$(f,g) = C(X) Longleftrightarrow Z(f) cap Z(g) = emptyset$
$begingroup$
Theorem:
a: If $I$is an ideal in $C(X)$ , then the family $Z[I] = { Z(f) : f in I } $ is a $z$-filter on $X$.
b: if $mathbf{F}$ is a $z$-filter on $X$, then the family $Z^{-1} [ mathbf{F} ] = { f : Z(f) in mathbf{F} }$ is an ideal in $C(X)$.
Remark:
$(f,g) neq C(X) $ if only if $Z(f) $meet $Z(g)$, hence if only if $f^{2}+g^{2}$ is not a unit of $C(X)$.
Can you give me more information about the above result?
Why can we say" $(f,g) neq C(X) Longleftrightarrow Z(f) cap Z(g) neq emptyset$ or
$(f,g) = C(X) Longleftrightarrow Z(f) cap Z(g) = emptyset$
general-topology filters
edited Dec 30 '18 at 16:53
Eric Wofsey
190k14216348
$endgroup$
add a comment |
$begingroup$
Theorem:
a: If $I$is an ideal in $C(X)$ , then the family $Z[I] = { Z(f) : f in I } $ is a $z$-filter on $X$.
b: if $mathbf{F}$ is a $z$-filter on $X$, then the family $Z^{-1} [ mathbf{F} ] = { f : Z(f) in mathbf{F} }$ is an ideal in $C(X)$.
Remark:
$(f,g) neq C(X) $ if only if $Z(f) $meet $Z(g)$, hence if only if $f^{2}+g^{2}$ is not a unit of $C(X)$.
Can you give me more information about the above result?
Why can we say" $(f,g) neq C(X) Longleftrightarrow Z(f) cap Z(g) neq emptyset$ or
$(f,g) = C(X) Longleftrightarrow Z(f) cap Z(g) = emptyset$
general-topology filters
edited Dec 30 '18 at 16:53
Eric Wofsey
190k14216348
$endgroup$
add a comment |
$begingroup$
Theorem:
a: If $I$is an ideal in $C(X)$ , then the family $Z[I] = { Z(f) : f in I } $ is a $z$-filter on $X$.
b: if $mathbf{F}$ is a $z$-filter on $X$, then the family $Z^{-1} [ mathbf{F} ] = { f : Z(f) in mathbf{F} }$ is an ideal in $C(X)$.
Remark:
$(f,g) neq C(X) $ if only if $Z(f) $meet $Z(g)$, hence if only if $f^{2}+g^{2}$ is not a unit of $C(X)$.
Can you give me more information about the above result?
Why can we say" $(f,g) neq C(X) Longleftrightarrow Z(f) cap Z(g) neq emptyset$ or
$(f,g) = C(X) Longleftrightarrow Z(f) cap Z(g) = emptyset$
general-topology filters
edited Dec 30 '18 at 16:53
Eric Wofsey
190k14216348
$endgroup$
Theorem:
a: If $I$is an ideal in $C(X)$ , then the family $Z[I] = { Z(f) : f in I } $ is a $z$-filter on $X$.
b: if $mathbf{F}$ is a $z$-filter on $X$, then the family $Z^{-1} [ mathbf{F} ] = { f : Z(f) in mathbf{F} }$ is an ideal in $C(X)$.
Remark:
$(f,g) neq C(X) $ if only if $Z(f) $meet $Z(g)$, hence if only if $f^{2}+g^{2}$ is not a unit of $C(X)$.
Can you give me more information about the above result?
Why can we say" $(f,g) neq C(X) Longleftrightarrow Z(f) cap Z(g) neq emptyset$ or
$(f,g) = C(X) Longleftrightarrow Z(f) cap Z(g) = emptyset$
general-topology filters
general-topology filters
edited Dec 30 '18 at 16:53
Eric Wofsey
190k14216348
edited Dec 30 '18 at 16:53
Eric Wofsey
190k14216348
edited Dec 30 '18 at 16:53
Eric Wofsey
190k14216348
edited Dec 30 '18 at 16:53
Eric Wofsey
190k14216348
edited Dec 30 '18 at 16:53
Eric Wofsey
190k14216348
190k14216348
asked Dec 30 '18 at 15:57
user387219
add a comment |
add a comment |
1 Answer
1
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$begingroup$
I am assuming $C(X)$ denotes the set of real valued (continuous) functions on $X$.
First of all, notice that the two equivalences are equivalent, because $A$ is equivalent to $B$ if and only if $neg A$ is equivalent to $neg B$ .
Then if $(f,g) = C(X)$, there are $h,k$ such that $fh+gk =1$. If $xin Z(f)cap Z(g)$, then $0=1$. Thus $Z(f)cap Z(g)=emptyset$.
Conversely, if $Z(f)cap Z(g)=emptyset$, then $f^2 + g^2$ never vanishes (because we're considering real valued functions), so it's invertible in $C(X)$, and is in $(f,g)$ , so $(f,g) = C(X)$ .
answered Dec 30 '18 at 16:12
MaxMax
15.5k11143
$endgroup$
add a comment |
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1 Answer
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$begingroup$
I am assuming $C(X)$ denotes the set of real valued (continuous) functions on $X$.
First of all, notice that the two equivalences are equivalent, because $A$ is equivalent to $B$ if and only if $neg A$ is equivalent to $neg B$ .
Then if $(f,g) = C(X)$, there are $h,k$ such that $fh+gk =1$. If $xin Z(f)cap Z(g)$, then $0=1$. Thus $Z(f)cap Z(g)=emptyset$.
Conversely, if $Z(f)cap Z(g)=emptyset$, then $f^2 + g^2$ never vanishes (because we're considering real valued functions), so it's invertible in $C(X)$, and is in $(f,g)$ , so $(f,g) = C(X)$ .
answered Dec 30 '18 at 16:12
MaxMax
15.5k11143
$endgroup$
add a comment |
$begingroup$
I am assuming $C(X)$ denotes the set of real valued (continuous) functions on $X$.
First of all, notice that the two equivalences are equivalent, because $A$ is equivalent to $B$ if and only if $neg A$ is equivalent to $neg B$ .
Then if $(f,g) = C(X)$, there are $h,k$ such that $fh+gk =1$. If $xin Z(f)cap Z(g)$, then $0=1$. Thus $Z(f)cap Z(g)=emptyset$.
Conversely, if $Z(f)cap Z(g)=emptyset$, then $f^2 + g^2$ never vanishes (because we're considering real valued functions), so it's invertible in $C(X)$, and is in $(f,g)$ , so $(f,g) = C(X)$ .
answered Dec 30 '18 at 16:12
MaxMax
15.5k11143
$endgroup$
add a comment |
$begingroup$
I am assuming $C(X)$ denotes the set of real valued (continuous) functions on $X$.
First of all, notice that the two equivalences are equivalent, because $A$ is equivalent to $B$ if and only if $neg A$ is equivalent to $neg B$ .
Then if $(f,g) = C(X)$, there are $h,k$ such that $fh+gk =1$. If $xin Z(f)cap Z(g)$, then $0=1$. Thus $Z(f)cap Z(g)=emptyset$.
Conversely, if $Z(f)cap Z(g)=emptyset$, then $f^2 + g^2$ never vanishes (because we're considering real valued functions), so it's invertible in $C(X)$, and is in $(f,g)$ , so $(f,g) = C(X)$ .
answered Dec 30 '18 at 16:12
MaxMax
15.5k11143
$endgroup$
I am assuming $C(X)$ denotes the set of real valued (continuous) functions on $X$.
First of all, notice that the two equivalences are equivalent, because $A$ is equivalent to $B$ if and only if $neg A$ is equivalent to $neg B$ .
Then if $(f,g) = C(X)$, there are $h,k$ such that $fh+gk =1$. If $xin Z(f)cap Z(g)$, then $0=1$. Thus $Z(f)cap Z(g)=emptyset$.
Conversely, if $Z(f)cap Z(g)=emptyset$, then $f^2 + g^2$ never vanishes (because we're considering real valued functions), so it's invertible in $C(X)$, and is in $(f,g)$ , so $(f,g) = C(X)$ .
answered Dec 30 '18 at 16:12
MaxMax
15.5k11143
answered Dec 30 '18 at 16:12
MaxMax
15.5k11143
answered Dec 30 '18 at 16:12
MaxMax
15.5k11143
answered Dec 30 '18 at 16:12
MaxMax
15.5k11143
15.5k11143
add a comment |
add a comment |
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