Queris related to “An arbitrary product $X=prod_{alpha in J} X_alpha$ is connected iff $X_{alpha }$ is...
$begingroup$
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
1,497927
$endgroup$
add a comment |
$begingroup$
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
1,497927
$endgroup$
1
$begingroup$
In response to 1: yes
$endgroup$
– mathworker21
Dec 12 '18 at 8:59
$begingroup$
The definition of $varphi$ is missing a subscript.
$endgroup$
– William Elliot
Dec 13 '18 at 2:38
$begingroup$
@William Elliot:see the edit
$endgroup$
– P.Styles
Dec 13 '18 at 3:31
$begingroup$
$pi and varphi$ appear to be in reverse order.
$endgroup$
– William Elliot
Dec 13 '18 at 7:49
$begingroup$
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
$endgroup$
– William Elliot
Dec 13 '18 at 7:51
add a comment |
$begingroup$
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
1,497927
$endgroup$
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
1,497927
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,497927
edited Dec 18 '18 at 17:22
P.Styles
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,497927
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,497927
asked Dec 12 '18 at 8:31
P.StylesP.Styles
1,497927
1,497927
1
$begingroup$
In response to 1: yes
$endgroup$
– mathworker21
Dec 12 '18 at 8:59
$begingroup$
The definition of $varphi$ is missing a subscript.
$endgroup$
– William Elliot
Dec 13 '18 at 2:38
$begingroup$
@William Elliot:see the edit
$endgroup$
– P.Styles
Dec 13 '18 at 3:31
$begingroup$
$pi and varphi$ appear to be in reverse order.
$endgroup$
– William Elliot
Dec 13 '18 at 7:49
$begingroup$
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
$endgroup$
– William Elliot
Dec 13 '18 at 7:51
add a comment |
1
$begingroup$
In response to 1: yes
$endgroup$
– mathworker21
Dec 12 '18 at 8:59
$begingroup$
The definition of $varphi$ is missing a subscript.
$endgroup$
– William Elliot
Dec 13 '18 at 2:38
$begingroup$
@William Elliot:see the edit
$endgroup$
– P.Styles
Dec 13 '18 at 3:31
$begingroup$
$pi and varphi$ appear to be in reverse order.
$endgroup$
– William Elliot
Dec 13 '18 at 7:49
$begingroup$
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
$endgroup$
– William Elliot
Dec 13 '18 at 7:51
1
1
$begingroup$
In response to 1: yes
$endgroup$
– mathworker21
Dec 12 '18 at 8:59
$begingroup$
In response to 1: yes
$endgroup$
– mathworker21
Dec 12 '18 at 8:59
$begingroup$
The definition of $varphi$ is missing a subscript.
$endgroup$
– William Elliot
Dec 13 '18 at 2:38
$begingroup$
The definition of $varphi$ is missing a subscript.
$endgroup$
– William Elliot
Dec 13 '18 at 2:38
$begingroup$
@William Elliot:see the edit
$endgroup$
– P.Styles
Dec 13 '18 at 3:31
$begingroup$
@William Elliot:see the edit
$endgroup$
– P.Styles
Dec 13 '18 at 3:31
$begingroup$
$pi and varphi$ appear to be in reverse order.
$endgroup$
– William Elliot
Dec 13 '18 at 7:49
$begingroup$
$pi and varphi$ appear to be in reverse order.
$endgroup$
– William Elliot
Dec 13 '18 at 7:49
$begingroup$
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
$endgroup$
– William Elliot
Dec 13 '18 at 7:51
$begingroup$
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
$endgroup$
– William Elliot
Dec 13 '18 at 7:51
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036410%2fqueris-related-to-an-arbitrary-product-x-prod-alpha-in-j-x-alpha-is-con%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036410%2fqueris-related-to-an-arbitrary-product-x-prod-alpha-in-j-x-alpha-is-con%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
In response to 1: yes
$endgroup$
– mathworker21
Dec 12 '18 at 8:59
$begingroup$
The definition of $varphi$ is missing a subscript.
$endgroup$
– William Elliot
Dec 13 '18 at 2:38
$begingroup$
@William Elliot:see the edit
$endgroup$
– P.Styles
Dec 13 '18 at 3:31
$begingroup$
$pi and varphi$ appear to be in reverse order.
$endgroup$
– William Elliot
Dec 13 '18 at 7:49
$begingroup$
2. The leap of faith likely depends upon showing finite products of connected sets are connected.
$endgroup$
– William Elliot
Dec 13 '18 at 7:51