Mixing
Algebra Chapter 0, exercise 1.3
$begingroup$
My question concerns the answer to exercise 1.3:
Given a partition $P$ on a set $S$, show how to define a relation $sim$ on $S$ such that $P$ is the corresponding partition.
My answer is:
We can define the relation $sim$ such that $P$ is the corresponding partition as let $X in P$, if $a in X$ and $b in X$ then $a sim b$.
Is that enough?
I found this online, but they show $P = P_sim$. Is that necessary?
Define, for $a,bin S$, $asim b$ if and only if there exists an
$Xinmathscr{P}$ such that $ain X$ and $bin X$. We will show that
$mathscr{P} = mathscr{P}_{sim}$.
($mathscr{P}subseteqmathscr{P}_{sim}$). Let $Xin mathscr{P}$; we want to show that $Xinmathscr{P}_{sim}$. We know that $X$ is
nonempty, so choose $ain X$ and consider
$[a]_{sim}inmathscr{P}_{sim}$. We need to show that
$X=[a]_{sim}$. Suppose $a'in X$ (it may be that $a'=a$.) Since
$a,a'in X$, $asim a'$, so $a'in[a]_{sim}$. Now, suppose $a'in
> [a]_{sim}$. We have $a'sim a$, so $a'in X$. Hence $X=[a]_{sim}in
> mathscr{P}_{sim}$, so $mathscr{P}subseteqmathscr{P}_{sim}$.
($mathscr{P}_{sim}subseteqmathscr{P}$). Let $[a]_{sim}inmathscr{P}_{sim}$. From exercise I.1.1 we know that
$[a]_{sim}$ is non-empty. Suppose $a'in[a]_{sim}$. By definition,
since $a'sim a$, there exists a set $X$ such that $a,a'in X$. Hence
$[a]_{sim}subseteq X$. Also, if $a,a'in X$ (not necessarily
distinct) then $asim a'$. Therefore,
$mathscr{P}_{sim}subseteqmathscr{P}$, and with 1. we get that the
sets $mathscr{P}$ and $mathscr{P}_{sim}$ are equal.
proof-verification relations equivalence-relations
edited Jan 27 at 20:14
Bernard
123k741117
asked Jan 27 at 20:08
avidavid
1082
$endgroup$
add a comment |
$begingroup$
My question concerns the answer to exercise 1.3:
Given a partition $P$ on a set $S$, show how to define a relation $sim$ on $S$ such that $P$ is the corresponding partition.
My answer is:
We can define the relation $sim$ such that $P$ is the corresponding partition as let $X in P$, if $a in X$ and $b in X$ then $a sim b$.
Is that enough?
I found this online, but they show $P = P_sim$. Is that necessary?
Define, for $a,bin S$, $asim b$ if and only if there exists an
$Xinmathscr{P}$ such that $ain X$ and $bin X$. We will show that
$mathscr{P} = mathscr{P}_{sim}$.
($mathscr{P}subseteqmathscr{P}_{sim}$). Let $Xin mathscr{P}$; we want to show that $Xinmathscr{P}_{sim}$. We know that $X$ is
nonempty, so choose $ain X$ and consider
$[a]_{sim}inmathscr{P}_{sim}$. We need to show that
$X=[a]_{sim}$. Suppose $a'in X$ (it may be that $a'=a$.) Since
$a,a'in X$, $asim a'$, so $a'in[a]_{sim}$. Now, suppose $a'in
> [a]_{sim}$. We have $a'sim a$, so $a'in X$. Hence $X=[a]_{sim}in
> mathscr{P}_{sim}$, so $mathscr{P}subseteqmathscr{P}_{sim}$.
($mathscr{P}_{sim}subseteqmathscr{P}$). Let $[a]_{sim}inmathscr{P}_{sim}$. From exercise I.1.1 we know that
$[a]_{sim}$ is non-empty. Suppose $a'in[a]_{sim}$. By definition,
since $a'sim a$, there exists a set $X$ such that $a,a'in X$. Hence
$[a]_{sim}subseteq X$. Also, if $a,a'in X$ (not necessarily
distinct) then $asim a'$. Therefore,
$mathscr{P}_{sim}subseteqmathscr{P}$, and with 1. we get that the
sets $mathscr{P}$ and $mathscr{P}_{sim}$ are equal.
proof-verification relations equivalence-relations
edited Jan 27 at 20:14
Bernard
123k741117
asked Jan 27 at 20:08
avidavid
1082
$endgroup$
$begingroup$
How does the text you're using define "corresponding partition" of a given relation?
$endgroup$
– coffeemath
Jan 27 at 20:13
add a comment |
$begingroup$
My question concerns the answer to exercise 1.3:
Given a partition $P$ on a set $S$, show how to define a relation $sim$ on $S$ such that $P$ is the corresponding partition.
My answer is:
We can define the relation $sim$ such that $P$ is the corresponding partition as let $X in P$, if $a in X$ and $b in X$ then $a sim b$.
Is that enough?
I found this online, but they show $P = P_sim$. Is that necessary?
Define, for $a,bin S$, $asim b$ if and only if there exists an
$Xinmathscr{P}$ such that $ain X$ and $bin X$. We will show that
$mathscr{P} = mathscr{P}_{sim}$.
($mathscr{P}subseteqmathscr{P}_{sim}$). Let $Xin mathscr{P}$; we want to show that $Xinmathscr{P}_{sim}$. We know that $X$ is
nonempty, so choose $ain X$ and consider
$[a]_{sim}inmathscr{P}_{sim}$. We need to show that
$X=[a]_{sim}$. Suppose $a'in X$ (it may be that $a'=a$.) Since
$a,a'in X$, $asim a'$, so $a'in[a]_{sim}$. Now, suppose $a'in
> [a]_{sim}$. We have $a'sim a$, so $a'in X$. Hence $X=[a]_{sim}in
> mathscr{P}_{sim}$, so $mathscr{P}subseteqmathscr{P}_{sim}$.
($mathscr{P}_{sim}subseteqmathscr{P}$). Let $[a]_{sim}inmathscr{P}_{sim}$. From exercise I.1.1 we know that
$[a]_{sim}$ is non-empty. Suppose $a'in[a]_{sim}$. By definition,
since $a'sim a$, there exists a set $X$ such that $a,a'in X$. Hence
$[a]_{sim}subseteq X$. Also, if $a,a'in X$ (not necessarily
distinct) then $asim a'$. Therefore,
$mathscr{P}_{sim}subseteqmathscr{P}$, and with 1. we get that the
sets $mathscr{P}$ and $mathscr{P}_{sim}$ are equal.
proof-verification relations equivalence-relations
edited Jan 27 at 20:14
Bernard
123k741117
asked Jan 27 at 20:08
avidavid
1082
$endgroup$
My question concerns the answer to exercise 1.3:
Given a partition $P$ on a set $S$, show how to define a relation $sim$ on $S$ such that $P$ is the corresponding partition.
My answer is:
We can define the relation $sim$ such that $P$ is the corresponding partition as let $X in P$, if $a in X$ and $b in X$ then $a sim b$.
Is that enough?
I found this online, but they show $P = P_sim$. Is that necessary?
Define, for $a,bin S$, $asim b$ if and only if there exists an
$Xinmathscr{P}$ such that $ain X$ and $bin X$. We will show that
$mathscr{P} = mathscr{P}_{sim}$.
($mathscr{P}subseteqmathscr{P}_{sim}$). Let $Xin mathscr{P}$; we want to show that $Xinmathscr{P}_{sim}$. We know that $X$ is
nonempty, so choose $ain X$ and consider
$[a]_{sim}inmathscr{P}_{sim}$. We need to show that
$X=[a]_{sim}$. Suppose $a'in X$ (it may be that $a'=a$.) Since
$a,a'in X$, $asim a'$, so $a'in[a]_{sim}$. Now, suppose $a'in
> [a]_{sim}$. We have $a'sim a$, so $a'in X$. Hence $X=[a]_{sim}in
> mathscr{P}_{sim}$, so $mathscr{P}subseteqmathscr{P}_{sim}$.
($mathscr{P}_{sim}subseteqmathscr{P}$). Let $[a]_{sim}inmathscr{P}_{sim}$. From exercise I.1.1 we know that
$[a]_{sim}$ is non-empty. Suppose $a'in[a]_{sim}$. By definition,
since $a'sim a$, there exists a set $X$ such that $a,a'in X$. Hence
$[a]_{sim}subseteq X$. Also, if $a,a'in X$ (not necessarily
distinct) then $asim a'$. Therefore,
$mathscr{P}_{sim}subseteqmathscr{P}$, and with 1. we get that the
sets $mathscr{P}$ and $mathscr{P}_{sim}$ are equal.
proof-verification relations equivalence-relations
proof-verification relations equivalence-relations
edited Jan 27 at 20:14
Bernard
123k741117
asked Jan 27 at 20:08
avidavid
1082
edited Jan 27 at 20:14
Bernard
123k741117
asked Jan 27 at 20:08
avidavid
1082
edited Jan 27 at 20:14
Bernard
123k741117
edited Jan 27 at 20:14
Bernard
123k741117
edited Jan 27 at 20:14
Bernard
123k741117
123k741117
asked Jan 27 at 20:08
avidavid
1082
asked Jan 27 at 20:08
avidavid
1082
asked Jan 27 at 20:08
avidavid
1082
1082
$begingroup$
How does the text you're using define "corresponding partition" of a given relation?
$endgroup$
– coffeemath
Jan 27 at 20:13
add a comment |
$begingroup$
How does the text you're using define "corresponding partition" of a given relation?
$endgroup$
– coffeemath
Jan 27 at 20:13
$begingroup$
How does the text you're using define "corresponding partition" of a given relation?
$endgroup$
– coffeemath
Jan 27 at 20:13
$begingroup$
How does the text you're using define "corresponding partition" of a given relation?
$endgroup$
– coffeemath
Jan 27 at 20:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
What you found online is just saying an easy thing in an overly complicated manner.
Given a partition $P$, the relation you are looking for is just: $a$ is equivalent to $b$ iff there exists some set in the partition such that both $a$ and $b$ belong to it.
Now the proof that this is actually the relation you are looking for goes this way:
Given an element $a$, it must belong to some set in the partition, call it $X_a$. Then, if $b$ is equivalent to $a$, it must belong to $X_a$. Conversely, every element in $X_a$ is equivalent to $a$. To conclude, you can observe that every set in the partition is non empty, by def of partition. So you have that each set in the partition is exactly the equivalence class of some element, and that each equivalence class belongs to the partition
answered Jan 27 at 20:47
GLeGLe
455
$endgroup$
add a comment |
$begingroup$
Your answer is correct, it's the correct partition, but you do need to prove that the partition arising from the equivalence relation is exactly equal to the given partition.
It may seem obvious, but the point of the exercise is to familiarize yourself with the partition-relation correspondence. So you need to make sure they have exactly the same equivalence classes, (i.e. there's an $X$ for every $[a]_sim$ and an $[a]_sim$ for every $X$)
answered Jan 27 at 20:23
Larry B.Larry B.
2,821828
$endgroup$
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%2f3090052%2falgebra-chapter-0-exercise-1-3%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What you found online is just saying an easy thing in an overly complicated manner.
Given a partition $P$, the relation you are looking for is just: $a$ is equivalent to $b$ iff there exists some set in the partition such that both $a$ and $b$ belong to it.
Now the proof that this is actually the relation you are looking for goes this way:
Given an element $a$, it must belong to some set in the partition, call it $X_a$. Then, if $b$ is equivalent to $a$, it must belong to $X_a$. Conversely, every element in $X_a$ is equivalent to $a$. To conclude, you can observe that every set in the partition is non empty, by def of partition. So you have that each set in the partition is exactly the equivalence class of some element, and that each equivalence class belongs to the partition
answered Jan 27 at 20:47
GLeGLe
455
$endgroup$
add a comment |
$begingroup$
What you found online is just saying an easy thing in an overly complicated manner.
Given a partition $P$, the relation you are looking for is just: $a$ is equivalent to $b$ iff there exists some set in the partition such that both $a$ and $b$ belong to it.
Now the proof that this is actually the relation you are looking for goes this way:
Given an element $a$, it must belong to some set in the partition, call it $X_a$. Then, if $b$ is equivalent to $a$, it must belong to $X_a$. Conversely, every element in $X_a$ is equivalent to $a$. To conclude, you can observe that every set in the partition is non empty, by def of partition. So you have that each set in the partition is exactly the equivalence class of some element, and that each equivalence class belongs to the partition
answered Jan 27 at 20:47
GLeGLe
455
$endgroup$
add a comment |
$begingroup$
What you found online is just saying an easy thing in an overly complicated manner.
Given a partition $P$, the relation you are looking for is just: $a$ is equivalent to $b$ iff there exists some set in the partition such that both $a$ and $b$ belong to it.
Now the proof that this is actually the relation you are looking for goes this way:
Given an element $a$, it must belong to some set in the partition, call it $X_a$. Then, if $b$ is equivalent to $a$, it must belong to $X_a$. Conversely, every element in $X_a$ is equivalent to $a$. To conclude, you can observe that every set in the partition is non empty, by def of partition. So you have that each set in the partition is exactly the equivalence class of some element, and that each equivalence class belongs to the partition
answered Jan 27 at 20:47
GLeGLe
455
$endgroup$
What you found online is just saying an easy thing in an overly complicated manner.
Given a partition $P$, the relation you are looking for is just: $a$ is equivalent to $b$ iff there exists some set in the partition such that both $a$ and $b$ belong to it.
Now the proof that this is actually the relation you are looking for goes this way:
Given an element $a$, it must belong to some set in the partition, call it $X_a$. Then, if $b$ is equivalent to $a$, it must belong to $X_a$. Conversely, every element in $X_a$ is equivalent to $a$. To conclude, you can observe that every set in the partition is non empty, by def of partition. So you have that each set in the partition is exactly the equivalence class of some element, and that each equivalence class belongs to the partition
answered Jan 27 at 20:47
GLeGLe
455
answered Jan 27 at 20:47
GLeGLe
455
answered Jan 27 at 20:47
GLeGLe
455
answered Jan 27 at 20:47
GLeGLe
455
455
add a comment |
add a comment |
$begingroup$
Your answer is correct, it's the correct partition, but you do need to prove that the partition arising from the equivalence relation is exactly equal to the given partition.
It may seem obvious, but the point of the exercise is to familiarize yourself with the partition-relation correspondence. So you need to make sure they have exactly the same equivalence classes, (i.e. there's an $X$ for every $[a]_sim$ and an $[a]_sim$ for every $X$)
answered Jan 27 at 20:23
Larry B.Larry B.
2,821828
$endgroup$
add a comment |
$begingroup$
Your answer is correct, it's the correct partition, but you do need to prove that the partition arising from the equivalence relation is exactly equal to the given partition.
It may seem obvious, but the point of the exercise is to familiarize yourself with the partition-relation correspondence. So you need to make sure they have exactly the same equivalence classes, (i.e. there's an $X$ for every $[a]_sim$ and an $[a]_sim$ for every $X$)
answered Jan 27 at 20:23
Larry B.Larry B.
2,821828
$endgroup$
add a comment |
$begingroup$
Your answer is correct, it's the correct partition, but you do need to prove that the partition arising from the equivalence relation is exactly equal to the given partition.
It may seem obvious, but the point of the exercise is to familiarize yourself with the partition-relation correspondence. So you need to make sure they have exactly the same equivalence classes, (i.e. there's an $X$ for every $[a]_sim$ and an $[a]_sim$ for every $X$)
answered Jan 27 at 20:23
Larry B.Larry B.
2,821828
$endgroup$
Your answer is correct, it's the correct partition, but you do need to prove that the partition arising from the equivalence relation is exactly equal to the given partition.
It may seem obvious, but the point of the exercise is to familiarize yourself with the partition-relation correspondence. So you need to make sure they have exactly the same equivalence classes, (i.e. there's an $X$ for every $[a]_sim$ and an $[a]_sim$ for every $X$)
answered Jan 27 at 20:23
Larry B.Larry B.
2,821828
answered Jan 27 at 20:23
Larry B.Larry B.
2,821828
answered Jan 27 at 20:23
Larry B.Larry B.
2,821828
answered Jan 27 at 20:23
Larry B.Larry B.
2,821828
2,821828
add a comment |
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%2f3090052%2falgebra-chapter-0-exercise-1-3%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
$begingroup$
How does the text you're using define "corresponding partition" of a given relation?
$endgroup$
– coffeemath
Jan 27 at 20:13