Mixing
Relation between components and path components of a topological space X.
$begingroup$
Theorem: If $X$ is a topological space, each path component of $X$ lies in a component of $X$. If $X$ is locally path connected, then the components and the path components of $X$ are the same.
Proof Attempt:
Let $X$ be a topological space and $P$ be a path component of $X$. Note $forall x_1,x_2 in P$, there exist a continuous function $f:[a,b]subsetBbb{R}to X$ such that $f(a)=x_1$ and $f(b)=x_2$. Since $[a,b]$ is connected, then $f([a,b])$ is also a connected subspace of $X$ and $x_1,x_2 in f([a,b])$. Thus, $Psubseteq f([a,b])$. Since connnected subspaces lie entirely within a component, let $f([a,b])subset C$ where $C$ is some component of $X$. Hence, $Psubset C$.
Moreover, what if we also assume that $X$ is locally path-connected. So far we have $x_1,x_2 in P$ and $f(a)=x_1, f(b)=x_2 in f([a,b])$. Let $x'in f([a,b])$. We want to show $f([a,b])subset P$. Then, by local path-connectedness of $X$, for every open set $O'$ in $X$ such that $x' in O'$ imply there exists a path-connected neighborhood $mathscr{P'}$ such that $x'in mathscr{P'}subset O'$. Since $x'in f([a,b])$, then $exists cin [a,b]subsetBbb{R}(f(c)=x')$. Since $f$ is continuous, then $g=f|_{[a,c]}$ and $h=f|_{[c,b]}$ are continuous and by definition of a path, $g(a)=x_1$, $g(c)=x'$, $h(c)=x'$, and $h(b)=x_2$ imply there exist paths between $x_1$ and $x'$ and also $x'$ and $x_2$; denoted, $x_1sim x'$ and $x'sim x_2$ where $sim$ relates two point linked by a path. Since $forallalphainmathscr{P'}(alphasim x')$, then, by transitivity of $sim$, $alphasim x_1$ and $alphasim x_2$. By the symmetry of $sim$, $mathscr{P'}subset P$, $Psubsetmathscr{P'}$, $f([a,b])subsetmathscr{P'}$, and $mathscr{P'}subset f([a,b])$. Therefore, $f([a,b])=P$.
general-topology proof-verification path-connected
asked Jan 7 at 16:13
TheLast CipherTheLast Cipher
691715
$endgroup$
add a comment |
$begingroup$
Theorem: If $X$ is a topological space, each path component of $X$ lies in a component of $X$. If $X$ is locally path connected, then the components and the path components of $X$ are the same.
Proof Attempt:
Let $X$ be a topological space and $P$ be a path component of $X$. Note $forall x_1,x_2 in P$, there exist a continuous function $f:[a,b]subsetBbb{R}to X$ such that $f(a)=x_1$ and $f(b)=x_2$. Since $[a,b]$ is connected, then $f([a,b])$ is also a connected subspace of $X$ and $x_1,x_2 in f([a,b])$. Thus, $Psubseteq f([a,b])$. Since connnected subspaces lie entirely within a component, let $f([a,b])subset C$ where $C$ is some component of $X$. Hence, $Psubset C$.
Moreover, what if we also assume that $X$ is locally path-connected. So far we have $x_1,x_2 in P$ and $f(a)=x_1, f(b)=x_2 in f([a,b])$. Let $x'in f([a,b])$. We want to show $f([a,b])subset P$. Then, by local path-connectedness of $X$, for every open set $O'$ in $X$ such that $x' in O'$ imply there exists a path-connected neighborhood $mathscr{P'}$ such that $x'in mathscr{P'}subset O'$. Since $x'in f([a,b])$, then $exists cin [a,b]subsetBbb{R}(f(c)=x')$. Since $f$ is continuous, then $g=f|_{[a,c]}$ and $h=f|_{[c,b]}$ are continuous and by definition of a path, $g(a)=x_1$, $g(c)=x'$, $h(c)=x'$, and $h(b)=x_2$ imply there exist paths between $x_1$ and $x'$ and also $x'$ and $x_2$; denoted, $x_1sim x'$ and $x'sim x_2$ where $sim$ relates two point linked by a path. Since $forallalphainmathscr{P'}(alphasim x')$, then, by transitivity of $sim$, $alphasim x_1$ and $alphasim x_2$. By the symmetry of $sim$, $mathscr{P'}subset P$, $Psubsetmathscr{P'}$, $f([a,b])subsetmathscr{P'}$, and $mathscr{P'}subset f([a,b])$. Therefore, $f([a,b])=P$.
general-topology proof-verification path-connected
asked Jan 7 at 16:13
TheLast CipherTheLast Cipher
691715
$endgroup$
add a comment |
$begingroup$
Theorem: If $X$ is a topological space, each path component of $X$ lies in a component of $X$. If $X$ is locally path connected, then the components and the path components of $X$ are the same.
Proof Attempt:
Let $X$ be a topological space and $P$ be a path component of $X$. Note $forall x_1,x_2 in P$, there exist a continuous function $f:[a,b]subsetBbb{R}to X$ such that $f(a)=x_1$ and $f(b)=x_2$. Since $[a,b]$ is connected, then $f([a,b])$ is also a connected subspace of $X$ and $x_1,x_2 in f([a,b])$. Thus, $Psubseteq f([a,b])$. Since connnected subspaces lie entirely within a component, let $f([a,b])subset C$ where $C$ is some component of $X$. Hence, $Psubset C$.
Moreover, what if we also assume that $X$ is locally path-connected. So far we have $x_1,x_2 in P$ and $f(a)=x_1, f(b)=x_2 in f([a,b])$. Let $x'in f([a,b])$. We want to show $f([a,b])subset P$. Then, by local path-connectedness of $X$, for every open set $O'$ in $X$ such that $x' in O'$ imply there exists a path-connected neighborhood $mathscr{P'}$ such that $x'in mathscr{P'}subset O'$. Since $x'in f([a,b])$, then $exists cin [a,b]subsetBbb{R}(f(c)=x')$. Since $f$ is continuous, then $g=f|_{[a,c]}$ and $h=f|_{[c,b]}$ are continuous and by definition of a path, $g(a)=x_1$, $g(c)=x'$, $h(c)=x'$, and $h(b)=x_2$ imply there exist paths between $x_1$ and $x'$ and also $x'$ and $x_2$; denoted, $x_1sim x'$ and $x'sim x_2$ where $sim$ relates two point linked by a path. Since $forallalphainmathscr{P'}(alphasim x')$, then, by transitivity of $sim$, $alphasim x_1$ and $alphasim x_2$. By the symmetry of $sim$, $mathscr{P'}subset P$, $Psubsetmathscr{P'}$, $f([a,b])subsetmathscr{P'}$, and $mathscr{P'}subset f([a,b])$. Therefore, $f([a,b])=P$.
general-topology proof-verification path-connected
asked Jan 7 at 16:13
TheLast CipherTheLast Cipher
691715
$endgroup$
Theorem: If $X$ is a topological space, each path component of $X$ lies in a component of $X$. If $X$ is locally path connected, then the components and the path components of $X$ are the same.
Proof Attempt:
Let $X$ be a topological space and $P$ be a path component of $X$. Note $forall x_1,x_2 in P$, there exist a continuous function $f:[a,b]subsetBbb{R}to X$ such that $f(a)=x_1$ and $f(b)=x_2$. Since $[a,b]$ is connected, then $f([a,b])$ is also a connected subspace of $X$ and $x_1,x_2 in f([a,b])$. Thus, $Psubseteq f([a,b])$. Since connnected subspaces lie entirely within a component, let $f([a,b])subset C$ where $C$ is some component of $X$. Hence, $Psubset C$.
Moreover, what if we also assume that $X$ is locally path-connected. So far we have $x_1,x_2 in P$ and $f(a)=x_1, f(b)=x_2 in f([a,b])$. Let $x'in f([a,b])$. We want to show $f([a,b])subset P$. Then, by local path-connectedness of $X$, for every open set $O'$ in $X$ such that $x' in O'$ imply there exists a path-connected neighborhood $mathscr{P'}$ such that $x'in mathscr{P'}subset O'$. Since $x'in f([a,b])$, then $exists cin [a,b]subsetBbb{R}(f(c)=x')$. Since $f$ is continuous, then $g=f|_{[a,c]}$ and $h=f|_{[c,b]}$ are continuous and by definition of a path, $g(a)=x_1$, $g(c)=x'$, $h(c)=x'$, and $h(b)=x_2$ imply there exist paths between $x_1$ and $x'$ and also $x'$ and $x_2$; denoted, $x_1sim x'$ and $x'sim x_2$ where $sim$ relates two point linked by a path. Since $forallalphainmathscr{P'}(alphasim x')$, then, by transitivity of $sim$, $alphasim x_1$ and $alphasim x_2$. By the symmetry of $sim$, $mathscr{P'}subset P$, $Psubsetmathscr{P'}$, $f([a,b])subsetmathscr{P'}$, and $mathscr{P'}subset f([a,b])$. Therefore, $f([a,b])=P$.
general-topology proof-verification path-connected
general-topology proof-verification path-connected
asked Jan 7 at 16:13
TheLast CipherTheLast Cipher
691715
asked Jan 7 at 16:13
TheLast CipherTheLast Cipher
691715
asked Jan 7 at 16:13
TheLast CipherTheLast Cipher
691715
asked Jan 7 at 16:13
TheLast CipherTheLast Cipher
691715
asked Jan 7 at 16:13
TheLast CipherTheLast Cipher
691715
691715
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Not acceptable.
Since X is locally connected, any component K of X is open.
Let a,b be in connected K. By theorem, there
is an overlapping chain of open base sets from a to b,
some open base sets U$_1$,... U$_n$ with a in U$_1,$ b in U$_n$
and each successive U's intersecting.
Thus, as each U is path connected, a path can be constructed from a to a point in $U_1 cap U_2$ and so on, step by step, to b.
Thus K is path connected.
answered Jan 8 at 3:56
William ElliotWilliam Elliot
7,8502720
$endgroup$
$begingroup$
Okay thanks I understand your argument. You are showing that components of a locally connected space $X$ is path connected which means the components of $X$ are the path components? Also, can you please help me find where I went wrong in the proof? I think I showed that connected spaces in $X$ are path components. Help me.
$endgroup$
– TheLast Cipher
Jan 8 at 8:17
$begingroup$
In the second part f pops up without explanation. Perhaps it is path from x1 to x2, in which case you have assumed the conclusion. Assuming f is continuous, f([a,b]) subset P is trivial as it is connected and contains a point of P. @TheLastCipher
$endgroup$
– William Elliot
Jan 8 at 21:32
add a comment |
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%2f3065173%2frelation-between-components-and-path-components-of-a-topological-space-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Not acceptable.
Since X is locally connected, any component K of X is open.
Let a,b be in connected K. By theorem, there
is an overlapping chain of open base sets from a to b,
some open base sets U$_1$,... U$_n$ with a in U$_1,$ b in U$_n$
and each successive U's intersecting.
Thus, as each U is path connected, a path can be constructed from a to a point in $U_1 cap U_2$ and so on, step by step, to b.
Thus K is path connected.
answered Jan 8 at 3:56
William ElliotWilliam Elliot
7,8502720
$endgroup$
$begingroup$
Okay thanks I understand your argument. You are showing that components of a locally connected space $X$ is path connected which means the components of $X$ are the path components? Also, can you please help me find where I went wrong in the proof? I think I showed that connected spaces in $X$ are path components. Help me.
$endgroup$
– TheLast Cipher
Jan 8 at 8:17
$begingroup$
In the second part f pops up without explanation. Perhaps it is path from x1 to x2, in which case you have assumed the conclusion. Assuming f is continuous, f([a,b]) subset P is trivial as it is connected and contains a point of P. @TheLastCipher
$endgroup$
– William Elliot
Jan 8 at 21:32
add a comment |
$begingroup$
Not acceptable.
Since X is locally connected, any component K of X is open.
Let a,b be in connected K. By theorem, there
is an overlapping chain of open base sets from a to b,
some open base sets U$_1$,... U$_n$ with a in U$_1,$ b in U$_n$
and each successive U's intersecting.
Thus, as each U is path connected, a path can be constructed from a to a point in $U_1 cap U_2$ and so on, step by step, to b.
Thus K is path connected.
answered Jan 8 at 3:56
William ElliotWilliam Elliot
7,8502720
$endgroup$
$begingroup$
Okay thanks I understand your argument. You are showing that components of a locally connected space $X$ is path connected which means the components of $X$ are the path components? Also, can you please help me find where I went wrong in the proof? I think I showed that connected spaces in $X$ are path components. Help me.
$endgroup$
– TheLast Cipher
Jan 8 at 8:17
$begingroup$
In the second part f pops up without explanation. Perhaps it is path from x1 to x2, in which case you have assumed the conclusion. Assuming f is continuous, f([a,b]) subset P is trivial as it is connected and contains a point of P. @TheLastCipher
$endgroup$
– William Elliot
Jan 8 at 21:32
add a comment |
$begingroup$
Not acceptable.
Since X is locally connected, any component K of X is open.
Let a,b be in connected K. By theorem, there
is an overlapping chain of open base sets from a to b,
some open base sets U$_1$,... U$_n$ with a in U$_1,$ b in U$_n$
and each successive U's intersecting.
Thus, as each U is path connected, a path can be constructed from a to a point in $U_1 cap U_2$ and so on, step by step, to b.
Thus K is path connected.
answered Jan 8 at 3:56
William ElliotWilliam Elliot
7,8502720
$endgroup$
Not acceptable.
Since X is locally connected, any component K of X is open.
Let a,b be in connected K. By theorem, there
is an overlapping chain of open base sets from a to b,
some open base sets U$_1$,... U$_n$ with a in U$_1,$ b in U$_n$
and each successive U's intersecting.
Thus, as each U is path connected, a path can be constructed from a to a point in $U_1 cap U_2$ and so on, step by step, to b.
Thus K is path connected.
answered Jan 8 at 3:56
William ElliotWilliam Elliot
7,8502720
answered Jan 8 at 3:56
William ElliotWilliam Elliot
7,8502720
answered Jan 8 at 3:56
William ElliotWilliam Elliot
7,8502720
answered Jan 8 at 3:56
William ElliotWilliam Elliot
7,8502720
7,8502720
$begingroup$
Okay thanks I understand your argument. You are showing that components of a locally connected space $X$ is path connected which means the components of $X$ are the path components? Also, can you please help me find where I went wrong in the proof? I think I showed that connected spaces in $X$ are path components. Help me.
$endgroup$
– TheLast Cipher
Jan 8 at 8:17
$begingroup$
In the second part f pops up without explanation. Perhaps it is path from x1 to x2, in which case you have assumed the conclusion. Assuming f is continuous, f([a,b]) subset P is trivial as it is connected and contains a point of P. @TheLastCipher
$endgroup$
– William Elliot
Jan 8 at 21:32
add a comment |
$begingroup$
Okay thanks I understand your argument. You are showing that components of a locally connected space $X$ is path connected which means the components of $X$ are the path components? Also, can you please help me find where I went wrong in the proof? I think I showed that connected spaces in $X$ are path components. Help me.
$endgroup$
– TheLast Cipher
Jan 8 at 8:17
$begingroup$
In the second part f pops up without explanation. Perhaps it is path from x1 to x2, in which case you have assumed the conclusion. Assuming f is continuous, f([a,b]) subset P is trivial as it is connected and contains a point of P. @TheLastCipher
$endgroup$
– William Elliot
Jan 8 at 21:32
$begingroup$
Okay thanks I understand your argument. You are showing that components of a locally connected space $X$ is path connected which means the components of $X$ are the path components? Also, can you please help me find where I went wrong in the proof? I think I showed that connected spaces in $X$ are path components. Help me.
$endgroup$
– TheLast Cipher
Jan 8 at 8:17
$begingroup$
Okay thanks I understand your argument. You are showing that components of a locally connected space $X$ is path connected which means the components of $X$ are the path components? Also, can you please help me find where I went wrong in the proof? I think I showed that connected spaces in $X$ are path components. Help me.
$endgroup$
– TheLast Cipher
Jan 8 at 8:17
$begingroup$
In the second part f pops up without explanation. Perhaps it is path from x1 to x2, in which case you have assumed the conclusion. Assuming f is continuous, f([a,b]) subset P is trivial as it is connected and contains a point of P. @TheLastCipher
$endgroup$
– William Elliot
Jan 8 at 21:32
$begingroup$
In the second part f pops up without explanation. Perhaps it is path from x1 to x2, in which case you have assumed the conclusion. Assuming f is continuous, f([a,b]) subset P is trivial as it is connected and contains a point of P. @TheLastCipher
$endgroup$
– William Elliot
Jan 8 at 21:32
add a comment |
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%2f3065173%2frelation-between-components-and-path-components-of-a-topological-space-x%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
