If ∀A ∈ PX , ∀B ∈ PY , |A| < |B| Then ∃A ∈ PX , ∃B ∈ PY , A ⊂ B
$begingroup$
X and Y are non-empty sets , X c Y (subset ) and Px is a partition of X and Py is a partition of Y .
How can I prove that this statement is true or false :
If ∀A ∈ PX, ∀B ∈ PY ,|A| < |B|
Then ∃A ∈ PX, ∃B ∈ PY , A ⊂ B
Thank you .
Ps: what I understood is that since XcY then A that belongs to X belongs to Y as well and since the cardinality of A is smaller than B then A is likely to be a subset of B . My guess is this statement is true but I have no idea how to prove it , which method to use ? I don’t know how to start my proof .
discrete-mathematics
$endgroup$
add a comment |
$begingroup$
X and Y are non-empty sets , X c Y (subset ) and Px is a partition of X and Py is a partition of Y .
How can I prove that this statement is true or false :
If ∀A ∈ PX, ∀B ∈ PY ,|A| < |B|
Then ∃A ∈ PX, ∃B ∈ PY , A ⊂ B
Thank you .
Ps: what I understood is that since XcY then A that belongs to X belongs to Y as well and since the cardinality of A is smaller than B then A is likely to be a subset of B . My guess is this statement is true but I have no idea how to prove it , which method to use ? I don’t know how to start my proof .
discrete-mathematics
$endgroup$
add a comment |
$begingroup$
X and Y are non-empty sets , X c Y (subset ) and Px is a partition of X and Py is a partition of Y .
How can I prove that this statement is true or false :
If ∀A ∈ PX, ∀B ∈ PY ,|A| < |B|
Then ∃A ∈ PX, ∃B ∈ PY , A ⊂ B
Thank you .
Ps: what I understood is that since XcY then A that belongs to X belongs to Y as well and since the cardinality of A is smaller than B then A is likely to be a subset of B . My guess is this statement is true but I have no idea how to prove it , which method to use ? I don’t know how to start my proof .
discrete-mathematics
$endgroup$
X and Y are non-empty sets , X c Y (subset ) and Px is a partition of X and Py is a partition of Y .
How can I prove that this statement is true or false :
If ∀A ∈ PX, ∀B ∈ PY ,|A| < |B|
Then ∃A ∈ PX, ∃B ∈ PY , A ⊂ B
Thank you .
Ps: what I understood is that since XcY then A that belongs to X belongs to Y as well and since the cardinality of A is smaller than B then A is likely to be a subset of B . My guess is this statement is true but I have no idea how to prove it , which method to use ? I don’t know how to start my proof .
discrete-mathematics
discrete-mathematics
edited Feb 1 at 21:40
soso xoxo
asked Feb 1 at 18:26


soso xoxososo xoxo
85
85
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
X = { 1,2,3,4,5,6 }.
Y = { 1,2,3,4,5,6,7 }.
PX = { {1,4}, {2,5}, {3,6} }.
PY = { {1,2,3}, {4,5,6,7} }.
What does that example show?
$endgroup$
$begingroup$
Thank you that helped me see that A is not necessary a subset of B
$endgroup$
– soso xoxo
Feb 2 at 2:52
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%2f3096568%2fif-%25e2%2588%2580a-%25e2%2588%2588-px-%25e2%2588%2580b-%25e2%2588%2588-py-a-b-then-%25e2%2588%2583a-%25e2%2588%2588-px-%25e2%2588%2583b-%25e2%2588%2588-py-a-%25e2%258a%2582-b%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
$begingroup$
X = { 1,2,3,4,5,6 }.
Y = { 1,2,3,4,5,6,7 }.
PX = { {1,4}, {2,5}, {3,6} }.
PY = { {1,2,3}, {4,5,6,7} }.
What does that example show?
$endgroup$
$begingroup$
Thank you that helped me see that A is not necessary a subset of B
$endgroup$
– soso xoxo
Feb 2 at 2:52
add a comment |
$begingroup$
X = { 1,2,3,4,5,6 }.
Y = { 1,2,3,4,5,6,7 }.
PX = { {1,4}, {2,5}, {3,6} }.
PY = { {1,2,3}, {4,5,6,7} }.
What does that example show?
$endgroup$
$begingroup$
Thank you that helped me see that A is not necessary a subset of B
$endgroup$
– soso xoxo
Feb 2 at 2:52
add a comment |
$begingroup$
X = { 1,2,3,4,5,6 }.
Y = { 1,2,3,4,5,6,7 }.
PX = { {1,4}, {2,5}, {3,6} }.
PY = { {1,2,3}, {4,5,6,7} }.
What does that example show?
$endgroup$
X = { 1,2,3,4,5,6 }.
Y = { 1,2,3,4,5,6,7 }.
PX = { {1,4}, {2,5}, {3,6} }.
PY = { {1,2,3}, {4,5,6,7} }.
What does that example show?
edited Feb 2 at 1:27
answered Feb 2 at 1:15
William ElliotWilliam Elliot
9,1562820
9,1562820
$begingroup$
Thank you that helped me see that A is not necessary a subset of B
$endgroup$
– soso xoxo
Feb 2 at 2:52
add a comment |
$begingroup$
Thank you that helped me see that A is not necessary a subset of B
$endgroup$
– soso xoxo
Feb 2 at 2:52
$begingroup$
Thank you that helped me see that A is not necessary a subset of B
$endgroup$
– soso xoxo
Feb 2 at 2:52
$begingroup$
Thank you that helped me see that A is not necessary a subset of B
$endgroup$
– soso xoxo
Feb 2 at 2:52
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%2f3096568%2fif-%25e2%2588%2580a-%25e2%2588%2588-px-%25e2%2588%2580b-%25e2%2588%2588-py-a-b-then-%25e2%2588%2583a-%25e2%2588%2588-px-%25e2%2588%2583b-%25e2%2588%2588-py-a-%25e2%258a%2582-b%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