$Ucap K$ is compact in the subspace topology












0















If $X,tau$ is a topological space and K is a closed subspace. If $U$ is a compact set of $X,tau$ then $Ucap K$ is compact.




Attempted proof:



Since $U$ is compact there is a finite sub covering such that $Usubset bigcup_limits{i=1}^{n}B_i$ $B_iin tauforall 1leqslant ileqslant n$



$Ucap Ksubset (bigcup_limits{i=1}^{n}B_i)cap K=bigcup_limits{i=1}^{n}(B_icap K)$



$B_icap K$ is open for the subspace topology for all $i$.



Hence $bigcup_limits{i=1}^{n}(B_icap K)$ is a finite sub covering of $Ucap K$ proving $Ucap K$ to be compact in the subspace topology.



Question:



Is my proof right? If not how should I correct it?



Thanks in advance!










share|cite|improve this question
























  • I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
    – Uncountable
    2 days ago










  • The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
    – Uncountable
    2 days ago
















0















If $X,tau$ is a topological space and K is a closed subspace. If $U$ is a compact set of $X,tau$ then $Ucap K$ is compact.




Attempted proof:



Since $U$ is compact there is a finite sub covering such that $Usubset bigcup_limits{i=1}^{n}B_i$ $B_iin tauforall 1leqslant ileqslant n$



$Ucap Ksubset (bigcup_limits{i=1}^{n}B_i)cap K=bigcup_limits{i=1}^{n}(B_icap K)$



$B_icap K$ is open for the subspace topology for all $i$.



Hence $bigcup_limits{i=1}^{n}(B_icap K)$ is a finite sub covering of $Ucap K$ proving $Ucap K$ to be compact in the subspace topology.



Question:



Is my proof right? If not how should I correct it?



Thanks in advance!










share|cite|improve this question
























  • I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
    – Uncountable
    2 days ago










  • The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
    – Uncountable
    2 days ago














0












0








0








If $X,tau$ is a topological space and K is a closed subspace. If $U$ is a compact set of $X,tau$ then $Ucap K$ is compact.




Attempted proof:



Since $U$ is compact there is a finite sub covering such that $Usubset bigcup_limits{i=1}^{n}B_i$ $B_iin tauforall 1leqslant ileqslant n$



$Ucap Ksubset (bigcup_limits{i=1}^{n}B_i)cap K=bigcup_limits{i=1}^{n}(B_icap K)$



$B_icap K$ is open for the subspace topology for all $i$.



Hence $bigcup_limits{i=1}^{n}(B_icap K)$ is a finite sub covering of $Ucap K$ proving $Ucap K$ to be compact in the subspace topology.



Question:



Is my proof right? If not how should I correct it?



Thanks in advance!










share|cite|improve this question
















If $X,tau$ is a topological space and K is a closed subspace. If $U$ is a compact set of $X,tau$ then $Ucap K$ is compact.




Attempted proof:



Since $U$ is compact there is a finite sub covering such that $Usubset bigcup_limits{i=1}^{n}B_i$ $B_iin tauforall 1leqslant ileqslant n$



$Ucap Ksubset (bigcup_limits{i=1}^{n}B_i)cap K=bigcup_limits{i=1}^{n}(B_icap K)$



$B_icap K$ is open for the subspace topology for all $i$.



Hence $bigcup_limits{i=1}^{n}(B_icap K)$ is a finite sub covering of $Ucap K$ proving $Ucap K$ to be compact in the subspace topology.



Question:



Is my proof right? If not how should I correct it?



Thanks in advance!







general-topology proof-verification compactness






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









José Carlos Santos

149k22117219




149k22117219










asked 2 days ago









Pedro Gomes

1,6792720




1,6792720












  • I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
    – Uncountable
    2 days ago










  • The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
    – Uncountable
    2 days ago


















  • I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
    – Uncountable
    2 days ago










  • The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
    – Uncountable
    2 days ago
















I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
– Uncountable
2 days ago




I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
– Uncountable
2 days ago












The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
– Uncountable
2 days ago




The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
– Uncountable
2 days ago










1 Answer
1






active

oldest

votes


















4














The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.



Let $(O_lambda)_{lambdainLambda}$ be an open cover of $Ucap K$. Add to this cover the set $Ucap K^complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $Ucap K^complement$ form a finite subcover of $Ucap K$.






share|cite|improve this answer























    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%2f3052348%2fu-cap-k-is-compact-in-the-subspace-topology%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.



    Let $(O_lambda)_{lambdainLambda}$ be an open cover of $Ucap K$. Add to this cover the set $Ucap K^complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $Ucap K^complement$ form a finite subcover of $Ucap K$.






    share|cite|improve this answer




























      4














      The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.



      Let $(O_lambda)_{lambdainLambda}$ be an open cover of $Ucap K$. Add to this cover the set $Ucap K^complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $Ucap K^complement$ form a finite subcover of $Ucap K$.






      share|cite|improve this answer


























        4












        4








        4






        The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.



        Let $(O_lambda)_{lambdainLambda}$ be an open cover of $Ucap K$. Add to this cover the set $Ucap K^complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $Ucap K^complement$ form a finite subcover of $Ucap K$.






        share|cite|improve this answer














        The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.



        Let $(O_lambda)_{lambdainLambda}$ be an open cover of $Ucap K$. Add to this cover the set $Ucap K^complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $Ucap K^complement$ form a finite subcover of $Ucap K$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        José Carlos Santos

        149k22117219




        149k22117219






























            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.





            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%2f3052348%2fu-cap-k-is-compact-in-the-subspace-topology%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

            Human spaceflight

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            張江高科駅