In a T1 space, derived set = closure
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How do I do the proof? My "proof" is more like a list of to me known facts.
Lemma. In a $T_1$ space, $bar{S}=S'$.
Proof.
We know that $bar{S}=Scup S'$ (already proven), thus $S'=bar{S}$ iff $Ssubseteq S'$. $xin S'$ iff $forall U_xin N(x)$, $U_xcap S-{x}nevarnothing$. Given any $xin S$, suppose $xnotin S'$, then, since $S'$ is closed (already proven, valid only in $T_1$), there is $Uin N(x)$ such that $Ucap S'=varnothing$. $x$ is an isolated point, thus there is $Vin N(x)$ such that $Scap V={x}$. I've also proved that in $T_1$ singletons are closed.
general-topology
asked Nov 21 at 9:30
PeptideChain
402310
add a comment |
up vote
0
down vote
favorite
How do I do the proof? My "proof" is more like a list of to me known facts.
Lemma. In a $T_1$ space, $bar{S}=S'$.
Proof.
We know that $bar{S}=Scup S'$ (already proven), thus $S'=bar{S}$ iff $Ssubseteq S'$. $xin S'$ iff $forall U_xin N(x)$, $U_xcap S-{x}nevarnothing$. Given any $xin S$, suppose $xnotin S'$, then, since $S'$ is closed (already proven, valid only in $T_1$), there is $Uin N(x)$ such that $Ucap S'=varnothing$. $x$ is an isolated point, thus there is $Vin N(x)$ such that $Scap V={x}$. I've also proved that in $T_1$ singletons are closed.
general-topology
asked Nov 21 at 9:30
PeptideChain
402310
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do I do the proof? My "proof" is more like a list of to me known facts.
Lemma. In a $T_1$ space, $bar{S}=S'$.
Proof.
We know that $bar{S}=Scup S'$ (already proven), thus $S'=bar{S}$ iff $Ssubseteq S'$. $xin S'$ iff $forall U_xin N(x)$, $U_xcap S-{x}nevarnothing$. Given any $xin S$, suppose $xnotin S'$, then, since $S'$ is closed (already proven, valid only in $T_1$), there is $Uin N(x)$ such that $Ucap S'=varnothing$. $x$ is an isolated point, thus there is $Vin N(x)$ such that $Scap V={x}$. I've also proved that in $T_1$ singletons are closed.
general-topology
asked Nov 21 at 9:30
PeptideChain
402310
How do I do the proof? My "proof" is more like a list of to me known facts.
Lemma. In a $T_1$ space, $bar{S}=S'$.
Proof.
We know that $bar{S}=Scup S'$ (already proven), thus $S'=bar{S}$ iff $Ssubseteq S'$. $xin S'$ iff $forall U_xin N(x)$, $U_xcap S-{x}nevarnothing$. Given any $xin S$, suppose $xnotin S'$, then, since $S'$ is closed (already proven, valid only in $T_1$), there is $Uin N(x)$ such that $Ucap S'=varnothing$. $x$ is an isolated point, thus there is $Vin N(x)$ such that $Scap V={x}$. I've also proved that in $T_1$ singletons are closed.
general-topology
general-topology
asked Nov 21 at 9:30
PeptideChain
402310
asked Nov 21 at 9:30
PeptideChain
402310
asked Nov 21 at 9:30
PeptideChain
402310
asked Nov 21 at 9:30
PeptideChain
402310
asked Nov 21 at 9:30
PeptideChain
402310
402310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The lemma is not true.
Let $S={x}$.
Then $S'=varnothing$ so cannot equalize the non-empty set $overline S$.
answered Nov 21 at 9:41
drhab
95.6k543126
so many wasted hours...and now I get why there was no way to find an answer in internet... tnx
– PeptideChain
Nov 21 at 9:44
You are welcome.
– drhab
Nov 21 at 9:45
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The lemma is not true.
Let $S={x}$.
Then $S'=varnothing$ so cannot equalize the non-empty set $overline S$.
answered Nov 21 at 9:41
drhab
95.6k543126
so many wasted hours...and now I get why there was no way to find an answer in internet... tnx
– PeptideChain
Nov 21 at 9:44
You are welcome.
– drhab
Nov 21 at 9:45
add a comment |
up vote
3
down vote
accepted
The lemma is not true.
Let $S={x}$.
Then $S'=varnothing$ so cannot equalize the non-empty set $overline S$.
answered Nov 21 at 9:41
drhab
95.6k543126
so many wasted hours...and now I get why there was no way to find an answer in internet... tnx
– PeptideChain
Nov 21 at 9:44
You are welcome.
– drhab
Nov 21 at 9:45
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The lemma is not true.
Let $S={x}$.
Then $S'=varnothing$ so cannot equalize the non-empty set $overline S$.
answered Nov 21 at 9:41
drhab
95.6k543126
The lemma is not true.
Let $S={x}$.
Then $S'=varnothing$ so cannot equalize the non-empty set $overline S$.
answered Nov 21 at 9:41
drhab
95.6k543126
answered Nov 21 at 9:41
drhab
95.6k543126
answered Nov 21 at 9:41
drhab
95.6k543126
answered Nov 21 at 9:41
drhab
95.6k543126
95.6k543126
so many wasted hours...and now I get why there was no way to find an answer in internet... tnx
– PeptideChain
Nov 21 at 9:44
You are welcome.
– drhab
Nov 21 at 9:45
add a comment |
so many wasted hours...and now I get why there was no way to find an answer in internet... tnx
– PeptideChain
Nov 21 at 9:44
You are welcome.
– drhab
Nov 21 at 9:45
so many wasted hours...and now I get why there was no way to find an answer in internet... tnx
– PeptideChain
Nov 21 at 9:44
so many wasted hours...and now I get why there was no way to find an answer in internet... tnx
– PeptideChain
Nov 21 at 9:44
You are welcome.
– drhab
Nov 21 at 9:45
You are welcome.
– drhab
Nov 21 at 9:45
add a comment |
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