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edited Jan 30 2011 at 0:14 |
EDIT: This is for the normal case....
$V \in H(1) \cup H(0) $, if and only if $V$ is a zero set of $X$. (this is by definition) So in particular there exists some $f_V \in C(X,[0,1])$ such that $f_V(x) = 0 \iff x \in V$. If $U \in H(2)$, then there exists some $V \in H(1) \cup H(0)$, and $g_U\in C(V,[0,1])$ such that $g_U(x) = 0 \iff x \in U$ Because $U$ a zero set of $V$, it will be closed, and so because $X$ is normal, we may apply Tietzes' Extension Theorem to produce a continuous map $h_{UV}:X\rightarrow[0,1]$ which extends $g_U$. Both the functions $h_{UV}$ and $f_V$ are defined on all of $X$, so we may add them to produce the new function $F:X\rightarrow[0,2]$ given by $F(x)=h_{UV}(x) + f_{V}(x)$. It follows that $U$ is the zero set of $F$, and we have that $U$ is a zero set of $X$.
By (1), (2), (3), (4) and (5), it follows that $U \in H(0) \cup H(1)$ and so $H(2) = H(1) \cup H(0)$. So noting that $X$ is always a zero set of $X$, we have that $H(2) = H(1)$.
Therefore, $\nu(X) = 1$
PS:
If you are looking for a hard problem in this form that deals with topology, may I suggest Alan Dows' "Sequential order" chapter from "Open Problems in Topology II"
EDIT
Edit: General Case (I think)
Fix two topologies $\tau_0$, $\tau_1$ defined over the same set $X$. Denote, the space with topology $\tau_k$ as $X_k$ and suppose the identity map $i:X_0\rightarrow X_1$ is continuous.
Then, for every $f \in C(X_1, [0,1])$, we have that $f \in C(X_0, [0,1])$ (as $f\circ i$ is continuous) and so $Z$ is a zero set for $X_0$ implies that $Z$ is a zero set for $X_1$, and every zero set of $X_1$ is contained in a zero set for $X_0$. It follows that $\nu(X)$ must increase as you refine the topology on $X$ (this is because you can only inject new witnesses attempt at general case removed due to the failure of 'zero set' being hereditary).
So noting that for the trivial topology $\nu(X) = 1$, and for the discrete topology $\nu(X) = 1$. The result followserror.
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edited Jan 30 2011 at 0:08 |
EDIT: This is for the normal case....
$V \in H(1) \cup H(0) $, if and only if $V$ is a zero set of $X$. (this is by definition) So in particular there exists some $f_V \in C(X,[0,1])$ such that $f_V(x) = 0 \iff x \in V$. If $U \in H(2)$, then there exists some $V \in H(1) \cup H(0)$, and $g_U\in C(V,[0,1])$ such that $g_U(x) = 0 \iff x \in U$ Because $U$ a zero set of $V$, it will be closed, and so because $X$ is normal, we may apply Tietzes' Extension Theorem to produce a continuous map $h_{UV}:X\rightarrow[0,1]$ which extends $g_U$. Both the functions $h_{UV}$ and $f_V$ are defined on all of $X$, so we may add them to produce the new function $F:X\rightarrow[0,2]$ given by $F(x)=h_{UV}(x) + f_{V}(x)$. It follows that $U$ is the zero set of $F$, and we have that $U$ is a zero set of $X$.
By (1), (2), (3), (4) and (5), it follows that $U \in H(0) \cup H(1)$ and so $H(2) = H(1) \cup H(0)$. So noting that $X$ is always a zero set of $X$, we have that $H(2) = H(1)$.
Therefore, $\nu(X) = 1$
PS:
If you are looking for a hard problem in this form that deals with topology, may I suggest Alan Dows' "Sequential order" chapter from "Open Problems in Topology II"
EDIT: General Case (I think)
Fix two topologies $\tau_0$, $\tau_1$ defined over the same set $X$. Denote, the space with topology $\tau_k$ as $X_k$ and suppose the identity map $i:X_0\rightarrow X_1$ is continuous.
Then, for every $f \in C(X_1, [0,1])$, we have that $f \in C(X_0, [0,1])$ (as $f\circ i$ is continuous) and so $Z$ is a zero set for $X_0$ implies that $Z$ is a zero set for $X_1$, and every zero set of $X_1$ is contained in a zero set for $X_0$. It follows that $\nu(X)$ must increase as you refine the topology on $X$ (this is because you can only inject new witnesses to the failure of 'zero set' being hereditary).
So noting that for the trivial topology $\nu(X) = 1$, and for the discrete topology $\nu(X) = 1$. The result follows.
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edited Jan 29 2011 at 23:49 |
EDIT: This is for the normal case....
$V \in H(1) \cup H(0) $, if and only if $V$ is a zero set of $X$. (this is by definition) So in particular there exists some $f_V \in C(X,[0,1])$ such that $f_V(x) = 0 \iff x \in V$. If $U \in H(2)$, then there exists some $V \in H(1) \cup H(0)$, and $g_U\in C(V,[0,1])$ such that $g_U(x) = 0 \iff x \in U$ Because $U$ a zero set of $V$, it will be closed, and so because $X$ is normal, we may apply Tietzes' Extension Theorem to produce a continuous map $h_{UV}:X\rightarrow[0,1]$ which extends $g_U$. Both the functions $h_{UV}$ and $f_V$ are defined on all of $X$, so we may add them to produce the new function $F:X\rightarrow[0,2]$ given by $F(x)=h_{UV}(x) + f_{V}(x)$. It follows that $U$ is the zero set of $F$, and we have that $U$ is a zero set of $X$.
By (1), (2), (3), (4) and (5), it follows that $U \in H(0) \cup H(1)$ and so $H(2) = H(1) \cup H(0)$. So noting that $X$ is always a zero set of $X$, we have that $H(2) = H(1)$.
Therefore, $\nu(X) = 1$
PS:
If you are looking for a hard problem in this form that deals with topology, may I suggest Alan Dows' "Sequential order" chapter from "Open Problems in Topology II"
EDIT: General Case (I think)
Fix two topologies $\tau_0$, $\tau_1$ defined over the same set $X$. Denote, the space with topology $\tau_k$ as $X_k$ and suppose the identity map $i:X_0\rightarrow X_1$ is continuous.
Then, for every $f \in C(X_1, [0,1])$, we have that $f \in C(X_0, [0,1])$ (as $f\circ i$ is continuous) and so $Z$ is a zero set for $X_0$ implies that $Z$ is a zero set for $X_1$, and every zero set of $X_1$ is contained in a zero set for $X_0$. It follows that $\nu(X)$ must increase as you refine the topology on $X$ (this is because you can only inject new witnesses to the failure of 'zero set' being hereditary).
So noting that for the trivial topology $\nu(X) = 1$, and for the discrete topology $\nu(X) = 1$. The result follows.
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edited Jan 29 2011 at 21:11 |
EDIT: Updated to reflect changes in Question. Made This is for the argument clearer, and fixed an off by one errornormal case....
$V \in H(1) \cup H(0) $, if and only if $V$ is a zero set of $X$. (this is by definition) So in particular there exists some $f_V \in C(X,[0,1])$ such that $f_V(x) = 0 \iff x \in V$. If $U \in H(2)$, then there exists some $V \in H(1) \cup H(0)$, and $g_U\in C(V,[0,1])$ such that $g_U(x) = 0 \iff x \in U$ Because $U$ a zero set of $V$, it will be closed, and so because $X$ is normal, we may apply Tietzes' Extension Theorem to produce a continuous map $h_{UV}:X\rightarrow[0,1]$ which extends $g_U$. Both the functions $h_{UV}$ and $f_V$ are defined on all of $X$, so we may add them to produce the new function $F:X\rightarrow[0,2]$ given by $F(x)=h_{UV}(x) + f_{V}(x)$. It follows that $U$ is the zero set of $F$, and we have that $U$ is a zero set of $X$.
By (1), (2), (3), (4) and (5), it follows that $U \in H(0) \cup H(1)$ and so $H(2) = H(1) \cup H(0)$. So noting that $X$ is always a zero set of $X$, we have that $H(2) = H(1)$.
Therefore, $\nu(X) = 1$
PS:
If you are looking for a hard problem in this form that deals with topology, may I suggest Alan Dows' "Sequential order" chapter from "Open Problems in Topology II"
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edited Jan 29 2011 at 18:55 |
EDIT: Updated to reflect changes in Question. Made the argument clearer, and fixed an off by one error.
$V \in H(1) \cup H(0) $, if and only if $V$ is a zero set of $X$. (this is by definition) So in particular there exists some $f_V \in C(X,[0,1])$ such that $f_V(x) = 0 \iff x \in V$. If $U \in H(2)$, then there exists some $V \in H(1) \cup H(0)$, and $g_U\in C(V,[0,1])$ such that $g_U(x) = 0 \iff x \in U$ Because $U$ a zero set of $V$, it will be closed, and so because $X$ is normal, we may apply Tietzes' Extension Theorem to produce a continuous map $h_{UV}:X\rightarrow[0,1]$ which extends $g_U$. Both the functions $h_{UV}$ and $f_V$ are defined on all of $X$, so we may add them to produce the new function $F:X\rightarrow[0,1]$ F:X\rightarrow[0,2]$ given by $F(x)=h_{UV}(x) + f_{V}(x)$. It follows that $U$ is the zero set of $F$, and we have that $U$ is a zero set of $X$.
By (1), (2), (3), (4) and (5), it follows that $U \in H(0) \cup H(1)$ and so $H(2) = H(1) \cup H(0)$. So noting that $X$ is always a zero set of $X$, we have that $H(2) = H(1)$.
Therefore, $\nu(X) = 1$
PS:
If you are looking for a hard problem in this form that deals with topology, may I suggest Alan Dows' "Sequential order" chapter from "Open Problems in Topology II"
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edited Jan 29 2011 at 18:46 |
If your intention was for each stage $H(\alpha+1)$ EDIT: Updated to be reflect changes in Question. Made the collection of every zero set from any set argument clearer, and fixed an off by one error. $V \in H(1) \cup H(0) $\bigcup_{\lambda\le\alpha} H(\lambda)$ which , if and only if $V$ is a subset zero set of $X$. Then the answer (this is by definition) So in particular there exists some $f_V \in C(X,[0,1])$ such that $\nu(X)$ f_V(x) = 0 \iff x \in V$. If $U \in H(2)$, then there exists some $V \in H(1) \cup H(0)$, and $g_U\in C(V,[0,1])$ such that $g_U(x) = 0 \iff x \in U$ Because $U$ a zero set of $V$, it will always be 2. The reason for this is restriction preserves continuityclosed, and so (because Tietzes' extension theorem applies in this situation$X$ is normal, and we need only consider may apply Tietzes' Extension Theorem to produce a continuous map $h_{UV}:X\rightarrow[0,1]$ which extends $g_U$. Both the functions into $[0,1]$ h_{UV}$ and not $f_V$ are defined on all of $\mathbb{R}$) X$, so we have may add them to produce the new function $F:X\rightarrow[0,1]$ given by $F(x)=h_{UV}(x) + f_{V}(x)$. It follows that any zero set from a subset of $X$ (which U$ is also a the zero set of $X$) F$, and we have that $U$ is also a zero set of $X$ X$. By (this 1), (2), (3), (4) and (5), it follows because we can rig up an extension which that $U \in H(0) \cup H(1)$ and so $H(2) = H(1) \cup H(0)$. So noting that $X$ is not always a zero outside set of the subset $X$, we are considering)have that $H(2) = H(1)$. Therefore, $\nu(X) = 1$
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edited Jan 29 2011 at 16:31 |
If your intention was for each stage $H(\alpha+1)$ to be the collection of every zero set from any set in $\bigcup_{\lambda\le\alpha} H(\lambda)$ which is a subset of $X$.
Then the answer is that $\nu(X)$ will always be 2. The reason for this is restriction preserves continuity, so (because Tietzes' extension theorem applies in this situation, and we need only consider continuous functions into $[0,1]$ and not all of $\mathbb{R}$) we have that any zero set from a subset of $X$ (equipped with the relative topology) which is also a zero set of $X$) is also a zero set of $X$ (this follows because we can rig the up an extension to which is not be zero outside of the subset we are considering)
PS:
If you are looking for a hard problem in this form that deals with topology, may I suggest Alan Dows' "Sequential order" chapter from "Open Problems in Topology II"
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edited Jan 29 2011 at 10:28 |
If your intention was for each stage $H(\alpha+1)$ to be the collection of every zero set from any set in $\bigcup_{\lambda\le\alpha} H(\lambda)$ which is a subset of $X$.
Then the answer is that $\nu(X)$ will always be 2. The reason for this is restriction preserves continuity, so (because Tietzes' extension theorem applies in this situation, and we need only consider continuous functions into $[0,1]$ and not all of $\mathbb{R}$) we have that any zero set from a subset of $X$ (equipped with the relative topology) is also a zero set of $X$.X$ (this follows because we can rig the extension to not be zero outside of the subset we are considering)
PS:
If you are looking for a hard problem in this form that deals with topology, may I suggest Alan Dows' "Sequential order" chapter from "Open Problems in Topology II"
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2
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edited Jan 29 2011 at 10:04 |
If your intention was for each stage $H(\alpha+1)$ to be the collection of every zero set from any set in $\bigcup_{\lambda\le\alpha} H(\lambda)$ which is a subset of $X$.
Then the answer is that $\nu(X)$ will always be 2. The reason for this is restriction preserves continuity, so (because Tietzes' extension theorem applies in this situation, and we need only consider continuous functions into $[0,1]$ and not all of $\mathbb{R}$) we have that any zero set from a subset of $X$ (equipped with the relative topology) is also a zero set of $X$.
PS:
If you are looking for a hard problem in this form that deals with topology, may I suggest Alan Dows' "Sequential order" chapter from "Open Problems in Topology II"
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answered Jan 29 2011 at 9:47 |
If your intention was for each stage $H(\alpha+1)$ to be the collection of every zero set from any set in $\bigcup_{\lambda\le\alpha} H(\lambda)$ which is a subset of $X$.
Then the answer is that $\nu(X)$ will always be 2. The reason for this is restriction preserves continuity, so (because Tietzes' extension theorem applies in this situation, and we need only consider continuous functions into $[0,1]$ and not all of $\mathbb{R}$) we have that any zero set from a subset of $X$ (equipped with the relative topology) is also a zero set of $X$.
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