Union of connected subsets is connected if intersection is nonempty












20












$begingroup$



Let $mathscr{F}$ be a collection of connected subsets of a metric space $M$ such that $bigcapmathscr{F}neemptyset$. Prove that $bigcupmathscr{F}$ is connected.




If $bigcupmathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty subsets $A,B$. Let $x$ be a point in $bigcapmathscr{F}$. Then either $xin A$ or $xin B$. I don't know where to go from here.










share|cite|improve this question











$endgroup$

















    20












    $begingroup$



    Let $mathscr{F}$ be a collection of connected subsets of a metric space $M$ such that $bigcapmathscr{F}neemptyset$. Prove that $bigcupmathscr{F}$ is connected.




    If $bigcupmathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty subsets $A,B$. Let $x$ be a point in $bigcapmathscr{F}$. Then either $xin A$ or $xin B$. I don't know where to go from here.










    share|cite|improve this question











    $endgroup$















      20












      20








      20


      9



      $begingroup$



      Let $mathscr{F}$ be a collection of connected subsets of a metric space $M$ such that $bigcapmathscr{F}neemptyset$. Prove that $bigcupmathscr{F}$ is connected.




      If $bigcupmathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty subsets $A,B$. Let $x$ be a point in $bigcapmathscr{F}$. Then either $xin A$ or $xin B$. I don't know where to go from here.










      share|cite|improve this question











      $endgroup$





      Let $mathscr{F}$ be a collection of connected subsets of a metric space $M$ such that $bigcapmathscr{F}neemptyset$. Prove that $bigcupmathscr{F}$ is connected.




      If $bigcupmathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty subsets $A,B$. Let $x$ be a point in $bigcapmathscr{F}$. Then either $xin A$ or $xin B$. I don't know where to go from here.







      general-topology connectedness






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 22 '17 at 5:57









      Martin Sleziak

      44.7k9117272




      44.7k9117272










      asked Jun 20 '13 at 0:39









      PJ MillerPJ Miller

      2,58322770




      2,58322770






















          5 Answers
          5






          active

          oldest

          votes


















          27












          $begingroup$

          HINT: You’re actually about halfway there, though you omitted an important qualification of $A$ and $B$: if $bigcupmathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty, relatively open subsets $A$ and $B$. Now fix $xinbigcapmathscr{F}$, and without loss of generality assume that $xin A$. $Bnevarnothing$, so pick any $yin B$. Then there is some $Finmathscr{F}$ such that $yin F$, and of course $xin F$. Thus, $xin Acap F$, and $yin Bcap F$, so $Acap Fnevarnothingne Bcap F$. Why is this a contradiction?






          share|cite|improve this answer









          $endgroup$





















            10












            $begingroup$

            Use that $X$ is connected if and only if the only continuous functions $f:Xto{0,1}$ are constant, where ${0,1}$ is endowed with the discrete topology.



            Now, you know each $F$ in $mathscr F$ is connected. Consider $f:bigcup mathscr Fto{0,1}$, $f$ continuous.



            Take $alpha inbigcapmathscr F$. Look at $f(alpha)$, and at $fmid_{F}:bigcup mathscr Fto{0,1}$ for any $Finmathscr F$.





            Since you mention metric spaces, I am not sure if you know about the first thing I mention, so let's prove it:




            THM Let $(X,mathscr T)$ be a metric (or more generally, a topological) space. Then $X$ is connected if and only if whenever $f:Xto{0,1}$ is continuous, it is constant. The space ${0,1}$ is endowed with the discrete metric (topology), that is, the open sets are $varnothing,{0},{1},{0,1}$.




            P First, suppose $X$ is disconnected, say by $A,B$, so $Acup B=X$ and $Acap B=varnothing$, $A,B$ open. Define $f:Xto{0,1}$ by $$f(x)=begin{cases}1& ; ; xin A\0&; ; xin Bend{cases}$$



            Then $f$ is continuous because $f^{-1}(G)$ is open for any open $G$ in ${0,1}$ (this is simply a case by case verification), yet it is not constant. Now suppose $f:Xto{0,1}$ is continuous but not constant. Set $A={x:f(x)=1}=f^{-1}({1})$ and $B={x:f(x)=0}=f^{-1}({0})$. By hypothesis, $A,Bneq varnothing$. Morover, both are open, since they are the preimage of open sets under a continuous map, and $Acup B=X$ and $Acap B=varnothing$. Thus $X$ is disconnected. $blacktriangle$






            share|cite|improve this answer











            $endgroup$





















              4












              $begingroup$

              I like "clopen" sets... it is a set that is at the same time closed and open. A topological space $X$ is not connected if, and only if, for any point $a in X$, you have a "non-trivial" clopen set $C$ containing $a$. That is, a clopen set such that $emptyset neq C subsetneq X$.



              Take $a in bigcap mathscr{F}$. Then, take a clopen set $C subset bigcup mathscr{F}$ containing $a$. In the relative topology, for any $X in mathscr{F}$, $C cap X$ is a clopen set. Since $X$ is connected, $C cap X = X$. That is, $X subset C$ for every $X in mathscr{F}$. Therefore, $bigcup mathscr{F} subset C$.



              Therefore, $bigcup mathscr{F}$ has no "non-trivial" clopen set, and so, it is connected.





              The proof admits some variants. For example, if you take any non-empty clopen set $C'$, it will intersect some $X in mathscr{F}$. From connectedness, we have that $X subset C'$. In particular, $a in C'$.






              share|cite|improve this answer











              $endgroup$





















                3












                $begingroup$

                The quickest way is using functions from $bigcupmathcal{F}$ to ${0,1}$, However you can do it directly.



                Let $A, B$ be a partition of $bigcupmathcal{F}$ into open, disjoint subsets. WLOG assume that $A$ is non-empty. Then there is some $xin A$ and hence some $Fin mathcal{F}$ so that $xin F$. As $F$ is connected it follows that $Fsubseteq A$, hence $bigcap mathcal{F} subseteq F subseteq A$. As this intersection is non-empty every $Gin mathcal{F}$ has a point in $A$ and by the connectness of each $G$ then $Gsubseteq A$. Hence every member of $mathcal{F}$ is a subset of $A$ and so $B$ is empty.






                share|cite|improve this answer









                $endgroup$





















                  0












                  $begingroup$

                  In the before last post, with the "clopen" sets, I believe there is a detail missing: it is not immediately clear (but true) that if $C$ is a clopen set with respect to the subset topology on $bigcupmathscr{F}$, that then $Ccap X$ is a clopen set in $X$, $Xinmathscr{F}$, with respect to the original topology, which is what you need.






                  share|cite|improve this answer











                  $endgroup$













                    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
                    });


                    }
                    });














                    draft saved

                    draft discarded


















                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f425007%2funion-of-connected-subsets-is-connected-if-intersection-is-nonempty%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown

























                    5 Answers
                    5






                    active

                    oldest

                    votes








                    5 Answers
                    5






                    active

                    oldest

                    votes









                    active

                    oldest

                    votes






                    active

                    oldest

                    votes









                    27












                    $begingroup$

                    HINT: You’re actually about halfway there, though you omitted an important qualification of $A$ and $B$: if $bigcupmathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty, relatively open subsets $A$ and $B$. Now fix $xinbigcapmathscr{F}$, and without loss of generality assume that $xin A$. $Bnevarnothing$, so pick any $yin B$. Then there is some $Finmathscr{F}$ such that $yin F$, and of course $xin F$. Thus, $xin Acap F$, and $yin Bcap F$, so $Acap Fnevarnothingne Bcap F$. Why is this a contradiction?






                    share|cite|improve this answer









                    $endgroup$


















                      27












                      $begingroup$

                      HINT: You’re actually about halfway there, though you omitted an important qualification of $A$ and $B$: if $bigcupmathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty, relatively open subsets $A$ and $B$. Now fix $xinbigcapmathscr{F}$, and without loss of generality assume that $xin A$. $Bnevarnothing$, so pick any $yin B$. Then there is some $Finmathscr{F}$ such that $yin F$, and of course $xin F$. Thus, $xin Acap F$, and $yin Bcap F$, so $Acap Fnevarnothingne Bcap F$. Why is this a contradiction?






                      share|cite|improve this answer









                      $endgroup$
















                        27












                        27








                        27





                        $begingroup$

                        HINT: You’re actually about halfway there, though you omitted an important qualification of $A$ and $B$: if $bigcupmathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty, relatively open subsets $A$ and $B$. Now fix $xinbigcapmathscr{F}$, and without loss of generality assume that $xin A$. $Bnevarnothing$, so pick any $yin B$. Then there is some $Finmathscr{F}$ such that $yin F$, and of course $xin F$. Thus, $xin Acap F$, and $yin Bcap F$, so $Acap Fnevarnothingne Bcap F$. Why is this a contradiction?






                        share|cite|improve this answer









                        $endgroup$



                        HINT: You’re actually about halfway there, though you omitted an important qualification of $A$ and $B$: if $bigcupmathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty, relatively open subsets $A$ and $B$. Now fix $xinbigcapmathscr{F}$, and without loss of generality assume that $xin A$. $Bnevarnothing$, so pick any $yin B$. Then there is some $Finmathscr{F}$ such that $yin F$, and of course $xin F$. Thus, $xin Acap F$, and $yin Bcap F$, so $Acap Fnevarnothingne Bcap F$. Why is this a contradiction?







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jun 20 '13 at 0:46









                        Brian M. ScottBrian M. Scott

                        456k38507908




                        456k38507908























                            10












                            $begingroup$

                            Use that $X$ is connected if and only if the only continuous functions $f:Xto{0,1}$ are constant, where ${0,1}$ is endowed with the discrete topology.



                            Now, you know each $F$ in $mathscr F$ is connected. Consider $f:bigcup mathscr Fto{0,1}$, $f$ continuous.



                            Take $alpha inbigcapmathscr F$. Look at $f(alpha)$, and at $fmid_{F}:bigcup mathscr Fto{0,1}$ for any $Finmathscr F$.





                            Since you mention metric spaces, I am not sure if you know about the first thing I mention, so let's prove it:




                            THM Let $(X,mathscr T)$ be a metric (or more generally, a topological) space. Then $X$ is connected if and only if whenever $f:Xto{0,1}$ is continuous, it is constant. The space ${0,1}$ is endowed with the discrete metric (topology), that is, the open sets are $varnothing,{0},{1},{0,1}$.




                            P First, suppose $X$ is disconnected, say by $A,B$, so $Acup B=X$ and $Acap B=varnothing$, $A,B$ open. Define $f:Xto{0,1}$ by $$f(x)=begin{cases}1& ; ; xin A\0&; ; xin Bend{cases}$$



                            Then $f$ is continuous because $f^{-1}(G)$ is open for any open $G$ in ${0,1}$ (this is simply a case by case verification), yet it is not constant. Now suppose $f:Xto{0,1}$ is continuous but not constant. Set $A={x:f(x)=1}=f^{-1}({1})$ and $B={x:f(x)=0}=f^{-1}({0})$. By hypothesis, $A,Bneq varnothing$. Morover, both are open, since they are the preimage of open sets under a continuous map, and $Acup B=X$ and $Acap B=varnothing$. Thus $X$ is disconnected. $blacktriangle$






                            share|cite|improve this answer











                            $endgroup$


















                              10












                              $begingroup$

                              Use that $X$ is connected if and only if the only continuous functions $f:Xto{0,1}$ are constant, where ${0,1}$ is endowed with the discrete topology.



                              Now, you know each $F$ in $mathscr F$ is connected. Consider $f:bigcup mathscr Fto{0,1}$, $f$ continuous.



                              Take $alpha inbigcapmathscr F$. Look at $f(alpha)$, and at $fmid_{F}:bigcup mathscr Fto{0,1}$ for any $Finmathscr F$.





                              Since you mention metric spaces, I am not sure if you know about the first thing I mention, so let's prove it:




                              THM Let $(X,mathscr T)$ be a metric (or more generally, a topological) space. Then $X$ is connected if and only if whenever $f:Xto{0,1}$ is continuous, it is constant. The space ${0,1}$ is endowed with the discrete metric (topology), that is, the open sets are $varnothing,{0},{1},{0,1}$.




                              P First, suppose $X$ is disconnected, say by $A,B$, so $Acup B=X$ and $Acap B=varnothing$, $A,B$ open. Define $f:Xto{0,1}$ by $$f(x)=begin{cases}1& ; ; xin A\0&; ; xin Bend{cases}$$



                              Then $f$ is continuous because $f^{-1}(G)$ is open for any open $G$ in ${0,1}$ (this is simply a case by case verification), yet it is not constant. Now suppose $f:Xto{0,1}$ is continuous but not constant. Set $A={x:f(x)=1}=f^{-1}({1})$ and $B={x:f(x)=0}=f^{-1}({0})$. By hypothesis, $A,Bneq varnothing$. Morover, both are open, since they are the preimage of open sets under a continuous map, and $Acup B=X$ and $Acap B=varnothing$. Thus $X$ is disconnected. $blacktriangle$






                              share|cite|improve this answer











                              $endgroup$
















                                10












                                10








                                10





                                $begingroup$

                                Use that $X$ is connected if and only if the only continuous functions $f:Xto{0,1}$ are constant, where ${0,1}$ is endowed with the discrete topology.



                                Now, you know each $F$ in $mathscr F$ is connected. Consider $f:bigcup mathscr Fto{0,1}$, $f$ continuous.



                                Take $alpha inbigcapmathscr F$. Look at $f(alpha)$, and at $fmid_{F}:bigcup mathscr Fto{0,1}$ for any $Finmathscr F$.





                                Since you mention metric spaces, I am not sure if you know about the first thing I mention, so let's prove it:




                                THM Let $(X,mathscr T)$ be a metric (or more generally, a topological) space. Then $X$ is connected if and only if whenever $f:Xto{0,1}$ is continuous, it is constant. The space ${0,1}$ is endowed with the discrete metric (topology), that is, the open sets are $varnothing,{0},{1},{0,1}$.




                                P First, suppose $X$ is disconnected, say by $A,B$, so $Acup B=X$ and $Acap B=varnothing$, $A,B$ open. Define $f:Xto{0,1}$ by $$f(x)=begin{cases}1& ; ; xin A\0&; ; xin Bend{cases}$$



                                Then $f$ is continuous because $f^{-1}(G)$ is open for any open $G$ in ${0,1}$ (this is simply a case by case verification), yet it is not constant. Now suppose $f:Xto{0,1}$ is continuous but not constant. Set $A={x:f(x)=1}=f^{-1}({1})$ and $B={x:f(x)=0}=f^{-1}({0})$. By hypothesis, $A,Bneq varnothing$. Morover, both are open, since they are the preimage of open sets under a continuous map, and $Acup B=X$ and $Acap B=varnothing$. Thus $X$ is disconnected. $blacktriangle$






                                share|cite|improve this answer











                                $endgroup$



                                Use that $X$ is connected if and only if the only continuous functions $f:Xto{0,1}$ are constant, where ${0,1}$ is endowed with the discrete topology.



                                Now, you know each $F$ in $mathscr F$ is connected. Consider $f:bigcup mathscr Fto{0,1}$, $f$ continuous.



                                Take $alpha inbigcapmathscr F$. Look at $f(alpha)$, and at $fmid_{F}:bigcup mathscr Fto{0,1}$ for any $Finmathscr F$.





                                Since you mention metric spaces, I am not sure if you know about the first thing I mention, so let's prove it:




                                THM Let $(X,mathscr T)$ be a metric (or more generally, a topological) space. Then $X$ is connected if and only if whenever $f:Xto{0,1}$ is continuous, it is constant. The space ${0,1}$ is endowed with the discrete metric (topology), that is, the open sets are $varnothing,{0},{1},{0,1}$.




                                P First, suppose $X$ is disconnected, say by $A,B$, so $Acup B=X$ and $Acap B=varnothing$, $A,B$ open. Define $f:Xto{0,1}$ by $$f(x)=begin{cases}1& ; ; xin A\0&; ; xin Bend{cases}$$



                                Then $f$ is continuous because $f^{-1}(G)$ is open for any open $G$ in ${0,1}$ (this is simply a case by case verification), yet it is not constant. Now suppose $f:Xto{0,1}$ is continuous but not constant. Set $A={x:f(x)=1}=f^{-1}({1})$ and $B={x:f(x)=0}=f^{-1}({0})$. By hypothesis, $A,Bneq varnothing$. Morover, both are open, since they are the preimage of open sets under a continuous map, and $Acup B=X$ and $Acap B=varnothing$. Thus $X$ is disconnected. $blacktriangle$







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Jun 2 '16 at 9:48

























                                answered Jun 20 '13 at 0:42









                                Pedro TamaroffPedro Tamaroff

                                96.5k10153297




                                96.5k10153297























                                    4












                                    $begingroup$

                                    I like "clopen" sets... it is a set that is at the same time closed and open. A topological space $X$ is not connected if, and only if, for any point $a in X$, you have a "non-trivial" clopen set $C$ containing $a$. That is, a clopen set such that $emptyset neq C subsetneq X$.



                                    Take $a in bigcap mathscr{F}$. Then, take a clopen set $C subset bigcup mathscr{F}$ containing $a$. In the relative topology, for any $X in mathscr{F}$, $C cap X$ is a clopen set. Since $X$ is connected, $C cap X = X$. That is, $X subset C$ for every $X in mathscr{F}$. Therefore, $bigcup mathscr{F} subset C$.



                                    Therefore, $bigcup mathscr{F}$ has no "non-trivial" clopen set, and so, it is connected.





                                    The proof admits some variants. For example, if you take any non-empty clopen set $C'$, it will intersect some $X in mathscr{F}$. From connectedness, we have that $X subset C'$. In particular, $a in C'$.






                                    share|cite|improve this answer











                                    $endgroup$


















                                      4












                                      $begingroup$

                                      I like "clopen" sets... it is a set that is at the same time closed and open. A topological space $X$ is not connected if, and only if, for any point $a in X$, you have a "non-trivial" clopen set $C$ containing $a$. That is, a clopen set such that $emptyset neq C subsetneq X$.



                                      Take $a in bigcap mathscr{F}$. Then, take a clopen set $C subset bigcup mathscr{F}$ containing $a$. In the relative topology, for any $X in mathscr{F}$, $C cap X$ is a clopen set. Since $X$ is connected, $C cap X = X$. That is, $X subset C$ for every $X in mathscr{F}$. Therefore, $bigcup mathscr{F} subset C$.



                                      Therefore, $bigcup mathscr{F}$ has no "non-trivial" clopen set, and so, it is connected.





                                      The proof admits some variants. For example, if you take any non-empty clopen set $C'$, it will intersect some $X in mathscr{F}$. From connectedness, we have that $X subset C'$. In particular, $a in C'$.






                                      share|cite|improve this answer











                                      $endgroup$
















                                        4












                                        4








                                        4





                                        $begingroup$

                                        I like "clopen" sets... it is a set that is at the same time closed and open. A topological space $X$ is not connected if, and only if, for any point $a in X$, you have a "non-trivial" clopen set $C$ containing $a$. That is, a clopen set such that $emptyset neq C subsetneq X$.



                                        Take $a in bigcap mathscr{F}$. Then, take a clopen set $C subset bigcup mathscr{F}$ containing $a$. In the relative topology, for any $X in mathscr{F}$, $C cap X$ is a clopen set. Since $X$ is connected, $C cap X = X$. That is, $X subset C$ for every $X in mathscr{F}$. Therefore, $bigcup mathscr{F} subset C$.



                                        Therefore, $bigcup mathscr{F}$ has no "non-trivial" clopen set, and so, it is connected.





                                        The proof admits some variants. For example, if you take any non-empty clopen set $C'$, it will intersect some $X in mathscr{F}$. From connectedness, we have that $X subset C'$. In particular, $a in C'$.






                                        share|cite|improve this answer











                                        $endgroup$



                                        I like "clopen" sets... it is a set that is at the same time closed and open. A topological space $X$ is not connected if, and only if, for any point $a in X$, you have a "non-trivial" clopen set $C$ containing $a$. That is, a clopen set such that $emptyset neq C subsetneq X$.



                                        Take $a in bigcap mathscr{F}$. Then, take a clopen set $C subset bigcup mathscr{F}$ containing $a$. In the relative topology, for any $X in mathscr{F}$, $C cap X$ is a clopen set. Since $X$ is connected, $C cap X = X$. That is, $X subset C$ for every $X in mathscr{F}$. Therefore, $bigcup mathscr{F} subset C$.



                                        Therefore, $bigcup mathscr{F}$ has no "non-trivial" clopen set, and so, it is connected.





                                        The proof admits some variants. For example, if you take any non-empty clopen set $C'$, it will intersect some $X in mathscr{F}$. From connectedness, we have that $X subset C'$. In particular, $a in C'$.







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Jun 20 '13 at 1:51

























                                        answered Jun 20 '13 at 1:38









                                        André CaldasAndré Caldas

                                        3,3971227




                                        3,3971227























                                            3












                                            $begingroup$

                                            The quickest way is using functions from $bigcupmathcal{F}$ to ${0,1}$, However you can do it directly.



                                            Let $A, B$ be a partition of $bigcupmathcal{F}$ into open, disjoint subsets. WLOG assume that $A$ is non-empty. Then there is some $xin A$ and hence some $Fin mathcal{F}$ so that $xin F$. As $F$ is connected it follows that $Fsubseteq A$, hence $bigcap mathcal{F} subseteq F subseteq A$. As this intersection is non-empty every $Gin mathcal{F}$ has a point in $A$ and by the connectness of each $G$ then $Gsubseteq A$. Hence every member of $mathcal{F}$ is a subset of $A$ and so $B$ is empty.






                                            share|cite|improve this answer









                                            $endgroup$


















                                              3












                                              $begingroup$

                                              The quickest way is using functions from $bigcupmathcal{F}$ to ${0,1}$, However you can do it directly.



                                              Let $A, B$ be a partition of $bigcupmathcal{F}$ into open, disjoint subsets. WLOG assume that $A$ is non-empty. Then there is some $xin A$ and hence some $Fin mathcal{F}$ so that $xin F$. As $F$ is connected it follows that $Fsubseteq A$, hence $bigcap mathcal{F} subseteq F subseteq A$. As this intersection is non-empty every $Gin mathcal{F}$ has a point in $A$ and by the connectness of each $G$ then $Gsubseteq A$. Hence every member of $mathcal{F}$ is a subset of $A$ and so $B$ is empty.






                                              share|cite|improve this answer









                                              $endgroup$
















                                                3












                                                3








                                                3





                                                $begingroup$

                                                The quickest way is using functions from $bigcupmathcal{F}$ to ${0,1}$, However you can do it directly.



                                                Let $A, B$ be a partition of $bigcupmathcal{F}$ into open, disjoint subsets. WLOG assume that $A$ is non-empty. Then there is some $xin A$ and hence some $Fin mathcal{F}$ so that $xin F$. As $F$ is connected it follows that $Fsubseteq A$, hence $bigcap mathcal{F} subseteq F subseteq A$. As this intersection is non-empty every $Gin mathcal{F}$ has a point in $A$ and by the connectness of each $G$ then $Gsubseteq A$. Hence every member of $mathcal{F}$ is a subset of $A$ and so $B$ is empty.






                                                share|cite|improve this answer









                                                $endgroup$



                                                The quickest way is using functions from $bigcupmathcal{F}$ to ${0,1}$, However you can do it directly.



                                                Let $A, B$ be a partition of $bigcupmathcal{F}$ into open, disjoint subsets. WLOG assume that $A$ is non-empty. Then there is some $xin A$ and hence some $Fin mathcal{F}$ so that $xin F$. As $F$ is connected it follows that $Fsubseteq A$, hence $bigcap mathcal{F} subseteq F subseteq A$. As this intersection is non-empty every $Gin mathcal{F}$ has a point in $A$ and by the connectness of each $G$ then $Gsubseteq A$. Hence every member of $mathcal{F}$ is a subset of $A$ and so $B$ is empty.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Jun 20 '13 at 0:49









                                                JamesJames

                                                4,2401821




                                                4,2401821























                                                    0












                                                    $begingroup$

                                                    In the before last post, with the "clopen" sets, I believe there is a detail missing: it is not immediately clear (but true) that if $C$ is a clopen set with respect to the subset topology on $bigcupmathscr{F}$, that then $Ccap X$ is a clopen set in $X$, $Xinmathscr{F}$, with respect to the original topology, which is what you need.






                                                    share|cite|improve this answer











                                                    $endgroup$


















                                                      0












                                                      $begingroup$

                                                      In the before last post, with the "clopen" sets, I believe there is a detail missing: it is not immediately clear (but true) that if $C$ is a clopen set with respect to the subset topology on $bigcupmathscr{F}$, that then $Ccap X$ is a clopen set in $X$, $Xinmathscr{F}$, with respect to the original topology, which is what you need.






                                                      share|cite|improve this answer











                                                      $endgroup$
















                                                        0












                                                        0








                                                        0





                                                        $begingroup$

                                                        In the before last post, with the "clopen" sets, I believe there is a detail missing: it is not immediately clear (but true) that if $C$ is a clopen set with respect to the subset topology on $bigcupmathscr{F}$, that then $Ccap X$ is a clopen set in $X$, $Xinmathscr{F}$, with respect to the original topology, which is what you need.






                                                        share|cite|improve this answer











                                                        $endgroup$



                                                        In the before last post, with the "clopen" sets, I believe there is a detail missing: it is not immediately clear (but true) that if $C$ is a clopen set with respect to the subset topology on $bigcupmathscr{F}$, that then $Ccap X$ is a clopen set in $X$, $Xinmathscr{F}$, with respect to the original topology, which is what you need.







                                                        share|cite|improve this answer














                                                        share|cite|improve this answer



                                                        share|cite|improve this answer








                                                        edited Apr 15 '17 at 20:06

























                                                        answered Apr 15 '17 at 16:03









                                                        Michiel VermeulenMichiel Vermeulen

                                                        315




                                                        315






























                                                            draft saved

                                                            draft discarded




















































                                                            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.




                                                            draft saved


                                                            draft discarded














                                                            StackExchange.ready(
                                                            function () {
                                                            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f425007%2funion-of-connected-subsets-is-connected-if-intersection-is-nonempty%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

                                                            Ellipse (mathématiques)

                                                            Quarter-circle Tiles

                                                            Mont Emei