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metric space is normal proof
Clash Royale CLAN TAG#URR8PPP
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I've seen this proof of the fact that a metric space is normal multiple times but I can't understand how it's valid.
Notation used: For $x in X$, and $Y$ a subset of $X$, define $D(x,Y)=inf d(x, y): y in Y$.
The above proof uses that if $Y$ is a closed set, then $D(x, Y)>0$, $forall x in X setminus Y $ and this in turn implies $space D(x, Y)> epsilon$, $forall x in X setminus Y $ for some $epsilon > 0$.
If the above proof was true, then we could also argue that for any $y in Y$ we have $B(y, fracepsilon3) subseteq Y$ (since $d(y, x)> epsilon, space forall x in X setminus Y Rightarrow B(y, fracepsilon3) cap X setminus Y = emptyset) $ so $Y$ is open which is obviously false.
I don't know what I'm missing becuase this kind of proof seems to appear everywhere.
Thank you!
metric-spaces topological-vector-spaces
asked yesterday
harlem
15716
 |Â
show 1 more comment
up vote
1
down vote
favorite
I've seen this proof of the fact that a metric space is normal multiple times but I can't understand how it's valid.
Notation used: For $x in X$, and $Y$ a subset of $X$, define $D(x,Y)=inf d(x, y): y in Y$.
The above proof uses that if $Y$ is a closed set, then $D(x, Y)>0$, $forall x in X setminus Y $ and this in turn implies $space D(x, Y)> epsilon$, $forall x in X setminus Y $ for some $epsilon > 0$.
If the above proof was true, then we could also argue that for any $y in Y$ we have $B(y, fracepsilon3) subseteq Y$ (since $d(y, x)> epsilon, space forall x in X setminus Y Rightarrow B(y, fracepsilon3) cap X setminus Y = emptyset) $ so $Y$ is open which is obviously false.
I don't know what I'm missing becuase this kind of proof seems to appear everywhere.
Thank you!
metric-spaces topological-vector-spaces
asked yesterday
harlem
15716
1
What do you mean by $inf d(x,Y)$ for all $x in X setminus Y$? What are you talking the $inf$ over?
– copper.hat
yesterday
I edited the question to avoid conusion.
– harlem
yesterday
It is not generally true that if $D(x,Y)>0$ for some $x$ that you have $D(z,Y) > epsilon >0$ for all $z notin Y$.
– copper.hat
yesterday
2
The proof seems to be incorrect unless you assume the metric space is compact. For example, in $mathbbR^2$, if $C_1 = (x, y) mid x = 0 $ and $C_2 = (x, y) mid xy = 1 $ then $d(C_1, C_2) = 0$ even though $C_1$ and $C_2$ are disjoint closed subsets.
– Daniel Schepler
yesterday
1
While indeed $Y$ closed implies $d(x, Y) > 0$ whenever $x in X setminus Y$, in general there is no uniform value of $epsilon > 0$ such that $d(x, Y) > epsilon$ for each $x in X setminus Y$.
– Daniel Schepler
yesterday
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've seen this proof of the fact that a metric space is normal multiple times but I can't understand how it's valid.
Notation used: For $x in X$, and $Y$ a subset of $X$, define $D(x,Y)=inf d(x, y): y in Y$.
The above proof uses that if $Y$ is a closed set, then $D(x, Y)>0$, $forall x in X setminus Y $ and this in turn implies $space D(x, Y)> epsilon$, $forall x in X setminus Y $ for some $epsilon > 0$.
If the above proof was true, then we could also argue that for any $y in Y$ we have $B(y, fracepsilon3) subseteq Y$ (since $d(y, x)> epsilon, space forall x in X setminus Y Rightarrow B(y, fracepsilon3) cap X setminus Y = emptyset) $ so $Y$ is open which is obviously false.
I don't know what I'm missing becuase this kind of proof seems to appear everywhere.
Thank you!
metric-spaces topological-vector-spaces
asked yesterday
harlem
15716
I've seen this proof of the fact that a metric space is normal multiple times but I can't understand how it's valid.
Notation used: For $x in X$, and $Y$ a subset of $X$, define $D(x,Y)=inf d(x, y): y in Y$.
The above proof uses that if $Y$ is a closed set, then $D(x, Y)>0$, $forall x in X setminus Y $ and this in turn implies $space D(x, Y)> epsilon$, $forall x in X setminus Y $ for some $epsilon > 0$.
If the above proof was true, then we could also argue that for any $y in Y$ we have $B(y, fracepsilon3) subseteq Y$ (since $d(y, x)> epsilon, space forall x in X setminus Y Rightarrow B(y, fracepsilon3) cap X setminus Y = emptyset) $ so $Y$ is open which is obviously false.
I don't know what I'm missing becuase this kind of proof seems to appear everywhere.
Thank you!
metric-spaces topological-vector-spaces
asked yesterday
harlem
15716
edited yesterday
asked yesterday
harlem
15716
asked yesterday
harlem
15716
asked yesterday
harlem
15716
15716
1
What do you mean by $inf d(x,Y)$ for all $x in X setminus Y$? What are you talking the $inf$ over?
– copper.hat
yesterday
I edited the question to avoid conusion.
– harlem
yesterday
It is not generally true that if $D(x,Y)>0$ for some $x$ that you have $D(z,Y) > epsilon >0$ for all $z notin Y$.
– copper.hat
yesterday
2
The proof seems to be incorrect unless you assume the metric space is compact. For example, in $mathbbR^2$, if $C_1 = (x, y) mid x = 0 $ and $C_2 = (x, y) mid xy = 1 $ then $d(C_1, C_2) = 0$ even though $C_1$ and $C_2$ are disjoint closed subsets.
– Daniel Schepler
yesterday
1
While indeed $Y$ closed implies $d(x, Y) > 0$ whenever $x in X setminus Y$, in general there is no uniform value of $epsilon > 0$ such that $d(x, Y) > epsilon$ for each $x in X setminus Y$.
– Daniel Schepler
yesterday
 |Â
show 1 more comment
1
What do you mean by $inf d(x,Y)$ for all $x in X setminus Y$? What are you talking the $inf$ over?
– copper.hat
yesterday
I edited the question to avoid conusion.
– harlem
yesterday
It is not generally true that if $D(x,Y)>0$ for some $x$ that you have $D(z,Y) > epsilon >0$ for all $z notin Y$.
– copper.hat
yesterday
2
The proof seems to be incorrect unless you assume the metric space is compact. For example, in $mathbbR^2$, if $C_1 = (x, y) mid x = 0 $ and $C_2 = (x, y) mid xy = 1 $ then $d(C_1, C_2) = 0$ even though $C_1$ and $C_2$ are disjoint closed subsets.
– Daniel Schepler
yesterday
1
While indeed $Y$ closed implies $d(x, Y) > 0$ whenever $x in X setminus Y$, in general there is no uniform value of $epsilon > 0$ such that $d(x, Y) > epsilon$ for each $x in X setminus Y$.
– Daniel Schepler
yesterday
1
1
What do you mean by $inf d(x,Y)$ for all $x in X setminus Y$? What are you talking the $inf$ over?
– copper.hat
yesterday
What do you mean by $inf d(x,Y)$ for all $x in X setminus Y$? What are you talking the $inf$ over?
– copper.hat
yesterday
I edited the question to avoid conusion.
– harlem
yesterday
I edited the question to avoid conusion.
– harlem
yesterday
It is not generally true that if $D(x,Y)>0$ for some $x$ that you have $D(z,Y) > epsilon >0$ for all $z notin Y$.
– copper.hat
yesterday
It is not generally true that if $D(x,Y)>0$ for some $x$ that you have $D(z,Y) > epsilon >0$ for all $z notin Y$.
– copper.hat
yesterday
2
2
The proof seems to be incorrect unless you assume the metric space is compact. For example, in $mathbbR^2$, if $C_1 = (x, y) mid x = 0 $ and $C_2 = (x, y) mid xy = 1 $ then $d(C_1, C_2) = 0$ even though $C_1$ and $C_2$ are disjoint closed subsets.
– Daniel Schepler
yesterday
The proof seems to be incorrect unless you assume the metric space is compact. For example, in $mathbbR^2$, if $C_1 = (x, y) mid x = 0 $ and $C_2 = (x, y) mid xy = 1 $ then $d(C_1, C_2) = 0$ even though $C_1$ and $C_2$ are disjoint closed subsets.
– Daniel Schepler
yesterday
1
1
While indeed $Y$ closed implies $d(x, Y) > 0$ whenever $x in X setminus Y$, in general there is no uniform value of $epsilon > 0$ such that $d(x, Y) > epsilon$ for each $x in X setminus Y$.
– Daniel Schepler
yesterday
While indeed $Y$ closed implies $d(x, Y) > 0$ whenever $x in X setminus Y$, in general there is no uniform value of $epsilon > 0$ such that $d(x, Y) > epsilon$ for each $x in X setminus Y$.
– Daniel Schepler
yesterday
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The proof in the image you linked to is not a valid proof.
It's not necessarily true that for all pairs $C_1,C_2$ of nonempty disjoint closed subsets of $X$, we have $d(C_1,C_2) > 0$.
For example, if $X=mathbbR^2$, and
beginalign*
C_1&= (a,0)mid ainmathbbR\[4pt]
C_2&=bigl(b,smallfrac1bbigr)mid b > 0\[4pt]
endalign*
then $C_1,C_2$ are nonempty disjoint closed subsets of $X$, but $d(C_1,C_2)=0$.
What is true is that if $C$ is a nonempty closed subset of $X$, and $xin X$, then $d(x,C)=0;$if and only if $xin C$.
Proof:$;$If $xin C$, then of course, $d(x,C)=0.;$Conversely, suppose $C$ is a nonempty closed subset of $X$, and $xin X$ is such that $d(x,C)=0.;$Then since $d(x,C)=0$, it follows that $B(x,r)cap C$ is nonempty, for all $r > 0,;$hence $x$ is in the closure of $C$, which is $C$.
Hence, if $C$ is a nonempty closed subset of $X$, then for all $xin XsetminusC$, we have $d(x,C) > 0$.
The proof can then be continued as follows . . .
- For each $xin C_1$, let $r=largefracd(x,C_2)3$, and let $U_x=B(x,r)$.$\[4pt]$
- For each $yin C_2$, let $s=largefracd(y,C_1)3$, and let $V_y=B(y,s)$.
Now let
beginalign*
U&=bigcup_xin C_1 U_x\[4pt]
V&=bigcup_yin C_2 V_y\[4pt]
endalign*
It's clear that $U,V$ are open subsets of $X$, with $C_1subseteq U$, and $C_2subseteq V$.
Suppose $Ucap Vnelargevarnothing$.
Let $zin Ucap V$.
Since $zin U$, we must have $zin U_x$, for some $xin C_1$, hence $d(x,z) < r$, where $r=largefracd(x,C_2)3$.
Since $zin V$, we must have $zin V_y$, for some $yin C_2$, hence $d(y,z) < s$, where $s=largefracd(y,C_1)3$.
Without loss of generality, assume $rge s$.$;$Then
$$3r=d(x,C_2)le d(x,y)le d(x,z)+d(y,z)< r+sle 2r$$
contradiction.
Therefore $Ucap V=largevarnothing$.
It follows that $X$ is normal.
answered yesterday
quasi
32.9k22258
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The proof in the image you linked to is not a valid proof.
It's not necessarily true that for all pairs $C_1,C_2$ of nonempty disjoint closed subsets of $X$, we have $d(C_1,C_2) > 0$.
For example, if $X=mathbbR^2$, and
beginalign*
C_1&= (a,0)mid ainmathbbR\[4pt]
C_2&=bigl(b,smallfrac1bbigr)mid b > 0\[4pt]
endalign*
then $C_1,C_2$ are nonempty disjoint closed subsets of $X$, but $d(C_1,C_2)=0$.
What is true is that if $C$ is a nonempty closed subset of $X$, and $xin X$, then $d(x,C)=0;$if and only if $xin C$.
Proof:$;$If $xin C$, then of course, $d(x,C)=0.;$Conversely, suppose $C$ is a nonempty closed subset of $X$, and $xin X$ is such that $d(x,C)=0.;$Then since $d(x,C)=0$, it follows that $B(x,r)cap C$ is nonempty, for all $r > 0,;$hence $x$ is in the closure of $C$, which is $C$.
Hence, if $C$ is a nonempty closed subset of $X$, then for all $xin XsetminusC$, we have $d(x,C) > 0$.
The proof can then be continued as follows . . .
- For each $xin C_1$, let $r=largefracd(x,C_2)3$, and let $U_x=B(x,r)$.$\[4pt]$
- For each $yin C_2$, let $s=largefracd(y,C_1)3$, and let $V_y=B(y,s)$.
Now let
beginalign*
U&=bigcup_xin C_1 U_x\[4pt]
V&=bigcup_yin C_2 V_y\[4pt]
endalign*
It's clear that $U,V$ are open subsets of $X$, with $C_1subseteq U$, and $C_2subseteq V$.
Suppose $Ucap Vnelargevarnothing$.
Let $zin Ucap V$.
Since $zin U$, we must have $zin U_x$, for some $xin C_1$, hence $d(x,z) < r$, where $r=largefracd(x,C_2)3$.
Since $zin V$, we must have $zin V_y$, for some $yin C_2$, hence $d(y,z) < s$, where $s=largefracd(y,C_1)3$.
Without loss of generality, assume $rge s$.$;$Then
$$3r=d(x,C_2)le d(x,y)le d(x,z)+d(y,z)< r+sle 2r$$
contradiction.
Therefore $Ucap V=largevarnothing$.
It follows that $X$ is normal.
answered yesterday
quasi
32.9k22258
add a comment |Â
up vote
2
down vote
accepted
The proof in the image you linked to is not a valid proof.
It's not necessarily true that for all pairs $C_1,C_2$ of nonempty disjoint closed subsets of $X$, we have $d(C_1,C_2) > 0$.
For example, if $X=mathbbR^2$, and
beginalign*
C_1&= (a,0)mid ainmathbbR\[4pt]
C_2&=bigl(b,smallfrac1bbigr)mid b > 0\[4pt]
endalign*
then $C_1,C_2$ are nonempty disjoint closed subsets of $X$, but $d(C_1,C_2)=0$.
What is true is that if $C$ is a nonempty closed subset of $X$, and $xin X$, then $d(x,C)=0;$if and only if $xin C$.
Proof:$;$If $xin C$, then of course, $d(x,C)=0.;$Conversely, suppose $C$ is a nonempty closed subset of $X$, and $xin X$ is such that $d(x,C)=0.;$Then since $d(x,C)=0$, it follows that $B(x,r)cap C$ is nonempty, for all $r > 0,;$hence $x$ is in the closure of $C$, which is $C$.
Hence, if $C$ is a nonempty closed subset of $X$, then for all $xin XsetminusC$, we have $d(x,C) > 0$.
The proof can then be continued as follows . . .
- For each $xin C_1$, let $r=largefracd(x,C_2)3$, and let $U_x=B(x,r)$.$\[4pt]$
- For each $yin C_2$, let $s=largefracd(y,C_1)3$, and let $V_y=B(y,s)$.
Now let
beginalign*
U&=bigcup_xin C_1 U_x\[4pt]
V&=bigcup_yin C_2 V_y\[4pt]
endalign*
It's clear that $U,V$ are open subsets of $X$, with $C_1subseteq U$, and $C_2subseteq V$.
Suppose $Ucap Vnelargevarnothing$.
Let $zin Ucap V$.
Since $zin U$, we must have $zin U_x$, for some $xin C_1$, hence $d(x,z) < r$, where $r=largefracd(x,C_2)3$.
Since $zin V$, we must have $zin V_y$, for some $yin C_2$, hence $d(y,z) < s$, where $s=largefracd(y,C_1)3$.
Without loss of generality, assume $rge s$.$;$Then
$$3r=d(x,C_2)le d(x,y)le d(x,z)+d(y,z)< r+sle 2r$$
contradiction.
Therefore $Ucap V=largevarnothing$.
It follows that $X$ is normal.
answered yesterday
quasi
32.9k22258
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The proof in the image you linked to is not a valid proof.
It's not necessarily true that for all pairs $C_1,C_2$ of nonempty disjoint closed subsets of $X$, we have $d(C_1,C_2) > 0$.
For example, if $X=mathbbR^2$, and
beginalign*
C_1&= (a,0)mid ainmathbbR\[4pt]
C_2&=bigl(b,smallfrac1bbigr)mid b > 0\[4pt]
endalign*
then $C_1,C_2$ are nonempty disjoint closed subsets of $X$, but $d(C_1,C_2)=0$.
What is true is that if $C$ is a nonempty closed subset of $X$, and $xin X$, then $d(x,C)=0;$if and only if $xin C$.
Proof:$;$If $xin C$, then of course, $d(x,C)=0.;$Conversely, suppose $C$ is a nonempty closed subset of $X$, and $xin X$ is such that $d(x,C)=0.;$Then since $d(x,C)=0$, it follows that $B(x,r)cap C$ is nonempty, for all $r > 0,;$hence $x$ is in the closure of $C$, which is $C$.
Hence, if $C$ is a nonempty closed subset of $X$, then for all $xin XsetminusC$, we have $d(x,C) > 0$.
The proof can then be continued as follows . . .
- For each $xin C_1$, let $r=largefracd(x,C_2)3$, and let $U_x=B(x,r)$.$\[4pt]$
- For each $yin C_2$, let $s=largefracd(y,C_1)3$, and let $V_y=B(y,s)$.
Now let
beginalign*
U&=bigcup_xin C_1 U_x\[4pt]
V&=bigcup_yin C_2 V_y\[4pt]
endalign*
It's clear that $U,V$ are open subsets of $X$, with $C_1subseteq U$, and $C_2subseteq V$.
Suppose $Ucap Vnelargevarnothing$.
Let $zin Ucap V$.
Since $zin U$, we must have $zin U_x$, for some $xin C_1$, hence $d(x,z) < r$, where $r=largefracd(x,C_2)3$.
Since $zin V$, we must have $zin V_y$, for some $yin C_2$, hence $d(y,z) < s$, where $s=largefracd(y,C_1)3$.
Without loss of generality, assume $rge s$.$;$Then
$$3r=d(x,C_2)le d(x,y)le d(x,z)+d(y,z)< r+sle 2r$$
contradiction.
Therefore $Ucap V=largevarnothing$.
It follows that $X$ is normal.
answered yesterday
quasi
32.9k22258
The proof in the image you linked to is not a valid proof.
It's not necessarily true that for all pairs $C_1,C_2$ of nonempty disjoint closed subsets of $X$, we have $d(C_1,C_2) > 0$.
For example, if $X=mathbbR^2$, and
beginalign*
C_1&= (a,0)mid ainmathbbR\[4pt]
C_2&=bigl(b,smallfrac1bbigr)mid b > 0\[4pt]
endalign*
then $C_1,C_2$ are nonempty disjoint closed subsets of $X$, but $d(C_1,C_2)=0$.
What is true is that if $C$ is a nonempty closed subset of $X$, and $xin X$, then $d(x,C)=0;$if and only if $xin C$.
Proof:$;$If $xin C$, then of course, $d(x,C)=0.;$Conversely, suppose $C$ is a nonempty closed subset of $X$, and $xin X$ is such that $d(x,C)=0.;$Then since $d(x,C)=0$, it follows that $B(x,r)cap C$ is nonempty, for all $r > 0,;$hence $x$ is in the closure of $C$, which is $C$.
Hence, if $C$ is a nonempty closed subset of $X$, then for all $xin XsetminusC$, we have $d(x,C) > 0$.
The proof can then be continued as follows . . .
- For each $xin C_1$, let $r=largefracd(x,C_2)3$, and let $U_x=B(x,r)$.$\[4pt]$
- For each $yin C_2$, let $s=largefracd(y,C_1)3$, and let $V_y=B(y,s)$.
Now let
beginalign*
U&=bigcup_xin C_1 U_x\[4pt]
V&=bigcup_yin C_2 V_y\[4pt]
endalign*
It's clear that $U,V$ are open subsets of $X$, with $C_1subseteq U$, and $C_2subseteq V$.
Suppose $Ucap Vnelargevarnothing$.
Let $zin Ucap V$.
Since $zin U$, we must have $zin U_x$, for some $xin C_1$, hence $d(x,z) < r$, where $r=largefracd(x,C_2)3$.
Since $zin V$, we must have $zin V_y$, for some $yin C_2$, hence $d(y,z) < s$, where $s=largefracd(y,C_1)3$.
Without loss of generality, assume $rge s$.$;$Then
$$3r=d(x,C_2)le d(x,y)le d(x,z)+d(y,z)< r+sle 2r$$
contradiction.
Therefore $Ucap V=largevarnothing$.
It follows that $X$ is normal.
answered yesterday
quasi
32.9k22258
edited yesterday
answered yesterday
quasi
32.9k22258
answered yesterday
quasi
32.9k22258
answered yesterday
quasi
32.9k22258
32.9k22258
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1
What do you mean by $inf d(x,Y)$ for all $x in X setminus Y$? What are you talking the $inf$ over?
– copper.hat
yesterday
I edited the question to avoid conusion.
– harlem
yesterday
It is not generally true that if $D(x,Y)>0$ for some $x$ that you have $D(z,Y) > epsilon >0$ for all $z notin Y$.
– copper.hat
yesterday
2
The proof seems to be incorrect unless you assume the metric space is compact. For example, in $mathbbR^2$, if $C_1 = (x, y) mid x = 0 $ and $C_2 = (x, y) mid xy = 1 $ then $d(C_1, C_2) = 0$ even though $C_1$ and $C_2$ are disjoint closed subsets.
– Daniel Schepler
yesterday
1
While indeed $Y$ closed implies $d(x, Y) > 0$ whenever $x in X setminus Y$, in general there is no uniform value of $epsilon > 0$ such that $d(x, Y) > epsilon$ for each $x in X setminus Y$.
– Daniel Schepler
yesterday