Mixing
Proof that there exist an open set with compact closure explanation
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I'm reading Rudins "Complex Analysis". There's a theorem that goes like this.
Theorem: Let $U$ be open, in a locally compact Hausdorff space. Let $Ksubset U$ and $K$ be compact. Then there exist an open set $V$ such that
$$Ksubset Vsubset overlineVsubset U $$
Proof:
Because every point of $K$ has a neighbourhood that's closure is compact, and we can cover $K$ with finite amount of such neighbourhoods, then there exist a set $G$, that has a compact closure, and is a subset of $K$. If $U=X$ then we can take $V=G$.
If not, define $C$ as the complement of $U$. From theorem $2.5$, for every point $pin C$, there exist an open set $W_p$, such that $Ksubset W_p$, and $pnotinoverlineW_p$.
And the proof goes on.
Theorem $2.5$: Suppose that $X$ is a Hausdorff space, $Ksubset X$ is a compact set, and $pin K^c$. Then there exist open sets $U$ and $W$, such that $pin U$, $Ksubset W$, and $U cap W = emptyset$.
My question is, how we can use this theorem in the proof, so that we can state the existence of such sets $W_p$. Mainly, where does $pnotinoverlineW_p$ come from. Thank you.
Another question that might help me. If $U$ and $V$ are open, $Ucap V=emptyset$ holds, does $overlineUcap V=emptyset$?
I've found a similar question here. In the comments, Andreas says that if $p$ was in closure of $W_p$, then every neighbourhood of $p$ would have to meet $W_p$. Why is that?
general-topology proof-explanation
asked Jul 22 at 16:36
Habrum
185
add a comment |Â
up vote
1
down vote
favorite
I'm reading Rudins "Complex Analysis". There's a theorem that goes like this.
Theorem: Let $U$ be open, in a locally compact Hausdorff space. Let $Ksubset U$ and $K$ be compact. Then there exist an open set $V$ such that
$$Ksubset Vsubset overlineVsubset U $$
Proof:
Because every point of $K$ has a neighbourhood that's closure is compact, and we can cover $K$ with finite amount of such neighbourhoods, then there exist a set $G$, that has a compact closure, and is a subset of $K$. If $U=X$ then we can take $V=G$.
If not, define $C$ as the complement of $U$. From theorem $2.5$, for every point $pin C$, there exist an open set $W_p$, such that $Ksubset W_p$, and $pnotinoverlineW_p$.
And the proof goes on.
Theorem $2.5$: Suppose that $X$ is a Hausdorff space, $Ksubset X$ is a compact set, and $pin K^c$. Then there exist open sets $U$ and $W$, such that $pin U$, $Ksubset W$, and $U cap W = emptyset$.
My question is, how we can use this theorem in the proof, so that we can state the existence of such sets $W_p$. Mainly, where does $pnotinoverlineW_p$ come from. Thank you.
Another question that might help me. If $U$ and $V$ are open, $Ucap V=emptyset$ holds, does $overlineUcap V=emptyset$?
I've found a similar question here. In the comments, Andreas says that if $p$ was in closure of $W_p$, then every neighbourhood of $p$ would have to meet $W_p$. Why is that?
general-topology proof-explanation
asked Jul 22 at 16:36
Habrum
185
Re: "Another question" : If $V$ is open then its complement is $V^c$ is closed and disjoint from $V $. So if $ V $ is open and is disjoint from $U$ then $Vcap overline Usubset V cap overline V^c=$ $V cap V^c=emptyset.$... Another way to see this is that if $V$ is open and $Ucap V= emptyset,$ then every $pin V $ has a nbhd (namely, $V $ ) that's disjoint from $U,$ so $pnot in overline U.$
– DanielWainfleet
Jul 22 at 19:10
The theorem, with $overline V$ compact, directly: (1). In a Hausdorff space if $C$ is compact and $pnotin C$ than $p, C$ are completely separated. (2). If $D$ is open and $ overline D$ is compact then $partial D= overline D$ $ D$ is compact. (3). We use (1) and (2) to prove that a compact $T_2$ space is regular..... Then for each $p in K$ let $V_p, U_p$ be open sets containing $p$ such that $overline V_p$ is compact and $overline U_psubset U. $ Now $A=V_pcap U_p: pin K$ is an open cover of $K.$ Let $B$ be a finite sub-cover of $A$ and let $V=cup B.$
– DanielWainfleet
Jul 22 at 19:28
ERRATUM: In my previous comment , (3) should say that a LOCALLY compact $T_2$ space is regular. I ran out of editing time .
– DanielWainfleet
Jul 22 at 19:35
1
@DanielWainfleet Thank you! I get it now.
– Habrum
Jul 22 at 20:03
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm reading Rudins "Complex Analysis". There's a theorem that goes like this.
Theorem: Let $U$ be open, in a locally compact Hausdorff space. Let $Ksubset U$ and $K$ be compact. Then there exist an open set $V$ such that
$$Ksubset Vsubset overlineVsubset U $$
Proof:
Because every point of $K$ has a neighbourhood that's closure is compact, and we can cover $K$ with finite amount of such neighbourhoods, then there exist a set $G$, that has a compact closure, and is a subset of $K$. If $U=X$ then we can take $V=G$.
If not, define $C$ as the complement of $U$. From theorem $2.5$, for every point $pin C$, there exist an open set $W_p$, such that $Ksubset W_p$, and $pnotinoverlineW_p$.
And the proof goes on.
Theorem $2.5$: Suppose that $X$ is a Hausdorff space, $Ksubset X$ is a compact set, and $pin K^c$. Then there exist open sets $U$ and $W$, such that $pin U$, $Ksubset W$, and $U cap W = emptyset$.
My question is, how we can use this theorem in the proof, so that we can state the existence of such sets $W_p$. Mainly, where does $pnotinoverlineW_p$ come from. Thank you.
Another question that might help me. If $U$ and $V$ are open, $Ucap V=emptyset$ holds, does $overlineUcap V=emptyset$?
I've found a similar question here. In the comments, Andreas says that if $p$ was in closure of $W_p$, then every neighbourhood of $p$ would have to meet $W_p$. Why is that?
general-topology proof-explanation
asked Jul 22 at 16:36
Habrum
185
I'm reading Rudins "Complex Analysis". There's a theorem that goes like this.
Theorem: Let $U$ be open, in a locally compact Hausdorff space. Let $Ksubset U$ and $K$ be compact. Then there exist an open set $V$ such that
$$Ksubset Vsubset overlineVsubset U $$
Proof:
Because every point of $K$ has a neighbourhood that's closure is compact, and we can cover $K$ with finite amount of such neighbourhoods, then there exist a set $G$, that has a compact closure, and is a subset of $K$. If $U=X$ then we can take $V=G$.
If not, define $C$ as the complement of $U$. From theorem $2.5$, for every point $pin C$, there exist an open set $W_p$, such that $Ksubset W_p$, and $pnotinoverlineW_p$.
And the proof goes on.
Theorem $2.5$: Suppose that $X$ is a Hausdorff space, $Ksubset X$ is a compact set, and $pin K^c$. Then there exist open sets $U$ and $W$, such that $pin U$, $Ksubset W$, and $U cap W = emptyset$.
My question is, how we can use this theorem in the proof, so that we can state the existence of such sets $W_p$. Mainly, where does $pnotinoverlineW_p$ come from. Thank you.
Another question that might help me. If $U$ and $V$ are open, $Ucap V=emptyset$ holds, does $overlineUcap V=emptyset$?
I've found a similar question here. In the comments, Andreas says that if $p$ was in closure of $W_p$, then every neighbourhood of $p$ would have to meet $W_p$. Why is that?
general-topology proof-explanation
asked Jul 22 at 16:36
Habrum
185
edited Jul 22 at 18:12
asked Jul 22 at 16:36
Habrum
185
asked Jul 22 at 16:36
Habrum
185
asked Jul 22 at 16:36
Habrum
185
185
Re: "Another question" : If $V$ is open then its complement is $V^c$ is closed and disjoint from $V $. So if $ V $ is open and is disjoint from $U$ then $Vcap overline Usubset V cap overline V^c=$ $V cap V^c=emptyset.$... Another way to see this is that if $V$ is open and $Ucap V= emptyset,$ then every $pin V $ has a nbhd (namely, $V $ ) that's disjoint from $U,$ so $pnot in overline U.$
– DanielWainfleet
Jul 22 at 19:10
The theorem, with $overline V$ compact, directly: (1). In a Hausdorff space if $C$ is compact and $pnotin C$ than $p, C$ are completely separated. (2). If $D$ is open and $ overline D$ is compact then $partial D= overline D$ $ D$ is compact. (3). We use (1) and (2) to prove that a compact $T_2$ space is regular..... Then for each $p in K$ let $V_p, U_p$ be open sets containing $p$ such that $overline V_p$ is compact and $overline U_psubset U. $ Now $A=V_pcap U_p: pin K$ is an open cover of $K.$ Let $B$ be a finite sub-cover of $A$ and let $V=cup B.$
– DanielWainfleet
Jul 22 at 19:28
ERRATUM: In my previous comment , (3) should say that a LOCALLY compact $T_2$ space is regular. I ran out of editing time .
– DanielWainfleet
Jul 22 at 19:35
1
@DanielWainfleet Thank you! I get it now.
– Habrum
Jul 22 at 20:03
add a comment |Â
Re: "Another question" : If $V$ is open then its complement is $V^c$ is closed and disjoint from $V $. So if $ V $ is open and is disjoint from $U$ then $Vcap overline Usubset V cap overline V^c=$ $V cap V^c=emptyset.$... Another way to see this is that if $V$ is open and $Ucap V= emptyset,$ then every $pin V $ has a nbhd (namely, $V $ ) that's disjoint from $U,$ so $pnot in overline U.$
– DanielWainfleet
Jul 22 at 19:10
The theorem, with $overline V$ compact, directly: (1). In a Hausdorff space if $C$ is compact and $pnotin C$ than $p, C$ are completely separated. (2). If $D$ is open and $ overline D$ is compact then $partial D= overline D$ $ D$ is compact. (3). We use (1) and (2) to prove that a compact $T_2$ space is regular..... Then for each $p in K$ let $V_p, U_p$ be open sets containing $p$ such that $overline V_p$ is compact and $overline U_psubset U. $ Now $A=V_pcap U_p: pin K$ is an open cover of $K.$ Let $B$ be a finite sub-cover of $A$ and let $V=cup B.$
– DanielWainfleet
Jul 22 at 19:28
ERRATUM: In my previous comment , (3) should say that a LOCALLY compact $T_2$ space is regular. I ran out of editing time .
– DanielWainfleet
Jul 22 at 19:35
1
@DanielWainfleet Thank you! I get it now.
– Habrum
Jul 22 at 20:03
Re: "Another question" : If $V$ is open then its complement is $V^c$ is closed and disjoint from $V $. So if $ V $ is open and is disjoint from $U$ then $Vcap overline Usubset V cap overline V^c=$ $V cap V^c=emptyset.$... Another way to see this is that if $V$ is open and $Ucap V= emptyset,$ then every $pin V $ has a nbhd (namely, $V $ ) that's disjoint from $U,$ so $pnot in overline U.$
– DanielWainfleet
Jul 22 at 19:10
Re: "Another question" : If $V$ is open then its complement is $V^c$ is closed and disjoint from $V $. So if $ V $ is open and is disjoint from $U$ then $Vcap overline Usubset V cap overline V^c=$ $V cap V^c=emptyset.$... Another way to see this is that if $V$ is open and $Ucap V= emptyset,$ then every $pin V $ has a nbhd (namely, $V $ ) that's disjoint from $U,$ so $pnot in overline U.$
– DanielWainfleet
Jul 22 at 19:10
The theorem, with $overline V$ compact, directly: (1). In a Hausdorff space if $C$ is compact and $pnotin C$ than $p, C$ are completely separated. (2). If $D$ is open and $ overline D$ is compact then $partial D= overline D$ $ D$ is compact. (3). We use (1) and (2) to prove that a compact $T_2$ space is regular..... Then for each $p in K$ let $V_p, U_p$ be open sets containing $p$ such that $overline V_p$ is compact and $overline U_psubset U. $ Now $A=V_pcap U_p: pin K$ is an open cover of $K.$ Let $B$ be a finite sub-cover of $A$ and let $V=cup B.$
– DanielWainfleet
Jul 22 at 19:28
The theorem, with $overline V$ compact, directly: (1). In a Hausdorff space if $C$ is compact and $pnotin C$ than $p, C$ are completely separated. (2). If $D$ is open and $ overline D$ is compact then $partial D= overline D$ $ D$ is compact. (3). We use (1) and (2) to prove that a compact $T_2$ space is regular..... Then for each $p in K$ let $V_p, U_p$ be open sets containing $p$ such that $overline V_p$ is compact and $overline U_psubset U. $ Now $A=V_pcap U_p: pin K$ is an open cover of $K.$ Let $B$ be a finite sub-cover of $A$ and let $V=cup B.$
– DanielWainfleet
Jul 22 at 19:28
ERRATUM: In my previous comment , (3) should say that a LOCALLY compact $T_2$ space is regular. I ran out of editing time .
– DanielWainfleet
Jul 22 at 19:35
ERRATUM: In my previous comment , (3) should say that a LOCALLY compact $T_2$ space is regular. I ran out of editing time .
– DanielWainfleet
Jul 22 at 19:35
1
1
@DanielWainfleet Thank you! I get it now.
– Habrum
Jul 22 at 20:03
@DanielWainfleet Thank you! I get it now.
– Habrum
Jul 22 at 20:03
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Use the Hausdorff condition. For each $ain K$ there are disjoint
open neighbourhoods $U_a$ and $V_a$ of $p$ and $a$ respectively.
Finitely many of the $V_a$ will cover $K$; let $W$ be their union,
and $U$ the intersection of the corresponding $U_a$.
answered Jul 22 at 16:46
Lord Shark the Unknown
85.2k950111
Yes, that's how the proof of Theorem $2.5$ is stated. My question is how to use this theorem
– Habrum
Jul 22 at 16:50
The point is that $U$ and $W$ are disjoint.
– Lord Shark the Unknown
Jul 22 at 17:02
$U$ and $W$ are disjoint, does that imply that, for example, $overlineU$ and $W$ are disjoint?
– Habrum
Jul 22 at 17:03
$W$ is open
– Lord Shark the Unknown
Jul 22 at 17:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Use the Hausdorff condition. For each $ain K$ there are disjoint
open neighbourhoods $U_a$ and $V_a$ of $p$ and $a$ respectively.
Finitely many of the $V_a$ will cover $K$; let $W$ be their union,
and $U$ the intersection of the corresponding $U_a$.
answered Jul 22 at 16:46
Lord Shark the Unknown
85.2k950111
Yes, that's how the proof of Theorem $2.5$ is stated. My question is how to use this theorem
– Habrum
Jul 22 at 16:50
The point is that $U$ and $W$ are disjoint.
– Lord Shark the Unknown
Jul 22 at 17:02
$U$ and $W$ are disjoint, does that imply that, for example, $overlineU$ and $W$ are disjoint?
– Habrum
Jul 22 at 17:03
$W$ is open
– Lord Shark the Unknown
Jul 22 at 17:04
add a comment |Â
up vote
0
down vote
Use the Hausdorff condition. For each $ain K$ there are disjoint
open neighbourhoods $U_a$ and $V_a$ of $p$ and $a$ respectively.
Finitely many of the $V_a$ will cover $K$; let $W$ be their union,
and $U$ the intersection of the corresponding $U_a$.
answered Jul 22 at 16:46
Lord Shark the Unknown
85.2k950111
Yes, that's how the proof of Theorem $2.5$ is stated. My question is how to use this theorem
– Habrum
Jul 22 at 16:50
The point is that $U$ and $W$ are disjoint.
– Lord Shark the Unknown
Jul 22 at 17:02
$U$ and $W$ are disjoint, does that imply that, for example, $overlineU$ and $W$ are disjoint?
– Habrum
Jul 22 at 17:03
$W$ is open
– Lord Shark the Unknown
Jul 22 at 17:04
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Use the Hausdorff condition. For each $ain K$ there are disjoint
open neighbourhoods $U_a$ and $V_a$ of $p$ and $a$ respectively.
Finitely many of the $V_a$ will cover $K$; let $W$ be their union,
and $U$ the intersection of the corresponding $U_a$.
answered Jul 22 at 16:46
Lord Shark the Unknown
85.2k950111
Use the Hausdorff condition. For each $ain K$ there are disjoint
open neighbourhoods $U_a$ and $V_a$ of $p$ and $a$ respectively.
Finitely many of the $V_a$ will cover $K$; let $W$ be their union,
and $U$ the intersection of the corresponding $U_a$.
answered Jul 22 at 16:46
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 16:46
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 16:46
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 16:46
Lord Shark the Unknown
85.2k950111
85.2k950111
Yes, that's how the proof of Theorem $2.5$ is stated. My question is how to use this theorem
– Habrum
Jul 22 at 16:50
The point is that $U$ and $W$ are disjoint.
– Lord Shark the Unknown
Jul 22 at 17:02
$U$ and $W$ are disjoint, does that imply that, for example, $overlineU$ and $W$ are disjoint?
– Habrum
Jul 22 at 17:03
$W$ is open
– Lord Shark the Unknown
Jul 22 at 17:04
add a comment |Â
Yes, that's how the proof of Theorem $2.5$ is stated. My question is how to use this theorem
– Habrum
Jul 22 at 16:50
The point is that $U$ and $W$ are disjoint.
– Lord Shark the Unknown
Jul 22 at 17:02
$U$ and $W$ are disjoint, does that imply that, for example, $overlineU$ and $W$ are disjoint?
– Habrum
Jul 22 at 17:03
$W$ is open
– Lord Shark the Unknown
Jul 22 at 17:04
Yes, that's how the proof of Theorem $2.5$ is stated. My question is how to use this theorem
– Habrum
Jul 22 at 16:50
Yes, that's how the proof of Theorem $2.5$ is stated. My question is how to use this theorem
– Habrum
Jul 22 at 16:50
The point is that $U$ and $W$ are disjoint.
– Lord Shark the Unknown
Jul 22 at 17:02
The point is that $U$ and $W$ are disjoint.
– Lord Shark the Unknown
Jul 22 at 17:02
$U$ and $W$ are disjoint, does that imply that, for example, $overlineU$ and $W$ are disjoint?
– Habrum
Jul 22 at 17:03
$U$ and $W$ are disjoint, does that imply that, for example, $overlineU$ and $W$ are disjoint?
– Habrum
Jul 22 at 17:03
$W$ is open
– Lord Shark the Unknown
Jul 22 at 17:04
$W$ is open
– Lord Shark the Unknown
Jul 22 at 17:04
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859558%2fproof-that-there-exist-an-open-set-with-compact-closure-explanation%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Re: "Another question" : If $V$ is open then its complement is $V^c$ is closed and disjoint from $V $. So if $ V $ is open and is disjoint from $U$ then $Vcap overline Usubset V cap overline V^c=$ $V cap V^c=emptyset.$... Another way to see this is that if $V$ is open and $Ucap V= emptyset,$ then every $pin V $ has a nbhd (namely, $V $ ) that's disjoint from $U,$ so $pnot in overline U.$
– DanielWainfleet
Jul 22 at 19:10
The theorem, with $overline V$ compact, directly: (1). In a Hausdorff space if $C$ is compact and $pnotin C$ than $p, C$ are completely separated. (2). If $D$ is open and $ overline D$ is compact then $partial D= overline D$ $ D$ is compact. (3). We use (1) and (2) to prove that a compact $T_2$ space is regular..... Then for each $p in K$ let $V_p, U_p$ be open sets containing $p$ such that $overline V_p$ is compact and $overline U_psubset U. $ Now $A=V_pcap U_p: pin K$ is an open cover of $K.$ Let $B$ be a finite sub-cover of $A$ and let $V=cup B.$
– DanielWainfleet
Jul 22 at 19:28
ERRATUM: In my previous comment , (3) should say that a LOCALLY compact $T_2$ space is regular. I ran out of editing time .
– DanielWainfleet
Jul 22 at 19:35
1
@DanielWainfleet Thank you! I get it now.
– Habrum
Jul 22 at 20:03