Is an open subset of an open set of a topological space open?












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If $X$ is an topological space and $ U subseteq V subseteq X$ with $V$ open in $X$ and $U$ open in $V$, is it true that $U$ is open in $X$?










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    0












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    If $X$ is an topological space and $ U subseteq V subseteq X$ with $V$ open in $X$ and $U$ open in $V$, is it true that $U$ is open in $X$?










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      $begingroup$


      If $X$ is an topological space and $ U subseteq V subseteq X$ with $V$ open in $X$ and $U$ open in $V$, is it true that $U$ is open in $X$?










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      If $X$ is an topological space and $ U subseteq V subseteq X$ with $V$ open in $X$ and $U$ open in $V$, is it true that $U$ is open in $X$?







      general-topology






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      edited Dec 5 '18 at 16:57









      Andrews

      3901317




      3901317










      asked Dec 5 '18 at 16:07









      gengen

      4532521




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          Yes. $U$ open in $V$ means that $U=Vcap W$ for some $W$ open in $X$. So $U$ is an intersection of two open sets in $X$.






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            Thank you (five more characters)
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            – gen
            Dec 5 '18 at 16:11



















          1












          $begingroup$

          Hint: $U $ is open in $V $ iff $U=Vcap S $ for some $S $ open in $X $.






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          • $begingroup$
            Is this just the definition of U being open in V?
            $endgroup$
            – gen
            Dec 5 '18 at 16:13










          • $begingroup$
            Yes. This is how a subspace topology is defined.
            $endgroup$
            – Thomas Shelby
            Dec 5 '18 at 16:15











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          2 Answers
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          2 Answers
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          $begingroup$

          Yes. $U$ open in $V$ means that $U=Vcap W$ for some $W$ open in $X$. So $U$ is an intersection of two open sets in $X$.






          share|cite|improve this answer









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          • $begingroup$
            Thank you (five more characters)
            $endgroup$
            – gen
            Dec 5 '18 at 16:11
















          2












          $begingroup$

          Yes. $U$ open in $V$ means that $U=Vcap W$ for some $W$ open in $X$. So $U$ is an intersection of two open sets in $X$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you (five more characters)
            $endgroup$
            – gen
            Dec 5 '18 at 16:11














          2












          2








          2





          $begingroup$

          Yes. $U$ open in $V$ means that $U=Vcap W$ for some $W$ open in $X$. So $U$ is an intersection of two open sets in $X$.






          share|cite|improve this answer









          $endgroup$



          Yes. $U$ open in $V$ means that $U=Vcap W$ for some $W$ open in $X$. So $U$ is an intersection of two open sets in $X$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 16:10









          ajotatxeajotatxe

          53.8k23890




          53.8k23890












          • $begingroup$
            Thank you (five more characters)
            $endgroup$
            – gen
            Dec 5 '18 at 16:11


















          • $begingroup$
            Thank you (five more characters)
            $endgroup$
            – gen
            Dec 5 '18 at 16:11
















          $begingroup$
          Thank you (five more characters)
          $endgroup$
          – gen
          Dec 5 '18 at 16:11




          $begingroup$
          Thank you (five more characters)
          $endgroup$
          – gen
          Dec 5 '18 at 16:11











          1












          $begingroup$

          Hint: $U $ is open in $V $ iff $U=Vcap S $ for some $S $ open in $X $.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is this just the definition of U being open in V?
            $endgroup$
            – gen
            Dec 5 '18 at 16:13










          • $begingroup$
            Yes. This is how a subspace topology is defined.
            $endgroup$
            – Thomas Shelby
            Dec 5 '18 at 16:15
















          1












          $begingroup$

          Hint: $U $ is open in $V $ iff $U=Vcap S $ for some $S $ open in $X $.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is this just the definition of U being open in V?
            $endgroup$
            – gen
            Dec 5 '18 at 16:13










          • $begingroup$
            Yes. This is how a subspace topology is defined.
            $endgroup$
            – Thomas Shelby
            Dec 5 '18 at 16:15














          1












          1








          1





          $begingroup$

          Hint: $U $ is open in $V $ iff $U=Vcap S $ for some $S $ open in $X $.






          share|cite|improve this answer











          $endgroup$



          Hint: $U $ is open in $V $ iff $U=Vcap S $ for some $S $ open in $X $.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '18 at 16:23

























          answered Dec 5 '18 at 16:12









          Thomas ShelbyThomas Shelby

          2,710421




          2,710421












          • $begingroup$
            Is this just the definition of U being open in V?
            $endgroup$
            – gen
            Dec 5 '18 at 16:13










          • $begingroup$
            Yes. This is how a subspace topology is defined.
            $endgroup$
            – Thomas Shelby
            Dec 5 '18 at 16:15


















          • $begingroup$
            Is this just the definition of U being open in V?
            $endgroup$
            – gen
            Dec 5 '18 at 16:13










          • $begingroup$
            Yes. This is how a subspace topology is defined.
            $endgroup$
            – Thomas Shelby
            Dec 5 '18 at 16:15
















          $begingroup$
          Is this just the definition of U being open in V?
          $endgroup$
          – gen
          Dec 5 '18 at 16:13




          $begingroup$
          Is this just the definition of U being open in V?
          $endgroup$
          – gen
          Dec 5 '18 at 16:13












          $begingroup$
          Yes. This is how a subspace topology is defined.
          $endgroup$
          – Thomas Shelby
          Dec 5 '18 at 16:15




          $begingroup$
          Yes. This is how a subspace topology is defined.
          $endgroup$
          – Thomas Shelby
          Dec 5 '18 at 16:15


















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