Using Zorn's lemma for a proof about irreducible continuous functions.
$begingroup$
Let $f:Xrightarrow Y$ onto and continuous between two Hausdorff compact spaces. We say that $f$ is irreducible iff for any closed proper subset $Hsubset X$, $f(H)neq Y$.
Now, consider an arbitrary, onto and continuous function $f:Xrightarrow Y$ where $X$ and $Y$ are Hausdorff compact spaces. I need to show that there exists a closed set $Zsubseteq X$ such that $fupharpoonright Z$ is irreducible.
My attempt: Let $Sigma={Zsubseteq X:Z text{is closed} wedge fupharpoonright Z text{is onto}}$. It is clear that $(Sigma,supset)$ is a partial order for $Sigma$ and $Sigmaneqemptyset$ because $X$ is one of its elements. Let $CsubseteqSigma$ a chain and as $X$ is a compact space, $bigcap Cneqemptyset$ (because $C$ has FIP)...
As you can see, I try to use Zorn's lemma in my proof. My problem is in the last part, I have many troubles when I try to show that $fupharpoonrightbigcap C$ is onto, I don't see as I can get it (well, I think that $bigcap C$ must be a lower bound for $C$ in $Sigma$). Can you give me a hint?
general-topology compactness
edited Jan 5 at 23:05
Asaf Karagila♦
306k33438769
asked Jan 5 at 22:24
GödelGödel
1,450319
$endgroup$
add a comment |
$begingroup$
Let $f:Xrightarrow Y$ onto and continuous between two Hausdorff compact spaces. We say that $f$ is irreducible iff for any closed proper subset $Hsubset X$, $f(H)neq Y$.
Now, consider an arbitrary, onto and continuous function $f:Xrightarrow Y$ where $X$ and $Y$ are Hausdorff compact spaces. I need to show that there exists a closed set $Zsubseteq X$ such that $fupharpoonright Z$ is irreducible.
My attempt: Let $Sigma={Zsubseteq X:Z text{is closed} wedge fupharpoonright Z text{is onto}}$. It is clear that $(Sigma,supset)$ is a partial order for $Sigma$ and $Sigmaneqemptyset$ because $X$ is one of its elements. Let $CsubseteqSigma$ a chain and as $X$ is a compact space, $bigcap Cneqemptyset$ (because $C$ has FIP)...
As you can see, I try to use Zorn's lemma in my proof. My problem is in the last part, I have many troubles when I try to show that $fupharpoonrightbigcap C$ is onto, I don't see as I can get it (well, I think that $bigcap C$ must be a lower bound for $C$ in $Sigma$). Can you give me a hint?
general-topology compactness
edited Jan 5 at 23:05
Asaf Karagila♦
306k33438769
asked Jan 5 at 22:24
GödelGödel
1,450319
$endgroup$
$begingroup$
I don't think this is really about Zorn's lemma as much as it is about continuous surjections and compact sets. But I'm happy to hear different opinions as to why this really fits under the axiom-of-choice tag.
$endgroup$
– Asaf Karagila♦
Jan 5 at 23:06
$begingroup$
@AsafKaragila maybe Zorn's lemma should have a tag of its own? There are probably quite a few questions on its use.
$endgroup$
– Henno Brandsma
Jan 5 at 23:09
$begingroup$
@Henno: (1) I don't think that there should be a tag for Zorn's lemma which is separate. (2) I still don't think this is about Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 6 at 0:25
add a comment |
$begingroup$
Let $f:Xrightarrow Y$ onto and continuous between two Hausdorff compact spaces. We say that $f$ is irreducible iff for any closed proper subset $Hsubset X$, $f(H)neq Y$.
Now, consider an arbitrary, onto and continuous function $f:Xrightarrow Y$ where $X$ and $Y$ are Hausdorff compact spaces. I need to show that there exists a closed set $Zsubseteq X$ such that $fupharpoonright Z$ is irreducible.
My attempt: Let $Sigma={Zsubseteq X:Z text{is closed} wedge fupharpoonright Z text{is onto}}$. It is clear that $(Sigma,supset)$ is a partial order for $Sigma$ and $Sigmaneqemptyset$ because $X$ is one of its elements. Let $CsubseteqSigma$ a chain and as $X$ is a compact space, $bigcap Cneqemptyset$ (because $C$ has FIP)...
As you can see, I try to use Zorn's lemma in my proof. My problem is in the last part, I have many troubles when I try to show that $fupharpoonrightbigcap C$ is onto, I don't see as I can get it (well, I think that $bigcap C$ must be a lower bound for $C$ in $Sigma$). Can you give me a hint?
general-topology compactness
edited Jan 5 at 23:05
Asaf Karagila♦
306k33438769
asked Jan 5 at 22:24
GödelGödel
1,450319
$endgroup$
Let $f:Xrightarrow Y$ onto and continuous between two Hausdorff compact spaces. We say that $f$ is irreducible iff for any closed proper subset $Hsubset X$, $f(H)neq Y$.
Now, consider an arbitrary, onto and continuous function $f:Xrightarrow Y$ where $X$ and $Y$ are Hausdorff compact spaces. I need to show that there exists a closed set $Zsubseteq X$ such that $fupharpoonright Z$ is irreducible.
My attempt: Let $Sigma={Zsubseteq X:Z text{is closed} wedge fupharpoonright Z text{is onto}}$. It is clear that $(Sigma,supset)$ is a partial order for $Sigma$ and $Sigmaneqemptyset$ because $X$ is one of its elements. Let $CsubseteqSigma$ a chain and as $X$ is a compact space, $bigcap Cneqemptyset$ (because $C$ has FIP)...
As you can see, I try to use Zorn's lemma in my proof. My problem is in the last part, I have many troubles when I try to show that $fupharpoonrightbigcap C$ is onto, I don't see as I can get it (well, I think that $bigcap C$ must be a lower bound for $C$ in $Sigma$). Can you give me a hint?
general-topology compactness
general-topology compactness
edited Jan 5 at 23:05
Asaf Karagila♦
306k33438769
asked Jan 5 at 22:24
GödelGödel
1,450319
edited Jan 5 at 23:05
Asaf Karagila♦
306k33438769
asked Jan 5 at 22:24
GödelGödel
1,450319
edited Jan 5 at 23:05
Asaf Karagila♦
306k33438769
edited Jan 5 at 23:05
Asaf Karagila♦
306k33438769
edited Jan 5 at 23:05
Asaf Karagila♦
306k33438769
306k33438769
asked Jan 5 at 22:24
GödelGödel
1,450319
asked Jan 5 at 22:24
GödelGödel
1,450319
asked Jan 5 at 22:24
GödelGödel
1,450319
1,450319
$begingroup$
I don't think this is really about Zorn's lemma as much as it is about continuous surjections and compact sets. But I'm happy to hear different opinions as to why this really fits under the axiom-of-choice tag.
$endgroup$
– Asaf Karagila♦
Jan 5 at 23:06
$begingroup$
@AsafKaragila maybe Zorn's lemma should have a tag of its own? There are probably quite a few questions on its use.
$endgroup$
– Henno Brandsma
Jan 5 at 23:09
$begingroup$
@Henno: (1) I don't think that there should be a tag for Zorn's lemma which is separate. (2) I still don't think this is about Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 6 at 0:25
add a comment |
$begingroup$
I don't think this is really about Zorn's lemma as much as it is about continuous surjections and compact sets. But I'm happy to hear different opinions as to why this really fits under the axiom-of-choice tag.
$endgroup$
– Asaf Karagila♦
Jan 5 at 23:06
$begingroup$
@AsafKaragila maybe Zorn's lemma should have a tag of its own? There are probably quite a few questions on its use.
$endgroup$
– Henno Brandsma
Jan 5 at 23:09
$begingroup$
@Henno: (1) I don't think that there should be a tag for Zorn's lemma which is separate. (2) I still don't think this is about Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 6 at 0:25
$begingroup$
I don't think this is really about Zorn's lemma as much as it is about continuous surjections and compact sets. But I'm happy to hear different opinions as to why this really fits under the axiom-of-choice tag.
$endgroup$
– Asaf Karagila♦
Jan 5 at 23:06
$begingroup$
I don't think this is really about Zorn's lemma as much as it is about continuous surjections and compact sets. But I'm happy to hear different opinions as to why this really fits under the axiom-of-choice tag.
$endgroup$
– Asaf Karagila♦
Jan 5 at 23:06
$begingroup$
@AsafKaragila maybe Zorn's lemma should have a tag of its own? There are probably quite a few questions on its use.
$endgroup$
– Henno Brandsma
Jan 5 at 23:09
$begingroup$
@AsafKaragila maybe Zorn's lemma should have a tag of its own? There are probably quite a few questions on its use.
$endgroup$
– Henno Brandsma
Jan 5 at 23:09
$begingroup$
@Henno: (1) I don't think that there should be a tag for Zorn's lemma which is separate. (2) I still don't think this is about Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 6 at 0:25
$begingroup$
@Henno: (1) I don't think that there should be a tag for Zorn's lemma which is separate. (2) I still don't think this is about Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 6 at 0:25
add a comment |
1 Answer
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$begingroup$
If your chain is denoted $C_i, i in I$ and we have that for all $i,j$ we either have $C_i subseteq C_j$ or reversely. (This is what being a chain means). We want to show that $C = bigcap_i C_i$ is the required upperbound in $Sigma$ and for this we need that $f$ maps $C$ onto $Y$. For this consider $y in Y$ and note that $D_i := C_i cap f^{-1}[{y}]$ consists of closed subsets of $X$ (here we need that $Y$ is $T_1$ but that follows from Hausdorffness, and that $f$ is continuous) that are all non-empty (as every $C_i$ has at least one point mapping to $y$, as $f|_{C_i}$ is onto). We still have a chain, and hence the FIP holds and we have some $x in bigcap_i D_i subseteq C$ by compactness. As $x in D_i$ (for any $i$), in particular $f(x) = y$ and as $y in Y$ was arbitrary, $f|_C$ is onto and the required Zorn chain-upperbound (as we have reverse inclusion as order).
So Zorn gives us a maximal element $M$ in $Sigma$ and this is by definition as required: being in $Sigma$ implies $f[M]=Y$ and if we have any closed $H subsetneq M$, it has to obey $f[M] neq Y$ or else it would be a member of $Sigma$ and thus a properly "larger" element than the maximal $M$, which cannot be.
answered Jan 5 at 22:45
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
add a comment |
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$begingroup$
If your chain is denoted $C_i, i in I$ and we have that for all $i,j$ we either have $C_i subseteq C_j$ or reversely. (This is what being a chain means). We want to show that $C = bigcap_i C_i$ is the required upperbound in $Sigma$ and for this we need that $f$ maps $C$ onto $Y$. For this consider $y in Y$ and note that $D_i := C_i cap f^{-1}[{y}]$ consists of closed subsets of $X$ (here we need that $Y$ is $T_1$ but that follows from Hausdorffness, and that $f$ is continuous) that are all non-empty (as every $C_i$ has at least one point mapping to $y$, as $f|_{C_i}$ is onto). We still have a chain, and hence the FIP holds and we have some $x in bigcap_i D_i subseteq C$ by compactness. As $x in D_i$ (for any $i$), in particular $f(x) = y$ and as $y in Y$ was arbitrary, $f|_C$ is onto and the required Zorn chain-upperbound (as we have reverse inclusion as order).
So Zorn gives us a maximal element $M$ in $Sigma$ and this is by definition as required: being in $Sigma$ implies $f[M]=Y$ and if we have any closed $H subsetneq M$, it has to obey $f[M] neq Y$ or else it would be a member of $Sigma$ and thus a properly "larger" element than the maximal $M$, which cannot be.
answered Jan 5 at 22:45
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
add a comment |
$begingroup$
If your chain is denoted $C_i, i in I$ and we have that for all $i,j$ we either have $C_i subseteq C_j$ or reversely. (This is what being a chain means). We want to show that $C = bigcap_i C_i$ is the required upperbound in $Sigma$ and for this we need that $f$ maps $C$ onto $Y$. For this consider $y in Y$ and note that $D_i := C_i cap f^{-1}[{y}]$ consists of closed subsets of $X$ (here we need that $Y$ is $T_1$ but that follows from Hausdorffness, and that $f$ is continuous) that are all non-empty (as every $C_i$ has at least one point mapping to $y$, as $f|_{C_i}$ is onto). We still have a chain, and hence the FIP holds and we have some $x in bigcap_i D_i subseteq C$ by compactness. As $x in D_i$ (for any $i$), in particular $f(x) = y$ and as $y in Y$ was arbitrary, $f|_C$ is onto and the required Zorn chain-upperbound (as we have reverse inclusion as order).
So Zorn gives us a maximal element $M$ in $Sigma$ and this is by definition as required: being in $Sigma$ implies $f[M]=Y$ and if we have any closed $H subsetneq M$, it has to obey $f[M] neq Y$ or else it would be a member of $Sigma$ and thus a properly "larger" element than the maximal $M$, which cannot be.
answered Jan 5 at 22:45
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
add a comment |
$begingroup$
If your chain is denoted $C_i, i in I$ and we have that for all $i,j$ we either have $C_i subseteq C_j$ or reversely. (This is what being a chain means). We want to show that $C = bigcap_i C_i$ is the required upperbound in $Sigma$ and for this we need that $f$ maps $C$ onto $Y$. For this consider $y in Y$ and note that $D_i := C_i cap f^{-1}[{y}]$ consists of closed subsets of $X$ (here we need that $Y$ is $T_1$ but that follows from Hausdorffness, and that $f$ is continuous) that are all non-empty (as every $C_i$ has at least one point mapping to $y$, as $f|_{C_i}$ is onto). We still have a chain, and hence the FIP holds and we have some $x in bigcap_i D_i subseteq C$ by compactness. As $x in D_i$ (for any $i$), in particular $f(x) = y$ and as $y in Y$ was arbitrary, $f|_C$ is onto and the required Zorn chain-upperbound (as we have reverse inclusion as order).
So Zorn gives us a maximal element $M$ in $Sigma$ and this is by definition as required: being in $Sigma$ implies $f[M]=Y$ and if we have any closed $H subsetneq M$, it has to obey $f[M] neq Y$ or else it would be a member of $Sigma$ and thus a properly "larger" element than the maximal $M$, which cannot be.
answered Jan 5 at 22:45
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
If your chain is denoted $C_i, i in I$ and we have that for all $i,j$ we either have $C_i subseteq C_j$ or reversely. (This is what being a chain means). We want to show that $C = bigcap_i C_i$ is the required upperbound in $Sigma$ and for this we need that $f$ maps $C$ onto $Y$. For this consider $y in Y$ and note that $D_i := C_i cap f^{-1}[{y}]$ consists of closed subsets of $X$ (here we need that $Y$ is $T_1$ but that follows from Hausdorffness, and that $f$ is continuous) that are all non-empty (as every $C_i$ has at least one point mapping to $y$, as $f|_{C_i}$ is onto). We still have a chain, and hence the FIP holds and we have some $x in bigcap_i D_i subseteq C$ by compactness. As $x in D_i$ (for any $i$), in particular $f(x) = y$ and as $y in Y$ was arbitrary, $f|_C$ is onto and the required Zorn chain-upperbound (as we have reverse inclusion as order).
So Zorn gives us a maximal element $M$ in $Sigma$ and this is by definition as required: being in $Sigma$ implies $f[M]=Y$ and if we have any closed $H subsetneq M$, it has to obey $f[M] neq Y$ or else it would be a member of $Sigma$ and thus a properly "larger" element than the maximal $M$, which cannot be.
answered Jan 5 at 22:45
Henno BrandsmaHenno Brandsma
113k348123
edited Jan 5 at 23:05
answered Jan 5 at 22:45
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 5 at 22:45
Henno BrandsmaHenno Brandsma
113k348123
answered Jan 5 at 22:45
Henno BrandsmaHenno Brandsma
113k348123
113k348123
add a comment |
add a comment |
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$begingroup$
I don't think this is really about Zorn's lemma as much as it is about continuous surjections and compact sets. But I'm happy to hear different opinions as to why this really fits under the axiom-of-choice tag.
$endgroup$
– Asaf Karagila♦
Jan 5 at 23:06
$begingroup$
@AsafKaragila maybe Zorn's lemma should have a tag of its own? There are probably quite a few questions on its use.
$endgroup$
– Henno Brandsma
Jan 5 at 23:09
$begingroup$
@Henno: (1) I don't think that there should be a tag for Zorn's lemma which is separate. (2) I still don't think this is about Zorn's lemma.
$endgroup$
– Asaf Karagila♦
Jan 6 at 0:25