
            2 hours ago










          • Thanks, i'll keep it in mind
            – Ladooscuro
            2 hours ago
















          Care to explain why that's not the right way? except for not assuming x neq y
          – Ladooscuro
          3 hours ago




          Care to explain why that's not the right way? except for not assuming x neq y
          – Ladooscuro
          3 hours ago




          1




          1




          @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.
          – egreg
          2 hours ago




          @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.
          – egreg
          2 hours ago












          Thanks, i'll keep it in mind
          – Ladooscuro
          2 hours ago




          Thanks, i'll keep it in mind
          – Ladooscuro
          2 hours ago










          Ladooscuro is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          Ladooscuro is a new contributor. Be nice, and check out our Code of Conduct.













          Ladooscuro is a new contributor. Be nice, and check out our Code of Conduct.












          Ladooscuro is a new contributor. Be nice, and check out our Code of Conduct.
















          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054317%2fa-space-is-hausdorff-iff%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          404 Error Contact Form 7 ajax form submitting

          Image
          0 The contact form does not work, and affects loading after one try. The chrome console shows the error message: POST http://example.com/wp-json/contact-form-7/v1/contact-forms/50/feedback 404 (Not Found) send @ jquery.js?ver=1.12.4:4 ajax @ jquery.js?ver=1.12.4:4 wpcf7.submit @ scripts.js?ver=5.0.3:346 (anonymous) @ scripts.js?ver=5.0.3:53 dispatch @ jquery.js?ver=1.12.4:3 r.handle @ jquery.js?ver=1.12.4:3 How to solve this issue? thanks in advance javascript wordpress contact-form-7 share | improve this question edited Jul 17 '18 at 5:39 Kiran Shahi 4,745 3 16 43 ...
          Read more

          How to know if a Active Directory user can login interactively

          Image
          1 I would like to know if and how is it possible to know if a AD user can login interactively (on a server) in a domain. I need to know if I can find it out using a LDAP search. Thanks. windows active-directory login user-permissions user-profile share | improve this question edited 1 hour ago yagmoth555 ♦ 11.6k 3 17 42 asked 1 hour ago Luigi Luigi 6 1 New contributor ...
          Read more

          Refactoring coordinates for Minecraft Pi buildings written in Python

          Image
          0 $begingroup$ I'm fairly new to coding and I wrote this script in Python for a house with a roof in Minecraft Pi. I've found different ways to write the coordinates to make the code base shorter, but I'm wondering if it is possible for me to refactor it any further so I don't have to list each coordinate line by line. import mcpi.minecraft as minecraft import mcpi.block as block mc = minecraft.Minecraft.create() p = mc.player.getTilePos() mc.setBlocks(p.x+1, p.y, p.z+1, #Main structure starting coordinates (DO NOT ALTER) p.x+19, p.y+4, p.z+6, #Structure size/coordinates (Alters cube size) block.STONE_BRICK) mc.setBlocks(p.x+2, p.y+1, p.z+2, #Hollowing out main structure interior p.x+18, p.y+3, p.z+5, block.AIR) mc.setBlocks(p.x, p.y...
          Read more