Understanding a proof about Hausdorff spaces.
$begingroup$
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
asked Jan 4 at 2:58
John KeeperJohn Keeper
541315
$endgroup$
add a comment |
$begingroup$
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
asked Jan 4 at 2:58
John KeeperJohn Keeper
541315
$endgroup$
add a comment |
$begingroup$
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
asked Jan 4 at 2:58
John KeeperJohn Keeper
541315
$endgroup$
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
general-topology proof-explanation
asked Jan 4 at 2:58
John KeeperJohn Keeper
541315
asked Jan 4 at 2:58
John KeeperJohn Keeper
541315
asked Jan 4 at 2:58
John KeeperJohn Keeper
541315
asked Jan 4 at 2:58
John KeeperJohn Keeper
541315
asked Jan 4 at 2:58
John KeeperJohn Keeper
541315
541315
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2 Answers
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$begingroup$
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.
edited Jan 4 at 5:39
Henno Brandsma
115k349125
answered Jan 4 at 3:03
Theo BenditTheo Bendit
20.1k12354
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add a comment |
$begingroup$
It shows that every point of $X-A$ is an interior point, hence $X-A$ is open so $A$ is closed.
answered Jan 4 at 3:02
RandallRandall
10.6k11431
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
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.
edited Jan 4 at 5:39
Henno Brandsma
115k349125
answered Jan 4 at 3:03
Theo BenditTheo Bendit
20.1k12354
$endgroup$
add a comment |
$begingroup$
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.
edited Jan 4 at 5:39
Henno Brandsma
115k349125
answered Jan 4 at 3:03
Theo BenditTheo Bendit
20.1k12354
$endgroup$
add a comment |
$begingroup$
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.
edited Jan 4 at 5:39
Henno Brandsma
115k349125
answered Jan 4 at 3:03
Theo BenditTheo Bendit
20.1k12354
$endgroup$
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.
edited Jan 4 at 5:39
Henno Brandsma
115k349125
answered Jan 4 at 3:03
Theo BenditTheo Bendit
20.1k12354
edited Jan 4 at 5:39
Henno Brandsma
115k349125
edited Jan 4 at 5:39
Henno Brandsma
115k349125
edited Jan 4 at 5:39
Henno Brandsma
115k349125
115k349125
answered Jan 4 at 3:03
Theo BenditTheo Bendit
20.1k12354
answered Jan 4 at 3:03
Theo BenditTheo Bendit
20.1k12354
answered Jan 4 at 3:03
Theo BenditTheo Bendit
20.1k12354
20.1k12354
add a comment |
add a comment |
$begingroup$
It shows that every point of $X-A$ is an interior point, hence $X-A$ is open so $A$ is closed.
answered Jan 4 at 3:02
RandallRandall
10.6k11431
$endgroup$
add a comment |
$begingroup$
It shows that every point of $X-A$ is an interior point, hence $X-A$ is open so $A$ is closed.
answered Jan 4 at 3:02
RandallRandall
10.6k11431
$endgroup$
add a comment |
$begingroup$
It shows that every point of $X-A$ is an interior point, hence $X-A$ is open so $A$ is closed.
answered Jan 4 at 3:02
RandallRandall
10.6k11431
$endgroup$
It shows that every point of $X-A$ is an interior point, hence $X-A$ is open so $A$ is closed.
answered Jan 4 at 3:02
RandallRandall
10.6k11431
answered Jan 4 at 3:02
RandallRandall
10.6k11431
answered Jan 4 at 3:02
RandallRandall
10.6k11431
answered Jan 4 at 3:02
RandallRandall
10.6k11431
10.6k11431
add a comment |
add a comment |
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