Is a closed $G_delta$ set in a Hausdorff space always a zero set?
I have been trying to prove that if $A$ is a closed set which is also an intersection of countably many open sets then $A$ is the zero set for some continuous real-valued function however have thus far failed. Is this even true?
general-topology functional-analysis
edited Feb 2 '13 at 21:36
Martin
6,8722145
asked Feb 2 '13 at 18:00
Thomas Martin
311
add a comment |
I have been trying to prove that if $A$ is a closed set which is also an intersection of countably many open sets then $A$ is the zero set for some continuous real-valued function however have thus far failed. Is this even true?
general-topology functional-analysis
edited Feb 2 '13 at 21:36
Martin
6,8722145
asked Feb 2 '13 at 18:00
Thomas Martin
311
add a comment |
I have been trying to prove that if $A$ is a closed set which is also an intersection of countably many open sets then $A$ is the zero set for some continuous real-valued function however have thus far failed. Is this even true?
general-topology functional-analysis
edited Feb 2 '13 at 21:36
Martin
6,8722145
asked Feb 2 '13 at 18:00
Thomas Martin
311
I have been trying to prove that if $A$ is a closed set which is also an intersection of countably many open sets then $A$ is the zero set for some continuous real-valued function however have thus far failed. Is this even true?
general-topology functional-analysis
general-topology functional-analysis
edited Feb 2 '13 at 21:36
Martin
6,8722145
asked Feb 2 '13 at 18:00
Thomas Martin
311
edited Feb 2 '13 at 21:36
Martin
6,8722145
asked Feb 2 '13 at 18:00
Thomas Martin
311
edited Feb 2 '13 at 21:36
Martin
6,8722145
edited Feb 2 '13 at 21:36
Martin
6,8722145
edited Feb 2 '13 at 21:36
Martin
6,8722145
6,8722145
asked Feb 2 '13 at 18:00
Thomas Martin
311
asked Feb 2 '13 at 18:00
Thomas Martin
311
asked Feb 2 '13 at 18:00
Thomas Martin
311
311
add a comment |
add a comment |
3 Answers
3
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oldest
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It is not true.
John Thomas, A regular space, not completely regular, Amer. Math. Monthly 76 (1969), 181-182, constructed a regular Hausdorff space $X$ with two points, $p$, and $q$, such that for each continuous $f:XtoBbb R$, $f(a)=f(b)$. Moreover, $X$ has countable local bases at $p$ and $q$. Thus, ${p}$ is a closed $G_delta$-set in $X$ that cannot be a zero-set: any zero-set containing $p$ must also contain $q$.
A. Mysior, A regular space which is not completely regular, Proc. Amer. Math. Soc. 81 (1981), 652-653, is freely available and has a regular Hausdorff space with a point $p$ and a closed set $F$ such that for each continuous $f:XtoBbb R$ and $xin F$, $f(x)=f(p)$. The point $p$ has a countable local base, so here again ${p}$ is a closed $G_delta$ that cannot be a zero-set.
answered Feb 2 '13 at 18:55
Brian M. Scott
455k38505907
add a comment |
Not even in completely regular Hausdorff spaces. In general we have
$$
text{compact $G_delta$}qquadLongrightarrowqquad
text{zero-set}qquadLongrightarrowqquad
text{closed $G_delta$}
$$
but none reversible.
answered Feb 2 '13 at 22:52
GEdgar
61.7k267168
A bit late too the party, but could you point me to a reference to the proof of compact $G_delta implies $ zero-set?
– John
Aug 29 '17 at 11:45
what can be example for completely regular space?
– Sushil
Mar 9 '18 at 18:49
add a comment |
Brian's answer covers the question fully. For fun, here's another example:
Bing's irrational slope space is a countable and connected Hausdorff space.
Now observe:
If $f colon X to mathbb{R}$ is continuous, then $f(X)$ is a countable and connected subset of $mathbb{R}$, hence it must be reduced to a point. Therefore all continuous functions $f colon X to mathbb{R}$ are constant, and the only zero sets are the empty set and the space itself.
Since $X$ is countable and $T_1$, every subset $F$ of $X$ is a $G_delta$-set: $F = bigcap_{x in X setminus F} X setminus {x}$, in particular, there is an abundance of closed $G_delta$ sets that are not zero sets.
answered Feb 2 '13 at 22:19
Martin
6,8722145
add a comment |
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3 Answers
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active
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3 Answers
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It is not true.
John Thomas, A regular space, not completely regular, Amer. Math. Monthly 76 (1969), 181-182, constructed a regular Hausdorff space $X$ with two points, $p$, and $q$, such that for each continuous $f:XtoBbb R$, $f(a)=f(b)$. Moreover, $X$ has countable local bases at $p$ and $q$. Thus, ${p}$ is a closed $G_delta$-set in $X$ that cannot be a zero-set: any zero-set containing $p$ must also contain $q$.
A. Mysior, A regular space which is not completely regular, Proc. Amer. Math. Soc. 81 (1981), 652-653, is freely available and has a regular Hausdorff space with a point $p$ and a closed set $F$ such that for each continuous $f:XtoBbb R$ and $xin F$, $f(x)=f(p)$. The point $p$ has a countable local base, so here again ${p}$ is a closed $G_delta$ that cannot be a zero-set.
answered Feb 2 '13 at 18:55
Brian M. Scott
455k38505907
add a comment |
It is not true.
John Thomas, A regular space, not completely regular, Amer. Math. Monthly 76 (1969), 181-182, constructed a regular Hausdorff space $X$ with two points, $p$, and $q$, such that for each continuous $f:XtoBbb R$, $f(a)=f(b)$. Moreover, $X$ has countable local bases at $p$ and $q$. Thus, ${p}$ is a closed $G_delta$-set in $X$ that cannot be a zero-set: any zero-set containing $p$ must also contain $q$.
A. Mysior, A regular space which is not completely regular, Proc. Amer. Math. Soc. 81 (1981), 652-653, is freely available and has a regular Hausdorff space with a point $p$ and a closed set $F$ such that for each continuous $f:XtoBbb R$ and $xin F$, $f(x)=f(p)$. The point $p$ has a countable local base, so here again ${p}$ is a closed $G_delta$ that cannot be a zero-set.
answered Feb 2 '13 at 18:55
Brian M. Scott
455k38505907
add a comment |
It is not true.
John Thomas, A regular space, not completely regular, Amer. Math. Monthly 76 (1969), 181-182, constructed a regular Hausdorff space $X$ with two points, $p$, and $q$, such that for each continuous $f:XtoBbb R$, $f(a)=f(b)$. Moreover, $X$ has countable local bases at $p$ and $q$. Thus, ${p}$ is a closed $G_delta$-set in $X$ that cannot be a zero-set: any zero-set containing $p$ must also contain $q$.
A. Mysior, A regular space which is not completely regular, Proc. Amer. Math. Soc. 81 (1981), 652-653, is freely available and has a regular Hausdorff space with a point $p$ and a closed set $F$ such that for each continuous $f:XtoBbb R$ and $xin F$, $f(x)=f(p)$. The point $p$ has a countable local base, so here again ${p}$ is a closed $G_delta$ that cannot be a zero-set.
answered Feb 2 '13 at 18:55
Brian M. Scott
455k38505907
It is not true.
John Thomas, A regular space, not completely regular, Amer. Math. Monthly 76 (1969), 181-182, constructed a regular Hausdorff space $X$ with two points, $p$, and $q$, such that for each continuous $f:XtoBbb R$, $f(a)=f(b)$. Moreover, $X$ has countable local bases at $p$ and $q$. Thus, ${p}$ is a closed $G_delta$-set in $X$ that cannot be a zero-set: any zero-set containing $p$ must also contain $q$.
A. Mysior, A regular space which is not completely regular, Proc. Amer. Math. Soc. 81 (1981), 652-653, is freely available and has a regular Hausdorff space with a point $p$ and a closed set $F$ such that for each continuous $f:XtoBbb R$ and $xin F$, $f(x)=f(p)$. The point $p$ has a countable local base, so here again ${p}$ is a closed $G_delta$ that cannot be a zero-set.
answered Feb 2 '13 at 18:55
Brian M. Scott
455k38505907
answered Feb 2 '13 at 18:55
Brian M. Scott
455k38505907
answered Feb 2 '13 at 18:55
Brian M. Scott
455k38505907
answered Feb 2 '13 at 18:55
Brian M. Scott
455k38505907
455k38505907
add a comment |
add a comment |
Not even in completely regular Hausdorff spaces. In general we have
$$
text{compact $G_delta$}qquadLongrightarrowqquad
text{zero-set}qquadLongrightarrowqquad
text{closed $G_delta$}
$$
but none reversible.
answered Feb 2 '13 at 22:52
GEdgar
61.7k267168
A bit late too the party, but could you point me to a reference to the proof of compact $G_delta implies $ zero-set?
– John
Aug 29 '17 at 11:45
what can be example for completely regular space?
– Sushil
Mar 9 '18 at 18:49
add a comment |
Not even in completely regular Hausdorff spaces. In general we have
$$
text{compact $G_delta$}qquadLongrightarrowqquad
text{zero-set}qquadLongrightarrowqquad
text{closed $G_delta$}
$$
but none reversible.
answered Feb 2 '13 at 22:52
GEdgar
61.7k267168
A bit late too the party, but could you point me to a reference to the proof of compact $G_delta implies $ zero-set?
– John
Aug 29 '17 at 11:45
what can be example for completely regular space?
– Sushil
Mar 9 '18 at 18:49
add a comment |
Not even in completely regular Hausdorff spaces. In general we have
$$
text{compact $G_delta$}qquadLongrightarrowqquad
text{zero-set}qquadLongrightarrowqquad
text{closed $G_delta$}
$$
but none reversible.
answered Feb 2 '13 at 22:52
GEdgar
61.7k267168
Not even in completely regular Hausdorff spaces. In general we have
$$
text{compact $G_delta$}qquadLongrightarrowqquad
text{zero-set}qquadLongrightarrowqquad
text{closed $G_delta$}
$$
but none reversible.
answered Feb 2 '13 at 22:52
GEdgar
61.7k267168
answered Feb 2 '13 at 22:52
GEdgar
61.7k267168
answered Feb 2 '13 at 22:52
GEdgar
61.7k267168
answered Feb 2 '13 at 22:52
GEdgar
61.7k267168
61.7k267168
A bit late too the party, but could you point me to a reference to the proof of compact $G_delta implies $ zero-set?
– John
Aug 29 '17 at 11:45
what can be example for completely regular space?
– Sushil
Mar 9 '18 at 18:49
add a comment |
A bit late too the party, but could you point me to a reference to the proof of compact $G_delta implies $ zero-set?
– John
Aug 29 '17 at 11:45
what can be example for completely regular space?
– Sushil
Mar 9 '18 at 18:49
A bit late too the party, but could you point me to a reference to the proof of compact $G_delta implies $ zero-set?
– John
Aug 29 '17 at 11:45
A bit late too the party, but could you point me to a reference to the proof of compact $G_delta implies $ zero-set?
– John
Aug 29 '17 at 11:45
what can be example for completely regular space?
– Sushil
Mar 9 '18 at 18:49
what can be example for completely regular space?
– Sushil
Mar 9 '18 at 18:49
add a comment |
Brian's answer covers the question fully. For fun, here's another example:
Bing's irrational slope space is a countable and connected Hausdorff space.
Now observe:
If $f colon X to mathbb{R}$ is continuous, then $f(X)$ is a countable and connected subset of $mathbb{R}$, hence it must be reduced to a point. Therefore all continuous functions $f colon X to mathbb{R}$ are constant, and the only zero sets are the empty set and the space itself.
Since $X$ is countable and $T_1$, every subset $F$ of $X$ is a $G_delta$-set: $F = bigcap_{x in X setminus F} X setminus {x}$, in particular, there is an abundance of closed $G_delta$ sets that are not zero sets.
answered Feb 2 '13 at 22:19
Martin
6,8722145
add a comment |
Brian's answer covers the question fully. For fun, here's another example:
Bing's irrational slope space is a countable and connected Hausdorff space.
Now observe:
If $f colon X to mathbb{R}$ is continuous, then $f(X)$ is a countable and connected subset of $mathbb{R}$, hence it must be reduced to a point. Therefore all continuous functions $f colon X to mathbb{R}$ are constant, and the only zero sets are the empty set and the space itself.
Since $X$ is countable and $T_1$, every subset $F$ of $X$ is a $G_delta$-set: $F = bigcap_{x in X setminus F} X setminus {x}$, in particular, there is an abundance of closed $G_delta$ sets that are not zero sets.
answered Feb 2 '13 at 22:19
Martin
6,8722145
add a comment |
Brian's answer covers the question fully. For fun, here's another example:
Bing's irrational slope space is a countable and connected Hausdorff space.
Now observe:
If $f colon X to mathbb{R}$ is continuous, then $f(X)$ is a countable and connected subset of $mathbb{R}$, hence it must be reduced to a point. Therefore all continuous functions $f colon X to mathbb{R}$ are constant, and the only zero sets are the empty set and the space itself.
Since $X$ is countable and $T_1$, every subset $F$ of $X$ is a $G_delta$-set: $F = bigcap_{x in X setminus F} X setminus {x}$, in particular, there is an abundance of closed $G_delta$ sets that are not zero sets.
answered Feb 2 '13 at 22:19
Martin
6,8722145
Brian's answer covers the question fully. For fun, here's another example:
Bing's irrational slope space is a countable and connected Hausdorff space.
Now observe:
If $f colon X to mathbb{R}$ is continuous, then $f(X)$ is a countable and connected subset of $mathbb{R}$, hence it must be reduced to a point. Therefore all continuous functions $f colon X to mathbb{R}$ are constant, and the only zero sets are the empty set and the space itself.
Since $X$ is countable and $T_1$, every subset $F$ of $X$ is a $G_delta$-set: $F = bigcap_{x in X setminus F} X setminus {x}$, in particular, there is an abundance of closed $G_delta$ sets that are not zero sets.
answered Feb 2 '13 at 22:19
Martin
6,8722145
edited Feb 2 '13 at 22:36
answered Feb 2 '13 at 22:19
Martin
6,8722145
answered Feb 2 '13 at 22:19
Martin
6,8722145
answered Feb 2 '13 at 22:19
Martin
6,8722145
6,8722145
add a comment |
add a comment |
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