For a closed $G_delta$ set $Fsubseteq X,$ does there exist a continuous function $f:Xto [0,1]$ such that...
(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 2 days ago
Paul Frost
9,4302631
asked 2 days ago
IdonknowIdonknow
2,345749113
|
show 4 more comments
(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 2 days ago
Paul Frost
9,4302631
asked 2 days ago
IdonknowIdonknow
2,345749113
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.
– Henno Brandsma
2 days ago
2
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
– Henno Brandsma
2 days ago
@HennoBrandsma yes, you are right. I have edited my post.
– Idonknow
2 days ago
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
– Idonknow
2 days ago
1
@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.
– Henno Brandsma
2 days ago
|
show 4 more comments
(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 2 days ago
Paul Frost
9,4302631
asked 2 days ago
IdonknowIdonknow
2,345749113
(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 2 days ago
Paul Frost
9,4302631
asked 2 days ago
IdonknowIdonknow
2,345749113
edited 2 days ago
Paul Frost
9,4302631
asked 2 days ago
IdonknowIdonknow
2,345749113
edited 2 days ago
Paul Frost
9,4302631
edited 2 days ago
Paul Frost
9,4302631
edited 2 days ago
Paul Frost
9,4302631
9,4302631
asked 2 days ago
IdonknowIdonknow
2,345749113
asked 2 days ago
IdonknowIdonknow
2,345749113
asked 2 days ago
IdonknowIdonknow
2,345749113
2,345749113
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.
– Henno Brandsma
2 days ago
2
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
– Henno Brandsma
2 days ago
@HennoBrandsma yes, you are right. I have edited my post.
– Idonknow
2 days ago
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
– Idonknow
2 days ago
1
@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.
– Henno Brandsma
2 days ago
|
show 4 more comments
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.
– Henno Brandsma
2 days ago
2
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
– Henno Brandsma
2 days ago
@HennoBrandsma yes, you are right. I have edited my post.
– Idonknow
2 days ago
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
– Idonknow
2 days ago
1
@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.
– Henno Brandsma
2 days ago
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.
– Henno Brandsma
2 days ago
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.
– Henno Brandsma
2 days ago
2
2
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
– Henno Brandsma
2 days ago
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
– Henno Brandsma
2 days ago
@HennoBrandsma yes, you are right. I have edited my post.
– Idonknow
2 days ago
@HennoBrandsma yes, you are right. I have edited my post.
– Idonknow
2 days ago
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
– Idonknow
2 days ago
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
– Idonknow
2 days ago
1
1
@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.
– Henno Brandsma
2 days ago
@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.
– Henno Brandsma
2 days ago
|
show 4 more comments
1 Answer
1
active
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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 18 hours ago
Alex RavskyAlex Ravsky
39.4k32181
add a comment |
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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 18 hours ago
Alex RavskyAlex Ravsky
39.4k32181
add a comment |
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 18 hours ago
Alex RavskyAlex Ravsky
39.4k32181
add a comment |
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 18 hours ago
Alex RavskyAlex Ravsky
39.4k32181
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 18 hours ago
Alex RavskyAlex Ravsky
39.4k32181
answered 18 hours ago
Alex RavskyAlex Ravsky
39.4k32181
answered 18 hours ago
Alex RavskyAlex Ravsky
39.4k32181
answered 18 hours ago
Alex RavskyAlex Ravsky
39.4k32181
39.4k32181
add a comment |
add a comment |
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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.
– Henno Brandsma
2 days ago
2
Note that this can only be true when $F$ is closed and $G_delta$. You cannot ask this for all closed sets.
– Henno Brandsma
2 days ago
@HennoBrandsma yes, you are right. I have edited my post.
– Idonknow
2 days ago
@HennoBrandsma by the way, if the converse holds, why would by question be useless?
– Idonknow
2 days ago
1
@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.
– Henno Brandsma
2 days ago