For a closed $G_delta$ set $Fsubseteq X,$ does there exist a continuous function $f:Xto [0,1]$ such that...
$begingroup$
(All spaces are Hausdorff.)
This question is a variant of my previous question.
Let $X$ be a completely regular space, that is, for every closed set $Fsubseteq X$ and $xnotin F,$ there exists a continuous function $g:Xto [0,1]$ such that $g(F) = {0}$ and $g(x) =1.$
Question: For every closed $G_delta$ set $Fsubseteq X,$ does there exist a continuous function $f:Xto [0,1]$ such that $f=0$ on $F$ and $fneq 0$ outside $F?$
A subset $Usubseteq X$ is a zero set if there exists a continuous function $g:Xto [0,1]$ such that $g^{-1}({0}) = U.$
It is well-known that if $X$ is normal then all closed $G_delta$ sets are zero sets.
However, I am not sure whether the same holds for completely regular space.
general-topology separation-axioms
edited Jan 5 at 18:32
Paul Frost
11.4k3934
asked Jan 5 at 12:51
IdonknowIdonknow
2,504850114
$endgroup$
|
show 5 more comments
$begingroup$
(All spaces are Hausdorff.)
This question is a variant of my previous question.
Let $X$ be a completely regular space, that is, for every closed set $Fsubseteq X$ and $xnotin F,$ there exists a continuous function $g:Xto [0,1]$ such that $g(F) = {0}$ and $g(x) =1.$
Question: For every closed $G_delta$ set $Fsubseteq X,$ does there exist a continuous function $f:Xto [0,1]$ such that $f=0$ on $F$ and $fneq 0$ outside $F?$
A subset $Usubseteq X$ is a zero set if there exists a continuous function $g:Xto [0,1]$ such that $g^{-1}({0}) = U.$
It is well-known that if $X$ is normal then all closed $G_delta$ sets are zero sets.
However, I am not sure whether the same holds for completely regular space.
general-topology separation-axioms
edited Jan 5 at 18:32
Paul Frost
11.4k3934
asked Jan 5 at 12:51
IdonknowIdonknow
2,504850114
$endgroup$
$begingroup$
I know that $X$ normal implies that all closed $G_delta$'s are zero-sets, but I believe the reverse does not hold. If it did your question would be useless. Do you have a reference ? "well-known" it is not, it's not mentioned in Engelking, a standard reference for such matters.
$endgroup$
– Henno Brandsma
Jan 5 at 14:26
2
$begingroup$
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
$endgroup$
– Henno Brandsma
Jan 5 at 14:27
$begingroup$
@HennoBrandsma yes, you are right. I have edited my post.
$endgroup$
– Idonknow
Jan 5 at 14:59
$begingroup$
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
$endgroup$
– Idonknow
Jan 5 at 15:00
1
$begingroup$
@PaulFrost that is the direction I agreed with, yes. The OP originally claimed that if all closed $G_delta$ sets are zero-sets then $X$ is normal. This I doubt the truth of, as said.
$endgroup$
– Henno Brandsma
Jan 5 at 18:46
|
show 5 more comments
$begingroup$
(All spaces are Hausdorff.)
This question is a variant of my previous question.
Let $X$ be a completely regular space, that is, for every closed set $Fsubseteq X$ and $xnotin F,$ there exists a continuous function $g:Xto [0,1]$ such that $g(F) = {0}$ and $g(x) =1.$
Question: For every closed $G_delta$ set $Fsubseteq X,$ does there exist a continuous function $f:Xto [0,1]$ such that $f=0$ on $F$ and $fneq 0$ outside $F?$
A subset $Usubseteq X$ is a zero set if there exists a continuous function $g:Xto [0,1]$ such that $g^{-1}({0}) = U.$
It is well-known that if $X$ is normal then all closed $G_delta$ sets are zero sets.
However, I am not sure whether the same holds for completely regular space.
general-topology separation-axioms
edited Jan 5 at 18:32
Paul Frost
11.4k3934
asked Jan 5 at 12:51
IdonknowIdonknow
2,504850114
$endgroup$
(All spaces are Hausdorff.)
This question is a variant of my previous question.
Let $X$ be a completely regular space, that is, for every closed set $Fsubseteq X$ and $xnotin F,$ there exists a continuous function $g:Xto [0,1]$ such that $g(F) = {0}$ and $g(x) =1.$
Question: For every closed $G_delta$ set $Fsubseteq X,$ does there exist a continuous function $f:Xto [0,1]$ such that $f=0$ on $F$ and $fneq 0$ outside $F?$
A subset $Usubseteq X$ is a zero set if there exists a continuous function $g:Xto [0,1]$ such that $g^{-1}({0}) = U.$
It is well-known that if $X$ is normal then all closed $G_delta$ sets are zero sets.
However, I am not sure whether the same holds for completely regular space.
general-topology separation-axioms
general-topology separation-axioms
edited Jan 5 at 18:32
Paul Frost
11.4k3934
asked Jan 5 at 12:51
IdonknowIdonknow
2,504850114
edited Jan 5 at 18:32
Paul Frost
11.4k3934
asked Jan 5 at 12:51
IdonknowIdonknow
2,504850114
edited Jan 5 at 18:32
Paul Frost
11.4k3934
edited Jan 5 at 18:32
Paul Frost
11.4k3934
edited Jan 5 at 18:32
Paul Frost
11.4k3934
11.4k3934
asked Jan 5 at 12:51
IdonknowIdonknow
2,504850114
asked Jan 5 at 12:51
IdonknowIdonknow
2,504850114
asked Jan 5 at 12:51
IdonknowIdonknow
2,504850114
2,504850114
$begingroup$
I know that $X$ normal implies that all closed $G_delta$'s are zero-sets, but I believe the reverse does not hold. If it did your question would be useless. Do you have a reference ? "well-known" it is not, it's not mentioned in Engelking, a standard reference for such matters.
$endgroup$
– Henno Brandsma
Jan 5 at 14:26
2
$begingroup$
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
$endgroup$
– Henno Brandsma
Jan 5 at 14:27
$begingroup$
@HennoBrandsma yes, you are right. I have edited my post.
$endgroup$
– Idonknow
Jan 5 at 14:59
$begingroup$
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
$endgroup$
– Idonknow
Jan 5 at 15:00
1
$begingroup$
@PaulFrost that is the direction I agreed with, yes. The OP originally claimed that if all closed $G_delta$ sets are zero-sets then $X$ is normal. This I doubt the truth of, as said.
$endgroup$
– Henno Brandsma
Jan 5 at 18:46
|
show 5 more comments
$begingroup$
I know that $X$ normal implies that all closed $G_delta$'s are zero-sets, but I believe the reverse does not hold. If it did your question would be useless. Do you have a reference ? "well-known" it is not, it's not mentioned in Engelking, a standard reference for such matters.
$endgroup$
– Henno Brandsma
Jan 5 at 14:26
2
$begingroup$
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
$endgroup$
– Henno Brandsma
Jan 5 at 14:27
$begingroup$
@HennoBrandsma yes, you are right. I have edited my post.
$endgroup$
– Idonknow
Jan 5 at 14:59
$begingroup$
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
$endgroup$
– Idonknow
Jan 5 at 15:00
1
$begingroup$
@PaulFrost that is the direction I agreed with, yes. The OP originally claimed that if all closed $G_delta$ sets are zero-sets then $X$ is normal. This I doubt the truth of, as said.
$endgroup$
– Henno Brandsma
Jan 5 at 18:46
$begingroup$
I know that $X$ normal implies that all closed $G_delta$'s are zero-sets, but I believe the reverse does not hold. If it did your question would be useless. Do you have a reference ? "well-known" it is not, it's not mentioned in Engelking, a standard reference for such matters.
$endgroup$
– Henno Brandsma
Jan 5 at 14:26
$begingroup$
I know that $X$ normal implies that all closed $G_delta$'s are zero-sets, but I believe the reverse does not hold. If it did your question would be useless. Do you have a reference ? "well-known" it is not, it's not mentioned in Engelking, a standard reference for such matters.
$endgroup$
– Henno Brandsma
Jan 5 at 14:26
2
2
$begingroup$
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
$endgroup$
– Henno Brandsma
Jan 5 at 14:27
$begingroup$
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
$endgroup$
– Henno Brandsma
Jan 5 at 14:27
$begingroup$
@HennoBrandsma yes, you are right. I have edited my post.
$endgroup$
– Idonknow
Jan 5 at 14:59
$begingroup$
@HennoBrandsma yes, you are right. I have edited my post.
$endgroup$
– Idonknow
Jan 5 at 14:59
$begingroup$
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
$endgroup$
– Idonknow
Jan 5 at 15:00
$begingroup$
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
$endgroup$
– Idonknow
Jan 5 at 15:00
1
1
$begingroup$
@PaulFrost that is the direction I agreed with, yes. The OP originally claimed that if all closed $G_delta$ sets are zero-sets then $X$ is normal. This I doubt the truth of, as said.
$endgroup$
– Henno Brandsma
Jan 5 at 18:46
$begingroup$
@PaulFrost that is the direction I agreed with, yes. The OP originally claimed that if all closed $G_delta$ sets are zero-sets then $X$ is normal. This I doubt the truth of, as said.
$endgroup$
– Henno Brandsma
Jan 5 at 18:46
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
There is a well-known counterexample. Let $S$ be the Sorgenfrey line, that is the real
line endowed with the Sorgenfrey topology (generated by the base consisting of half-intervals $[a,b)$, $a<b$). It is well-known that a product $X=Stimes S$ is a Tychonoff but non-normal space (see, for instance, Examples 1.4.4 and 2.3.12 in [Eng]). Let $D={(x,-x)in Stimes S: xin S}$. Then any subset $Y$ of $D$ is closed in $X$, and we have $2^{|D|}=2^{frak c}$ such subsets. Moreover, since $Y=bigcap_{ninBbb N} Y+[0,1/n)times [0,1/n)$, $Y$ is a $G_delta$-subset of $X$.
On the other hand, let $C$ be a countable dense subset of $X$. By Theorem 2.1.9 in [Eng], each continuous real-valued function $f$ on $X$ is uniquely determined by its restriction on $C$, so there are at most $frak c^omega=frak c$ such functions. Thus most subsets of $D$ are not zero-sets.
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 7 at 7:14
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
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1 Answer
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$begingroup$
There is a well-known counterexample. Let $S$ be the Sorgenfrey line, that is the real
line endowed with the Sorgenfrey topology (generated by the base consisting of half-intervals $[a,b)$, $a<b$). It is well-known that a product $X=Stimes S$ is a Tychonoff but non-normal space (see, for instance, Examples 1.4.4 and 2.3.12 in [Eng]). Let $D={(x,-x)in Stimes S: xin S}$. Then any subset $Y$ of $D$ is closed in $X$, and we have $2^{|D|}=2^{frak c}$ such subsets. Moreover, since $Y=bigcap_{ninBbb N} Y+[0,1/n)times [0,1/n)$, $Y$ is a $G_delta$-subset of $X$.
On the other hand, let $C$ be a countable dense subset of $X$. By Theorem 2.1.9 in [Eng], each continuous real-valued function $f$ on $X$ is uniquely determined by its restriction on $C$, so there are at most $frak c^omega=frak c$ such functions. Thus most subsets of $D$ are not zero-sets.
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 7 at 7:14
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
$begingroup$
There is a well-known counterexample. Let $S$ be the Sorgenfrey line, that is the real
line endowed with the Sorgenfrey topology (generated by the base consisting of half-intervals $[a,b)$, $a<b$). It is well-known that a product $X=Stimes S$ is a Tychonoff but non-normal space (see, for instance, Examples 1.4.4 and 2.3.12 in [Eng]). Let $D={(x,-x)in Stimes S: xin S}$. Then any subset $Y$ of $D$ is closed in $X$, and we have $2^{|D|}=2^{frak c}$ such subsets. Moreover, since $Y=bigcap_{ninBbb N} Y+[0,1/n)times [0,1/n)$, $Y$ is a $G_delta$-subset of $X$.
On the other hand, let $C$ be a countable dense subset of $X$. By Theorem 2.1.9 in [Eng], each continuous real-valued function $f$ on $X$ is uniquely determined by its restriction on $C$, so there are at most $frak c^omega=frak c$ such functions. Thus most subsets of $D$ are not zero-sets.
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 7 at 7:14
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
$begingroup$
There is a well-known counterexample. Let $S$ be the Sorgenfrey line, that is the real
line endowed with the Sorgenfrey topology (generated by the base consisting of half-intervals $[a,b)$, $a<b$). It is well-known that a product $X=Stimes S$ is a Tychonoff but non-normal space (see, for instance, Examples 1.4.4 and 2.3.12 in [Eng]). Let $D={(x,-x)in Stimes S: xin S}$. Then any subset $Y$ of $D$ is closed in $X$, and we have $2^{|D|}=2^{frak c}$ such subsets. Moreover, since $Y=bigcap_{ninBbb N} Y+[0,1/n)times [0,1/n)$, $Y$ is a $G_delta$-subset of $X$.
On the other hand, let $C$ be a countable dense subset of $X$. By Theorem 2.1.9 in [Eng], each continuous real-valued function $f$ on $X$ is uniquely determined by its restriction on $C$, so there are at most $frak c^omega=frak c$ such functions. Thus most subsets of $D$ are not zero-sets.
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 7 at 7:14
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
There is a well-known counterexample. Let $S$ be the Sorgenfrey line, that is the real
line endowed with the Sorgenfrey topology (generated by the base consisting of half-intervals $[a,b)$, $a<b$). It is well-known that a product $X=Stimes S$ is a Tychonoff but non-normal space (see, for instance, Examples 1.4.4 and 2.3.12 in [Eng]). Let $D={(x,-x)in Stimes S: xin S}$. Then any subset $Y$ of $D$ is closed in $X$, and we have $2^{|D|}=2^{frak c}$ such subsets. Moreover, since $Y=bigcap_{ninBbb N} Y+[0,1/n)times [0,1/n)$, $Y$ is a $G_delta$-subset of $X$.
On the other hand, let $C$ be a countable dense subset of $X$. By Theorem 2.1.9 in [Eng], each continuous real-valued function $f$ on $X$ is uniquely determined by its restriction on $C$, so there are at most $frak c^omega=frak c$ such functions. Thus most subsets of $D$ are not zero-sets.
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 7 at 7:14
Alex RavskyAlex Ravsky
42.4k32383
answered Jan 7 at 7:14
Alex RavskyAlex Ravsky
42.4k32383
answered Jan 7 at 7:14
Alex RavskyAlex Ravsky
42.4k32383
answered Jan 7 at 7:14
Alex RavskyAlex Ravsky
42.4k32383
42.4k32383
add a comment |
add a comment |
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$begingroup$
I know that $X$ normal implies that all closed $G_delta$'s are zero-sets, but I believe the reverse does not hold. If it did your question would be useless. Do you have a reference ? "well-known" it is not, it's not mentioned in Engelking, a standard reference for such matters.
$endgroup$
– Henno Brandsma
Jan 5 at 14:26
2
$begingroup$
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
$endgroup$
– Henno Brandsma
Jan 5 at 14:27
$begingroup$
@HennoBrandsma yes, you are right. I have edited my post.
$endgroup$
– Idonknow
Jan 5 at 14:59
$begingroup$
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
$endgroup$
– Idonknow
Jan 5 at 15:00
1
$begingroup$
@PaulFrost that is the direction I agreed with, yes. The OP originally claimed that if all closed $G_delta$ sets are zero-sets then $X$ is normal. This I doubt the truth of, as said.
$endgroup$
– Henno Brandsma
Jan 5 at 18:46