Queris related to “An arbitrary product $X=prod_{alpha in J} X_alpha$ is connected iff $X_{alpha }$ is...
The following proof of
An arbitrary product $X=prod_{alpha in J} X_alpha$ is connected iff $X_{alpha }$ is connected for each $alphain J$ is provided by
my professor.I've some queries in this proof.
Let X is connected then we shall show that each $X_{alpha}$ is connected,where $alphain J,$some index set.
Consider $pi_{alpha}:Xrightarrow X_{alpha},{alpha}in J$.
Since $pi_{alpha}$ is a projection map so it is continuous & $X$ is connected by assumption.
Using the result:"Continuous image of a connected set is connected."
we have $X_{alpha}$ is connected.
Conversely,
Result Used:
X is connected iff every continuous function $f:Xrightarrow ${$0,1$}
is constant.
Fix $f:Xrightarrow ${$0,1$} a continuous function.
Fix $(a_{alpha})in X$ and for each $alphain J,$define $varphi_{alpha}:X_{alpha}rightarrow X$ by $varphi_{alpha}opi_{beta}(x) =
begin{cases}
a_{beta}, & text{if $betaneq alpha$} \[2ex]
x, & text{if $beta=alpha$ }
end{cases}$.
Let $g_{alpha}=fovarphi_{alpha}:X_{alpha}rightarrow${$0,1$} for each $alphain J.$Then each $g_{alpha}$ is constant since $X$ is connected.
Hence,for each $alphain J, f$ is constant on $varphi [X_{alpha}].$
Say,Without loss of generality,for some $alpha_0in J,f[varphi_{alpha_0[X_{alpha}]}]=${$0$}.Since,$(a_{alpha})in varphi_{beta}[X_{beta}]$ for each $beta in J,$ it must be that $forall alphain J,f[varphi_{alpha_0[X_{alpha}]}]=${$0$}.
Then $f^{-1}$[{$0$}] is an open subset of $X$ by continuity which contains all of these elements.$tag{*}$
1.(Please clarify how $f^{-1}$[{$0$}] is open subset of $X$ :Is it because {0} is open in {0,1},$f$ is continuous,so inverse image of open set is open? )
Let $Usubset f^{-1}$[{$0$}] be a basis element of $X$ containing $(a_{alpha})$ so that $pi_{alpha}[U]=X_{alpha}$ for all but finitely many $alpha in J$.Let K be the finite subset of $J$ comprised of these $alpha $ and let $(x_{alpha})in U$ be that element such that $x_{alpha}=a_{alpha} $ if $alpha in K.$
Changing the coordinates of $(x_{alpha})$ indexed by $K$ one at a time,it follows from ($*$) that each time we change it,it remains in $f^{-1}$[{$0$}].Since $pi_{alpha}[U]=X_{alpha}$ for all but finitely many $alpha in J$.
We have shown that $prod_{alpha in J} X_alpha=f^{-1}$[{$0$}].(##2.I'm not getting this sudden transition.Please explain!!)
If there is some scope of refinement in the above proof,feel free to express your views.Also check this proof critically...
real-analysis general-topology proof-verification connectedness open-map
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,425726
add a comment |
The following proof of
An arbitrary product $X=prod_{alpha in J} X_alpha$ is connected iff $X_{alpha }$ is connected for each $alphain J$ is provided by
my professor.I've some queries in this proof.
Let X is connected then we shall show that each $X_{alpha}$ is connected,where $alphain J,$some index set.
Consider $pi_{alpha}:Xrightarrow X_{alpha},{alpha}in J$.
Since $pi_{alpha}$ is a projection map so it is continuous & $X$ is connected by assumption.
Using the result:"Continuous image of a connected set is connected."
we have $X_{alpha}$ is connected.
Conversely,
Result Used:
X is connected iff every continuous function $f:Xrightarrow ${$0,1$}
is constant.
Fix $f:Xrightarrow ${$0,1$} a continuous function.
Fix $(a_{alpha})in X$ and for each $alphain J,$define $varphi_{alpha}:X_{alpha}rightarrow X$ by $varphi_{alpha}opi_{beta}(x) =
begin{cases}
a_{beta}, & text{if $betaneq alpha$} \[2ex]
x, & text{if $beta=alpha$ }
end{cases}$.
Let $g_{alpha}=fovarphi_{alpha}:X_{alpha}rightarrow${$0,1$} for each $alphain J.$Then each $g_{alpha}$ is constant since $X$ is connected.
Hence,for each $alphain J, f$ is constant on $varphi [X_{alpha}].$
Say,Without loss of generality,for some $alpha_0in J,f[varphi_{alpha_0[X_{alpha}]}]=${$0$}.Since,$(a_{alpha})in varphi_{beta}[X_{beta}]$ for each $beta in J,$ it must be that $forall alphain J,f[varphi_{alpha_0[X_{alpha}]}]=${$0$}.
Then $f^{-1}$[{$0$}] is an open subset of $X$ by continuity which contains all of these elements.$tag{*}$
1.(Please clarify how $f^{-1}$[{$0$}] is open subset of $X$ :Is it because {0} is open in {0,1},$f$ is continuous,so inverse image of open set is open? )
Let $Usubset f^{-1}$[{$0$}] be a basis element of $X$ containing $(a_{alpha})$ so that $pi_{alpha}[U]=X_{alpha}$ for all but finitely many $alpha in J$.Let K be the finite subset of $J$ comprised of these $alpha $ and let $(x_{alpha})in U$ be that element such that $x_{alpha}=a_{alpha} $ if $alpha in K.$
Changing the coordinates of $(x_{alpha})$ indexed by $K$ one at a time,it follows from ($*$) that each time we change it,it remains in $f^{-1}$[{$0$}].Since $pi_{alpha}[U]=X_{alpha}$ for all but finitely many $alpha in J$.
We have shown that $prod_{alpha in J} X_alpha=f^{-1}$[{$0$}].(##2.I'm not getting this sudden transition.Please explain!!)
If there is some scope of refinement in the above proof,feel free to express your views.Also check this proof critically...
real-analysis general-topology proof-verification connectedness open-map
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,425726
1
In response to 1: yes
– mathworker21
Dec 12 '18 at 8:59
The definition of $varphi$ is missing a subscript.
– William Elliot
Dec 13 '18 at 2:38
@William Elliot:see the edit
– P.Styles
Dec 13 '18 at 3:31
$pi and varphi$ appear to be in reverse order.
– William Elliot
Dec 13 '18 at 7:49
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
– William Elliot
Dec 13 '18 at 7:51
add a comment |
The following proof of
An arbitrary product $X=prod_{alpha in J} X_alpha$ is connected iff $X_{alpha }$ is connected for each $alphain J$ is provided by
my professor.I've some queries in this proof.
Let X is connected then we shall show that each $X_{alpha}$ is connected,where $alphain J,$some index set.
Consider $pi_{alpha}:Xrightarrow X_{alpha},{alpha}in J$.
Since $pi_{alpha}$ is a projection map so it is continuous & $X$ is connected by assumption.
Using the result:"Continuous image of a connected set is connected."
we have $X_{alpha}$ is connected.
Conversely,
Result Used:
X is connected iff every continuous function $f:Xrightarrow ${$0,1$}
is constant.
Fix $f:Xrightarrow ${$0,1$} a continuous function.
Fix $(a_{alpha})in X$ and for each $alphain J,$define $varphi_{alpha}:X_{alpha}rightarrow X$ by $varphi_{alpha}opi_{beta}(x) =
begin{cases}
a_{beta}, & text{if $betaneq alpha$} \[2ex]
x, & text{if $beta=alpha$ }
end{cases}$.
Let $g_{alpha}=fovarphi_{alpha}:X_{alpha}rightarrow${$0,1$} for each $alphain J.$Then each $g_{alpha}$ is constant since $X$ is connected.
Hence,for each $alphain J, f$ is constant on $varphi [X_{alpha}].$
Say,Without loss of generality,for some $alpha_0in J,f[varphi_{alpha_0[X_{alpha}]}]=${$0$}.Since,$(a_{alpha})in varphi_{beta}[X_{beta}]$ for each $beta in J,$ it must be that $forall alphain J,f[varphi_{alpha_0[X_{alpha}]}]=${$0$}.
Then $f^{-1}$[{$0$}] is an open subset of $X$ by continuity which contains all of these elements.$tag{*}$
1.(Please clarify how $f^{-1}$[{$0$}] is open subset of $X$ :Is it because {0} is open in {0,1},$f$ is continuous,so inverse image of open set is open? )
Let $Usubset f^{-1}$[{$0$}] be a basis element of $X$ containing $(a_{alpha})$ so that $pi_{alpha}[U]=X_{alpha}$ for all but finitely many $alpha in J$.Let K be the finite subset of $J$ comprised of these $alpha $ and let $(x_{alpha})in U$ be that element such that $x_{alpha}=a_{alpha} $ if $alpha in K.$
Changing the coordinates of $(x_{alpha})$ indexed by $K$ one at a time,it follows from ($*$) that each time we change it,it remains in $f^{-1}$[{$0$}].Since $pi_{alpha}[U]=X_{alpha}$ for all but finitely many $alpha in J$.
We have shown that $prod_{alpha in J} X_alpha=f^{-1}$[{$0$}].(##2.I'm not getting this sudden transition.Please explain!!)
If there is some scope of refinement in the above proof,feel free to express your views.Also check this proof critically...
real-analysis general-topology proof-verification connectedness open-map
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,425726
The following proof of
An arbitrary product $X=prod_{alpha in J} X_alpha$ is connected iff $X_{alpha }$ is connected for each $alphain J$ is provided by
my professor.I've some queries in this proof.
Let X is connected then we shall show that each $X_{alpha}$ is connected,where $alphain J,$some index set.
Consider $pi_{alpha}:Xrightarrow X_{alpha},{alpha}in J$.
Since $pi_{alpha}$ is a projection map so it is continuous & $X$ is connected by assumption.
Using the result:"Continuous image of a connected set is connected."
we have $X_{alpha}$ is connected.
Conversely,
Result Used:
X is connected iff every continuous function $f:Xrightarrow ${$0,1$}
is constant.
Fix $f:Xrightarrow ${$0,1$} a continuous function.
Fix $(a_{alpha})in X$ and for each $alphain J,$define $varphi_{alpha}:X_{alpha}rightarrow X$ by $varphi_{alpha}opi_{beta}(x) =
begin{cases}
a_{beta}, & text{if $betaneq alpha$} \[2ex]
x, & text{if $beta=alpha$ }
end{cases}$.
Let $g_{alpha}=fovarphi_{alpha}:X_{alpha}rightarrow${$0,1$} for each $alphain J.$Then each $g_{alpha}$ is constant since $X$ is connected.
Hence,for each $alphain J, f$ is constant on $varphi [X_{alpha}].$
Say,Without loss of generality,for some $alpha_0in J,f[varphi_{alpha_0[X_{alpha}]}]=${$0$}.Since,$(a_{alpha})in varphi_{beta}[X_{beta}]$ for each $beta in J,$ it must be that $forall alphain J,f[varphi_{alpha_0[X_{alpha}]}]=${$0$}.
Then $f^{-1}$[{$0$}] is an open subset of $X$ by continuity which contains all of these elements.$tag{*}$
1.(Please clarify how $f^{-1}$[{$0$}] is open subset of $X$ :Is it because {0} is open in {0,1},$f$ is continuous,so inverse image of open set is open? )
Let $Usubset f^{-1}$[{$0$}] be a basis element of $X$ containing $(a_{alpha})$ so that $pi_{alpha}[U]=X_{alpha}$ for all but finitely many $alpha in J$.Let K be the finite subset of $J$ comprised of these $alpha $ and let $(x_{alpha})in U$ be that element such that $x_{alpha}=a_{alpha} $ if $alpha in K.$
Changing the coordinates of $(x_{alpha})$ indexed by $K$ one at a time,it follows from ($*$) that each time we change it,it remains in $f^{-1}$[{$0$}].Since $pi_{alpha}[U]=X_{alpha}$ for all but finitely many $alpha in J$.
We have shown that $prod_{alpha in J} X_alpha=f^{-1}$[{$0$}].(##2.I'm not getting this sudden transition.Please explain!!)
If there is some scope of refinement in the above proof,feel free to express your views.Also check this proof critically...
real-analysis general-topology proof-verification connectedness open-map
real-analysis general-topology proof-verification connectedness open-map
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,425726
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,425726
edited Dec 18 '18 at 17:22
P.Styles
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,425726
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,425726
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,425726
1,425726
1
In response to 1: yes
– mathworker21
Dec 12 '18 at 8:59
The definition of $varphi$ is missing a subscript.
– William Elliot
Dec 13 '18 at 2:38
@William Elliot:see the edit
– P.Styles
Dec 13 '18 at 3:31
$pi and varphi$ appear to be in reverse order.
– William Elliot
Dec 13 '18 at 7:49
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
– William Elliot
Dec 13 '18 at 7:51
add a comment |
1
In response to 1: yes
– mathworker21
Dec 12 '18 at 8:59
The definition of $varphi$ is missing a subscript.
– William Elliot
Dec 13 '18 at 2:38
@William Elliot:see the edit
– P.Styles
Dec 13 '18 at 3:31
$pi and varphi$ appear to be in reverse order.
– William Elliot
Dec 13 '18 at 7:49
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
– William Elliot
Dec 13 '18 at 7:51
1
1
In response to 1: yes
– mathworker21
Dec 12 '18 at 8:59
In response to 1: yes
– mathworker21
Dec 12 '18 at 8:59
The definition of $varphi$ is missing a subscript.
– William Elliot
Dec 13 '18 at 2:38
The definition of $varphi$ is missing a subscript.
– William Elliot
Dec 13 '18 at 2:38
@William Elliot:see the edit
– P.Styles
Dec 13 '18 at 3:31
@William Elliot:see the edit
– P.Styles
Dec 13 '18 at 3:31
$pi and varphi$ appear to be in reverse order.
– William Elliot
Dec 13 '18 at 7:49
$pi and varphi$ appear to be in reverse order.
– William Elliot
Dec 13 '18 at 7:49
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
– William Elliot
Dec 13 '18 at 7:51
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
– William Elliot
Dec 13 '18 at 7:51
add a comment |
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1
In response to 1: yes
– mathworker21
Dec 12 '18 at 8:59
The definition of $varphi$ is missing a subscript.
– William Elliot
Dec 13 '18 at 2:38
@William Elliot:see the edit
– P.Styles
Dec 13 '18 at 3:31
$pi and varphi$ appear to be in reverse order.
– William Elliot
Dec 13 '18 at 7:49
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
– William Elliot
Dec 13 '18 at 7:51