Understanding a proof about Hausdorff spaces.












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If $X$ is a Hausdorff space and $Asubset X$ is compact then $A$ is closed in $X$.



Proof: Suppose $A neq emptyset,X$ Let $xnotin A$ We want to find an open $U$ so that $xin U$ and $ Ucap A=emptyset$. For every $ain A$ applying the Hausdorff property on x,a we will have two open disjoint sets $U_a$, $V_a$ so that $xin U_a$ and $ain V_a$. Taking all the open sets $V_a$ for $ain A$, we will have a cover of $A$. Because $A$ is compact, this cover will have a finite subcover $Asubset V_{a_1}cup...cup V_{a_n} $. Therefore the open $U:= U_{a_1}cap...cap U_{a_n}$ is the open subset that we wanted.



I don't undersand why finding an open $U$ so that $xin U$ and $ Ucap A=emptyset$ proves that $A$ is closed in $X$.










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    If $X$ is a Hausdorff space and $Asubset X$ is compact then $A$ is closed in $X$.



    Proof: Suppose $A neq emptyset,X$ Let $xnotin A$ We want to find an open $U$ so that $xin U$ and $ Ucap A=emptyset$. For every $ain A$ applying the Hausdorff property on x,a we will have two open disjoint sets $U_a$, $V_a$ so that $xin U_a$ and $ain V_a$. Taking all the open sets $V_a$ for $ain A$, we will have a cover of $A$. Because $A$ is compact, this cover will have a finite subcover $Asubset V_{a_1}cup...cup V_{a_n} $. Therefore the open $U:= U_{a_1}cap...cap U_{a_n}$ is the open subset that we wanted.



    I don't undersand why finding an open $U$ so that $xin U$ and $ Ucap A=emptyset$ proves that $A$ is closed in $X$.










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      If $X$ is a Hausdorff space and $Asubset X$ is compact then $A$ is closed in $X$.



      Proof: Suppose $A neq emptyset,X$ Let $xnotin A$ We want to find an open $U$ so that $xin U$ and $ Ucap A=emptyset$. For every $ain A$ applying the Hausdorff property on x,a we will have two open disjoint sets $U_a$, $V_a$ so that $xin U_a$ and $ain V_a$. Taking all the open sets $V_a$ for $ain A$, we will have a cover of $A$. Because $A$ is compact, this cover will have a finite subcover $Asubset V_{a_1}cup...cup V_{a_n} $. Therefore the open $U:= U_{a_1}cap...cap U_{a_n}$ is the open subset that we wanted.



      I don't undersand why finding an open $U$ so that $xin U$ and $ Ucap A=emptyset$ proves that $A$ is closed in $X$.










      share|cite|improve this question













      If $X$ is a Hausdorff space and $Asubset X$ is compact then $A$ is closed in $X$.



      Proof: Suppose $A neq emptyset,X$ Let $xnotin A$ We want to find an open $U$ so that $xin U$ and $ Ucap A=emptyset$. For every $ain A$ applying the Hausdorff property on x,a we will have two open disjoint sets $U_a$, $V_a$ so that $xin U_a$ and $ain V_a$. Taking all the open sets $V_a$ for $ain A$, we will have a cover of $A$. Because $A$ is compact, this cover will have a finite subcover $Asubset V_{a_1}cup...cup V_{a_n} $. Therefore the open $U:= U_{a_1}cap...cap U_{a_n}$ is the open subset that we wanted.



      I don't undersand why finding an open $U$ so that $xin U$ and $ Ucap A=emptyset$ proves that $A$ is closed in $X$.







      general-topology proof-explanation






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      asked Jan 4 at 2:58









      John KeeperJohn Keeper

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          To say $A$ is closed is to say $X setminus A$ is open. You're starting with an arbitrary point $x in X setminus A$, and finding an open set $U_x$ that contains $x$, but fails to intersect $A$. In other words,
          $$x subseteq U_x subseteq X setminus A.$$
          Around every point in the set $X setminus A$, you've found an open neighbourhood in $X setminus A$.



          This implies $X setminus A$ is open, and hence $A$ is closed. Why? Consider the set
          $$B = bigcup_{x in X setminus A} U_x.$$
          Note that, if $x in X setminus A$, then $x in U_x$, hence $x in B$, so
          $$X setminus A subseteq B.$$
          On the other hand, each $U_x subseteq X setminus A$, so $B subseteq X setminus A$. Hence,
          $$X setminus A = B.$$
          This means $X setminus A$ is a union of open sets, which is therefore open.






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            It shows that every point of $X-A$ is an interior point, hence $X-A$ is open so $A$ is closed.






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              2 Answers
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              To say $A$ is closed is to say $X setminus A$ is open. You're starting with an arbitrary point $x in X setminus A$, and finding an open set $U_x$ that contains $x$, but fails to intersect $A$. In other words,
              $$x subseteq U_x subseteq X setminus A.$$
              Around every point in the set $X setminus A$, you've found an open neighbourhood in $X setminus A$.



              This implies $X setminus A$ is open, and hence $A$ is closed. Why? Consider the set
              $$B = bigcup_{x in X setminus A} U_x.$$
              Note that, if $x in X setminus A$, then $x in U_x$, hence $x in B$, so
              $$X setminus A subseteq B.$$
              On the other hand, each $U_x subseteq X setminus A$, so $B subseteq X setminus A$. Hence,
              $$X setminus A = B.$$
              This means $X setminus A$ is a union of open sets, which is therefore open.






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                1














                To say $A$ is closed is to say $X setminus A$ is open. You're starting with an arbitrary point $x in X setminus A$, and finding an open set $U_x$ that contains $x$, but fails to intersect $A$. In other words,
                $$x subseteq U_x subseteq X setminus A.$$
                Around every point in the set $X setminus A$, you've found an open neighbourhood in $X setminus A$.



                This implies $X setminus A$ is open, and hence $A$ is closed. Why? Consider the set
                $$B = bigcup_{x in X setminus A} U_x.$$
                Note that, if $x in X setminus A$, then $x in U_x$, hence $x in B$, so
                $$X setminus A subseteq B.$$
                On the other hand, each $U_x subseteq X setminus A$, so $B subseteq X setminus A$. Hence,
                $$X setminus A = B.$$
                This means $X setminus A$ is a union of open sets, which is therefore open.






                share|cite|improve this answer


























                  1












                  1








                  1






                  To say $A$ is closed is to say $X setminus A$ is open. You're starting with an arbitrary point $x in X setminus A$, and finding an open set $U_x$ that contains $x$, but fails to intersect $A$. In other words,
                  $$x subseteq U_x subseteq X setminus A.$$
                  Around every point in the set $X setminus A$, you've found an open neighbourhood in $X setminus A$.



                  This implies $X setminus A$ is open, and hence $A$ is closed. Why? Consider the set
                  $$B = bigcup_{x in X setminus A} U_x.$$
                  Note that, if $x in X setminus A$, then $x in U_x$, hence $x in B$, so
                  $$X setminus A subseteq B.$$
                  On the other hand, each $U_x subseteq X setminus A$, so $B subseteq X setminus A$. Hence,
                  $$X setminus A = B.$$
                  This means $X setminus A$ is a union of open sets, which is therefore open.






                  share|cite|improve this answer














                  To say $A$ is closed is to say $X setminus A$ is open. You're starting with an arbitrary point $x in X setminus A$, and finding an open set $U_x$ that contains $x$, but fails to intersect $A$. In other words,
                  $$x subseteq U_x subseteq X setminus A.$$
                  Around every point in the set $X setminus A$, you've found an open neighbourhood in $X setminus A$.



                  This implies $X setminus A$ is open, and hence $A$ is closed. Why? Consider the set
                  $$B = bigcup_{x in X setminus A} U_x.$$
                  Note that, if $x in X setminus A$, then $x in U_x$, hence $x in B$, so
                  $$X setminus A subseteq B.$$
                  On the other hand, each $U_x subseteq X setminus A$, so $B subseteq X setminus A$. Hence,
                  $$X setminus A = B.$$
                  This means $X setminus A$ is a union of open sets, which is therefore open.







                  share|cite|improve this answer














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                  edited Jan 4 at 5:39









                  Henno Brandsma

                  105k347114




                  105k347114










                  answered Jan 4 at 3:03









                  Theo BenditTheo Bendit

                  16.7k12148




                  16.7k12148























                      1














                      It shows that every point of $X-A$ is an interior point, hence $X-A$ is open so $A$ is closed.






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                        1














                        It shows that every point of $X-A$ is an interior point, hence $X-A$ is open so $A$ is closed.






                        share|cite|improve this answer
























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                          It shows that every point of $X-A$ is an interior point, hence $X-A$ is open so $A$ is closed.






                          share|cite|improve this answer












                          It shows that every point of $X-A$ is an interior point, hence $X-A$ is open so $A$ is closed.







                          share|cite|improve this answer












                          share|cite|improve this answer



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                          answered Jan 4 at 3:02









                          RandallRandall

                          9,14611129




                          9,14611129






























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