Show that $I_{Z(I)}=I$
Problem
Let $X$ be a compact Hausdorff space with $C(X)$ the ring of continuous functions on $X$ mapped to $Bbb R$. If $A$ is closed subset of $X$, define
$$I_{A}={f in C(X)| f|_{A} = 0 }.$$
If $Isubseteq C(X)$ is an ideal, define
$$Z(I):={x in X|f(x)=0 text{for all } f in I }.$$
Given a proper, non-zero ideal $Isubseteq C(X)$ that is closed in $C(X)$ (where we give $C(X)$ the sup-norm topology), is it true that $I_{Z(I)} = I$?
Work
If $f in I$, then $f$ is zero for any $x in Z(I)$, but the other direction I am not sure. If it were true, it would mean that for any function $g$ which is zero on $Z(I)$ would have to also be part of the ideal $I$. Any tips appreciated.
general-topology ring-theory
asked Nov 24 at 3:34
IntegrateThis
1,7131717
|
show 1 more comment
Problem
Let $X$ be a compact Hausdorff space with $C(X)$ the ring of continuous functions on $X$ mapped to $Bbb R$. If $A$ is closed subset of $X$, define
$$I_{A}={f in C(X)| f|_{A} = 0 }.$$
If $Isubseteq C(X)$ is an ideal, define
$$Z(I):={x in X|f(x)=0 text{for all } f in I }.$$
Given a proper, non-zero ideal $Isubseteq C(X)$ that is closed in $C(X)$ (where we give $C(X)$ the sup-norm topology), is it true that $I_{Z(I)} = I$?
Work
If $f in I$, then $f$ is zero for any $x in Z(I)$, but the other direction I am not sure. If it were true, it would mean that for any function $g$ which is zero on $Z(I)$ would have to also be part of the ideal $I$. Any tips appreciated.
general-topology ring-theory
asked Nov 24 at 3:34
IntegrateThis
1,7131717
I see a downvote, if something is not clear I can expand. Also I would like to give more of an attempt but I am genuinely stumped.
– IntegrateThis
Nov 24 at 4:12
I voted up. Fairly clear question (though what explicitly is $I$?) and some effort was demonstrated.
– Santana Afton
Nov 24 at 4:20
@SantanaAfton added.
– IntegrateThis
Nov 24 at 4:21
@SantanaAfton answered, sorry for forgetting the details lol.
– IntegrateThis
Nov 24 at 4:29
Maybe you need to add $f:Xrightarrow mathbb{C}$, to include that $f$ is a complex-valued continuous function.
– i707107
Nov 24 at 4:31
|
show 1 more comment
Problem
Let $X$ be a compact Hausdorff space with $C(X)$ the ring of continuous functions on $X$ mapped to $Bbb R$. If $A$ is closed subset of $X$, define
$$I_{A}={f in C(X)| f|_{A} = 0 }.$$
If $Isubseteq C(X)$ is an ideal, define
$$Z(I):={x in X|f(x)=0 text{for all } f in I }.$$
Given a proper, non-zero ideal $Isubseteq C(X)$ that is closed in $C(X)$ (where we give $C(X)$ the sup-norm topology), is it true that $I_{Z(I)} = I$?
Work
If $f in I$, then $f$ is zero for any $x in Z(I)$, but the other direction I am not sure. If it were true, it would mean that for any function $g$ which is zero on $Z(I)$ would have to also be part of the ideal $I$. Any tips appreciated.
general-topology ring-theory
asked Nov 24 at 3:34
IntegrateThis
1,7131717
Problem
Let $X$ be a compact Hausdorff space with $C(X)$ the ring of continuous functions on $X$ mapped to $Bbb R$. If $A$ is closed subset of $X$, define
$$I_{A}={f in C(X)| f|_{A} = 0 }.$$
If $Isubseteq C(X)$ is an ideal, define
$$Z(I):={x in X|f(x)=0 text{for all } f in I }.$$
Given a proper, non-zero ideal $Isubseteq C(X)$ that is closed in $C(X)$ (where we give $C(X)$ the sup-norm topology), is it true that $I_{Z(I)} = I$?
Work
If $f in I$, then $f$ is zero for any $x in Z(I)$, but the other direction I am not sure. If it were true, it would mean that for any function $g$ which is zero on $Z(I)$ would have to also be part of the ideal $I$. Any tips appreciated.
general-topology ring-theory
general-topology ring-theory
asked Nov 24 at 3:34
IntegrateThis
1,7131717
asked Nov 24 at 3:34
IntegrateThis
1,7131717
edited Nov 24 at 4:35
asked Nov 24 at 3:34
IntegrateThis
1,7131717
asked Nov 24 at 3:34
IntegrateThis
1,7131717
asked Nov 24 at 3:34
IntegrateThis
1,7131717
1,7131717
I see a downvote, if something is not clear I can expand. Also I would like to give more of an attempt but I am genuinely stumped.
– IntegrateThis
Nov 24 at 4:12
I voted up. Fairly clear question (though what explicitly is $I$?) and some effort was demonstrated.
– Santana Afton
Nov 24 at 4:20
@SantanaAfton added.
– IntegrateThis
Nov 24 at 4:21
@SantanaAfton answered, sorry for forgetting the details lol.
– IntegrateThis
Nov 24 at 4:29
Maybe you need to add $f:Xrightarrow mathbb{C}$, to include that $f$ is a complex-valued continuous function.
– i707107
Nov 24 at 4:31
|
show 1 more comment
I see a downvote, if something is not clear I can expand. Also I would like to give more of an attempt but I am genuinely stumped.
– IntegrateThis
Nov 24 at 4:12
I voted up. Fairly clear question (though what explicitly is $I$?) and some effort was demonstrated.
– Santana Afton
Nov 24 at 4:20
@SantanaAfton added.
– IntegrateThis
Nov 24 at 4:21
@SantanaAfton answered, sorry for forgetting the details lol.
– IntegrateThis
Nov 24 at 4:29
Maybe you need to add $f:Xrightarrow mathbb{C}$, to include that $f$ is a complex-valued continuous function.
– i707107
Nov 24 at 4:31
I see a downvote, if something is not clear I can expand. Also I would like to give more of an attempt but I am genuinely stumped.
– IntegrateThis
Nov 24 at 4:12
I see a downvote, if something is not clear I can expand. Also I would like to give more of an attempt but I am genuinely stumped.
– IntegrateThis
Nov 24 at 4:12
I voted up. Fairly clear question (though what explicitly is $I$?) and some effort was demonstrated.
– Santana Afton
Nov 24 at 4:20
I voted up. Fairly clear question (though what explicitly is $I$?) and some effort was demonstrated.
– Santana Afton
Nov 24 at 4:20
@SantanaAfton added.
– IntegrateThis
Nov 24 at 4:21
@SantanaAfton added.
– IntegrateThis
Nov 24 at 4:21
@SantanaAfton answered, sorry for forgetting the details lol.
– IntegrateThis
Nov 24 at 4:29
@SantanaAfton answered, sorry for forgetting the details lol.
– IntegrateThis
Nov 24 at 4:29
Maybe you need to add $f:Xrightarrow mathbb{C}$, to include that $f$ is a complex-valued continuous function.
– i707107
Nov 24 at 4:31
Maybe you need to add $f:Xrightarrow mathbb{C}$, to include that $f$ is a complex-valued continuous function.
– i707107
Nov 24 at 4:31
|
show 1 more comment
1 Answer
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oldest
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This is true and follows from a characterization of closed ideals of $C(X)$: every closed ideal is of the form $I_A$ for some closed $A$.
So $I$ in your problem = $I_A$ for some closed $A$. And $Z(I_A) = A$, which finishes it.
answered Nov 24 at 5:49
user25959
1,563816
add a comment |
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1 Answer
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1 Answer
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votes
This is true and follows from a characterization of closed ideals of $C(X)$: every closed ideal is of the form $I_A$ for some closed $A$.
So $I$ in your problem = $I_A$ for some closed $A$. And $Z(I_A) = A$, which finishes it.
answered Nov 24 at 5:49
user25959
1,563816
add a comment |
This is true and follows from a characterization of closed ideals of $C(X)$: every closed ideal is of the form $I_A$ for some closed $A$.
So $I$ in your problem = $I_A$ for some closed $A$. And $Z(I_A) = A$, which finishes it.
answered Nov 24 at 5:49
user25959
1,563816
add a comment |
This is true and follows from a characterization of closed ideals of $C(X)$: every closed ideal is of the form $I_A$ for some closed $A$.
So $I$ in your problem = $I_A$ for some closed $A$. And $Z(I_A) = A$, which finishes it.
answered Nov 24 at 5:49
user25959
1,563816
This is true and follows from a characterization of closed ideals of $C(X)$: every closed ideal is of the form $I_A$ for some closed $A$.
So $I$ in your problem = $I_A$ for some closed $A$. And $Z(I_A) = A$, which finishes it.
answered Nov 24 at 5:49
user25959
1,563816
answered Nov 24 at 5:49
user25959
1,563816
answered Nov 24 at 5:49
user25959
1,563816
answered Nov 24 at 5:49
user25959
1,563816
1,563816
add a comment |
add a comment |
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I see a downvote, if something is not clear I can expand. Also I would like to give more of an attempt but I am genuinely stumped.
– IntegrateThis
Nov 24 at 4:12
I voted up. Fairly clear question (though what explicitly is $I$?) and some effort was demonstrated.
– Santana Afton
Nov 24 at 4:20
@SantanaAfton added.
– IntegrateThis
Nov 24 at 4:21
@SantanaAfton answered, sorry for forgetting the details lol.
– IntegrateThis
Nov 24 at 4:29
Maybe you need to add $f:Xrightarrow mathbb{C}$, to include that $f$ is a complex-valued continuous function.
– i707107
Nov 24 at 4:31