Product of connected spaces - Proof
$begingroup$
I'm working on the "iff"-relation given by:
$X=prod_{iin I}X_i$ is connected iff each $X_i$ non-empty is connected for all $iin I$.
I could prove the "$Rightarrow$"-direction very easyly. I also proved that a finite product of connected spaces is connected. Now i want to prove the following:
- Choose $z=(z_i)inprod_{iin I}X_i$. For every finite subset $Jsubset I$ is the set $X_J:=left{xin X:x_i=z_i forall I-Jright}$ connected.
I have given the follwoing proof: This set is homeomorphic with a finite product $prod_{jin J}X_j$ given by the map defined by: $x=(x_j)_{jin J}$ mapped on $y=(y_i)_{iin I}$ such that $y_j=x_j$ if $jin J$ and $y_j=z_j$ if $jnotin J$. This mapping is continuous and injective (and also the inverse is continous since it is the projection map). But then we know that $X_J$ is connected since every finite product is connected (if the components are connected).
Is this proof correct? The only thing i have to prove now is that $Y=cup_{Jsubset I,J finite}X_J$ is dense in X. How do I do that? Can someone help? Thank you
general-topology connectedness
edited Oct 10 '15 at 11:50
kroner
1,2793826
asked Dec 13 '12 at 15:40
Cut-pointCut-point
462
$endgroup$
add a comment |
$begingroup$
I'm working on the "iff"-relation given by:
$X=prod_{iin I}X_i$ is connected iff each $X_i$ non-empty is connected for all $iin I$.
I could prove the "$Rightarrow$"-direction very easyly. I also proved that a finite product of connected spaces is connected. Now i want to prove the following:
- Choose $z=(z_i)inprod_{iin I}X_i$. For every finite subset $Jsubset I$ is the set $X_J:=left{xin X:x_i=z_i forall I-Jright}$ connected.
I have given the follwoing proof: This set is homeomorphic with a finite product $prod_{jin J}X_j$ given by the map defined by: $x=(x_j)_{jin J}$ mapped on $y=(y_i)_{iin I}$ such that $y_j=x_j$ if $jin J$ and $y_j=z_j$ if $jnotin J$. This mapping is continuous and injective (and also the inverse is continous since it is the projection map). But then we know that $X_J$ is connected since every finite product is connected (if the components are connected).
Is this proof correct? The only thing i have to prove now is that $Y=cup_{Jsubset I,J finite}X_J$ is dense in X. How do I do that? Can someone help? Thank you
general-topology connectedness
edited Oct 10 '15 at 11:50
kroner
1,2793826
asked Dec 13 '12 at 15:40
Cut-pointCut-point
462
$endgroup$
add a comment |
$begingroup$
I'm working on the "iff"-relation given by:
$X=prod_{iin I}X_i$ is connected iff each $X_i$ non-empty is connected for all $iin I$.
I could prove the "$Rightarrow$"-direction very easyly. I also proved that a finite product of connected spaces is connected. Now i want to prove the following:
- Choose $z=(z_i)inprod_{iin I}X_i$. For every finite subset $Jsubset I$ is the set $X_J:=left{xin X:x_i=z_i forall I-Jright}$ connected.
I have given the follwoing proof: This set is homeomorphic with a finite product $prod_{jin J}X_j$ given by the map defined by: $x=(x_j)_{jin J}$ mapped on $y=(y_i)_{iin I}$ such that $y_j=x_j$ if $jin J$ and $y_j=z_j$ if $jnotin J$. This mapping is continuous and injective (and also the inverse is continous since it is the projection map). But then we know that $X_J$ is connected since every finite product is connected (if the components are connected).
Is this proof correct? The only thing i have to prove now is that $Y=cup_{Jsubset I,J finite}X_J$ is dense in X. How do I do that? Can someone help? Thank you
general-topology connectedness
edited Oct 10 '15 at 11:50
kroner
1,2793826
asked Dec 13 '12 at 15:40
Cut-pointCut-point
462
$endgroup$
I'm working on the "iff"-relation given by:
$X=prod_{iin I}X_i$ is connected iff each $X_i$ non-empty is connected for all $iin I$.
I could prove the "$Rightarrow$"-direction very easyly. I also proved that a finite product of connected spaces is connected. Now i want to prove the following:
- Choose $z=(z_i)inprod_{iin I}X_i$. For every finite subset $Jsubset I$ is the set $X_J:=left{xin X:x_i=z_i forall I-Jright}$ connected.
I have given the follwoing proof: This set is homeomorphic with a finite product $prod_{jin J}X_j$ given by the map defined by: $x=(x_j)_{jin J}$ mapped on $y=(y_i)_{iin I}$ such that $y_j=x_j$ if $jin J$ and $y_j=z_j$ if $jnotin J$. This mapping is continuous and injective (and also the inverse is continous since it is the projection map). But then we know that $X_J$ is connected since every finite product is connected (if the components are connected).
Is this proof correct? The only thing i have to prove now is that $Y=cup_{Jsubset I,J finite}X_J$ is dense in X. How do I do that? Can someone help? Thank you
general-topology connectedness
general-topology connectedness
edited Oct 10 '15 at 11:50
kroner
1,2793826
asked Dec 13 '12 at 15:40
Cut-pointCut-point
462
edited Oct 10 '15 at 11:50
kroner
1,2793826
asked Dec 13 '12 at 15:40
Cut-pointCut-point
462
edited Oct 10 '15 at 11:50
kroner
1,2793826
edited Oct 10 '15 at 11:50
kroner
1,2793826
edited Oct 10 '15 at 11:50
kroner
1,2793826
1,2793826
asked Dec 13 '12 at 15:40
Cut-pointCut-point
462
asked Dec 13 '12 at 15:40
Cut-pointCut-point
462
asked Dec 13 '12 at 15:40
Cut-pointCut-point
462
462
add a comment |
add a comment |
1 Answer
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What you've done so far looks good. As you note correctly, it suffices to show that the set $Y=bigcup_{J⊂I, Jtext{ finite}}X_J$, which can also be described as ${yin X; y(i)=z(i)textrm{ for almost all }iin I}$ is dense, since this means that $X=overline Y$ and is thus connected, being the closure of the connected set $Y$. In order to do this, let $U=p^{-1}_{j_1}(U_{j_1})capdotscap p^{-1}_{j_n}(U_{j_n})$ be an open basis set. Now choose a $y$ such that $y(j)=p_j(y)in U_j forall jin{j_1,dots,j_n}$ and otherwise $y(i)=z(i)$. Then $yin Ycap U.$
answered Dec 13 '12 at 21:52
Stefan HamckeStefan Hamcke
21.6k42877
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
What you've done so far looks good. As you note correctly, it suffices to show that the set $Y=bigcup_{J⊂I, Jtext{ finite}}X_J$, which can also be described as ${yin X; y(i)=z(i)textrm{ for almost all }iin I}$ is dense, since this means that $X=overline Y$ and is thus connected, being the closure of the connected set $Y$. In order to do this, let $U=p^{-1}_{j_1}(U_{j_1})capdotscap p^{-1}_{j_n}(U_{j_n})$ be an open basis set. Now choose a $y$ such that $y(j)=p_j(y)in U_j forall jin{j_1,dots,j_n}$ and otherwise $y(i)=z(i)$. Then $yin Ycap U.$
answered Dec 13 '12 at 21:52
Stefan HamckeStefan Hamcke
21.6k42877
$endgroup$
add a comment |
$begingroup$
What you've done so far looks good. As you note correctly, it suffices to show that the set $Y=bigcup_{J⊂I, Jtext{ finite}}X_J$, which can also be described as ${yin X; y(i)=z(i)textrm{ for almost all }iin I}$ is dense, since this means that $X=overline Y$ and is thus connected, being the closure of the connected set $Y$. In order to do this, let $U=p^{-1}_{j_1}(U_{j_1})capdotscap p^{-1}_{j_n}(U_{j_n})$ be an open basis set. Now choose a $y$ such that $y(j)=p_j(y)in U_j forall jin{j_1,dots,j_n}$ and otherwise $y(i)=z(i)$. Then $yin Ycap U.$
answered Dec 13 '12 at 21:52
Stefan HamckeStefan Hamcke
21.6k42877
$endgroup$
add a comment |
$begingroup$
What you've done so far looks good. As you note correctly, it suffices to show that the set $Y=bigcup_{J⊂I, Jtext{ finite}}X_J$, which can also be described as ${yin X; y(i)=z(i)textrm{ for almost all }iin I}$ is dense, since this means that $X=overline Y$ and is thus connected, being the closure of the connected set $Y$. In order to do this, let $U=p^{-1}_{j_1}(U_{j_1})capdotscap p^{-1}_{j_n}(U_{j_n})$ be an open basis set. Now choose a $y$ such that $y(j)=p_j(y)in U_j forall jin{j_1,dots,j_n}$ and otherwise $y(i)=z(i)$. Then $yin Ycap U.$
answered Dec 13 '12 at 21:52
Stefan HamckeStefan Hamcke
21.6k42877
$endgroup$
What you've done so far looks good. As you note correctly, it suffices to show that the set $Y=bigcup_{J⊂I, Jtext{ finite}}X_J$, which can also be described as ${yin X; y(i)=z(i)textrm{ for almost all }iin I}$ is dense, since this means that $X=overline Y$ and is thus connected, being the closure of the connected set $Y$. In order to do this, let $U=p^{-1}_{j_1}(U_{j_1})capdotscap p^{-1}_{j_n}(U_{j_n})$ be an open basis set. Now choose a $y$ such that $y(j)=p_j(y)in U_j forall jin{j_1,dots,j_n}$ and otherwise $y(i)=z(i)$. Then $yin Ycap U.$
answered Dec 13 '12 at 21:52
Stefan HamckeStefan Hamcke
21.6k42877
edited Nov 26 '13 at 15:55
answered Dec 13 '12 at 21:52
Stefan HamckeStefan Hamcke
21.6k42877
answered Dec 13 '12 at 21:52
Stefan HamckeStefan Hamcke
21.6k42877
answered Dec 13 '12 at 21:52
Stefan HamckeStefan Hamcke
21.6k42877
21.6k42877
add a comment |
add a comment |
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