How can I prove that for any $A, B$ if $Asubseteq B$ and $Bsubseteq C$, then $(C-A)cup (B-A)subseteq C$?
$begingroup$
How can I prove that for any $A, B$ if $Asubseteq B$ and $Bsubseteq C$, then $(C-A)cup (B-A)subseteq C$?
I've been working on this question and I haven't really made much progress with it. I know that I can rewrite it as $(C cap A^c) cup (Bcap A^c)$. I'm pretty sure that if $A subseteq B$ and $B subseteq C$ then $A subseteq C$. If this is the case then wouldn't $(C cap A^c) = emptyset$ and $(B cap A^c) = emptyset$ or am not understanding something with set theory? Thank you for the help.
discrete-mathematics elementary-set-theory
edited Dec 9 '18 at 22:32
Shaun
8,932113681
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
$endgroup$
add a comment |
$begingroup$
How can I prove that for any $A, B$ if $Asubseteq B$ and $Bsubseteq C$, then $(C-A)cup (B-A)subseteq C$?
I've been working on this question and I haven't really made much progress with it. I know that I can rewrite it as $(C cap A^c) cup (Bcap A^c)$. I'm pretty sure that if $A subseteq B$ and $B subseteq C$ then $A subseteq C$. If this is the case then wouldn't $(C cap A^c) = emptyset$ and $(B cap A^c) = emptyset$ or am not understanding something with set theory? Thank you for the help.
discrete-mathematics elementary-set-theory
edited Dec 9 '18 at 22:32
Shaun
8,932113681
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
$endgroup$
1
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09
add a comment |
$begingroup$
How can I prove that for any $A, B$ if $Asubseteq B$ and $Bsubseteq C$, then $(C-A)cup (B-A)subseteq C$?
I've been working on this question and I haven't really made much progress with it. I know that I can rewrite it as $(C cap A^c) cup (Bcap A^c)$. I'm pretty sure that if $A subseteq B$ and $B subseteq C$ then $A subseteq C$. If this is the case then wouldn't $(C cap A^c) = emptyset$ and $(B cap A^c) = emptyset$ or am not understanding something with set theory? Thank you for the help.
discrete-mathematics elementary-set-theory
edited Dec 9 '18 at 22:32
Shaun
8,932113681
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
$endgroup$
How can I prove that for any $A, B$ if $Asubseteq B$ and $Bsubseteq C$, then $(C-A)cup (B-A)subseteq C$?
I've been working on this question and I haven't really made much progress with it. I know that I can rewrite it as $(C cap A^c) cup (Bcap A^c)$. I'm pretty sure that if $A subseteq B$ and $B subseteq C$ then $A subseteq C$. If this is the case then wouldn't $(C cap A^c) = emptyset$ and $(B cap A^c) = emptyset$ or am not understanding something with set theory? Thank you for the help.
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
edited Dec 9 '18 at 22:32
Shaun
8,932113681
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
edited Dec 9 '18 at 22:32
Shaun
8,932113681
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
edited Dec 9 '18 at 22:32
Shaun
8,932113681
edited Dec 9 '18 at 22:32
Shaun
8,932113681
edited Dec 9 '18 at 22:32
Shaun
8,932113681
8,932113681
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
asked Dec 9 '18 at 22:29
Dominic MartireDominic Martire
62
62
1
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09
add a comment |
1
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09
1
1
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Quite simply: by definition, $C-Asubseteq Cˆ$ and $B-Asubseteq B$. Further, $Bsubseteq C$...
answered Dec 9 '18 at 22:38
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
It's not the case that $X subseteq Y$ implies $Y-X = emptyset$. Indeed if $X = {0}$ and $Y = {0,1}$ then $Y - X = {1}$.
We do have, however, that $X subseteq Y$ implies $Y - X subseteq Y$. Applying this to what you have so far, conclude that $C- A subseteq C$ and $B - A subseteq B subseteq C$. Hence $(C-A) cup (B-A) subseteq C$ too.
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,501722
$endgroup$
add a comment |
$begingroup$
For every $xin (B-A)cup (C-A)$, either $xin B-A$ or $xin C-A$ (or both). For every $xin B-A$, $xin B$ hence $xin C$. For every $xin C-A$, obviously $xin C$. Hence if $xin (B-A)cup (C-A)$, surely $xin C$. By definition, this means that it is a subset of $C$.
answered Dec 10 '18 at 0:13
YiFanYiFan
2,8091422
$endgroup$
add a comment |
$begingroup$
Draw Venn Diagrams. This should be obvious.
Now just prove it with Element chasing or definitions or whatever you like.
answered Dec 9 '18 at 23:10
fleabloodfleablood
69.3k22685
$endgroup$
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%2f3033128%2fhow-can-i-prove-that-for-any-a-b-if-a-subseteq-b-and-b-subseteq-c-then%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Quite simply: by definition, $C-Asubseteq Cˆ$ and $B-Asubseteq B$. Further, $Bsubseteq C$...
answered Dec 9 '18 at 22:38
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
Quite simply: by definition, $C-Asubseteq Cˆ$ and $B-Asubseteq B$. Further, $Bsubseteq C$...
answered Dec 9 '18 at 22:38
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
Quite simply: by definition, $C-Asubseteq Cˆ$ and $B-Asubseteq B$. Further, $Bsubseteq C$...
answered Dec 9 '18 at 22:38
BernardBernard
119k740113
$endgroup$
Quite simply: by definition, $C-Asubseteq Cˆ$ and $B-Asubseteq B$. Further, $Bsubseteq C$...
answered Dec 9 '18 at 22:38
BernardBernard
119k740113
answered Dec 9 '18 at 22:38
BernardBernard
119k740113
answered Dec 9 '18 at 22:38
BernardBernard
119k740113
answered Dec 9 '18 at 22:38
BernardBernard
119k740113
119k740113
add a comment |
add a comment |
$begingroup$
It's not the case that $X subseteq Y$ implies $Y-X = emptyset$. Indeed if $X = {0}$ and $Y = {0,1}$ then $Y - X = {1}$.
We do have, however, that $X subseteq Y$ implies $Y - X subseteq Y$. Applying this to what you have so far, conclude that $C- A subseteq C$ and $B - A subseteq B subseteq C$. Hence $(C-A) cup (B-A) subseteq C$ too.
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,501722
$endgroup$
add a comment |
$begingroup$
It's not the case that $X subseteq Y$ implies $Y-X = emptyset$. Indeed if $X = {0}$ and $Y = {0,1}$ then $Y - X = {1}$.
We do have, however, that $X subseteq Y$ implies $Y - X subseteq Y$. Applying this to what you have so far, conclude that $C- A subseteq C$ and $B - A subseteq B subseteq C$. Hence $(C-A) cup (B-A) subseteq C$ too.
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,501722
$endgroup$
add a comment |
$begingroup$
It's not the case that $X subseteq Y$ implies $Y-X = emptyset$. Indeed if $X = {0}$ and $Y = {0,1}$ then $Y - X = {1}$.
We do have, however, that $X subseteq Y$ implies $Y - X subseteq Y$. Applying this to what you have so far, conclude that $C- A subseteq C$ and $B - A subseteq B subseteq C$. Hence $(C-A) cup (B-A) subseteq C$ too.
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,501722
$endgroup$
It's not the case that $X subseteq Y$ implies $Y-X = emptyset$. Indeed if $X = {0}$ and $Y = {0,1}$ then $Y - X = {1}$.
We do have, however, that $X subseteq Y$ implies $Y - X subseteq Y$. Applying this to what you have so far, conclude that $C- A subseteq C$ and $B - A subseteq B subseteq C$. Hence $(C-A) cup (B-A) subseteq C$ too.
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,501722
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,501722
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,501722
answered Dec 9 '18 at 22:42
Badam BaplanBadam Baplan
4,501722
4,501722
add a comment |
add a comment |
$begingroup$
For every $xin (B-A)cup (C-A)$, either $xin B-A$ or $xin C-A$ (or both). For every $xin B-A$, $xin B$ hence $xin C$. For every $xin C-A$, obviously $xin C$. Hence if $xin (B-A)cup (C-A)$, surely $xin C$. By definition, this means that it is a subset of $C$.
answered Dec 10 '18 at 0:13
YiFanYiFan
2,8091422
$endgroup$
add a comment |
$begingroup$
For every $xin (B-A)cup (C-A)$, either $xin B-A$ or $xin C-A$ (or both). For every $xin B-A$, $xin B$ hence $xin C$. For every $xin C-A$, obviously $xin C$. Hence if $xin (B-A)cup (C-A)$, surely $xin C$. By definition, this means that it is a subset of $C$.
answered Dec 10 '18 at 0:13
YiFanYiFan
2,8091422
$endgroup$
add a comment |
$begingroup$
For every $xin (B-A)cup (C-A)$, either $xin B-A$ or $xin C-A$ (or both). For every $xin B-A$, $xin B$ hence $xin C$. For every $xin C-A$, obviously $xin C$. Hence if $xin (B-A)cup (C-A)$, surely $xin C$. By definition, this means that it is a subset of $C$.
answered Dec 10 '18 at 0:13
YiFanYiFan
2,8091422
$endgroup$
For every $xin (B-A)cup (C-A)$, either $xin B-A$ or $xin C-A$ (or both). For every $xin B-A$, $xin B$ hence $xin C$. For every $xin C-A$, obviously $xin C$. Hence if $xin (B-A)cup (C-A)$, surely $xin C$. By definition, this means that it is a subset of $C$.
answered Dec 10 '18 at 0:13
YiFanYiFan
2,8091422
answered Dec 10 '18 at 0:13
YiFanYiFan
2,8091422
answered Dec 10 '18 at 0:13
YiFanYiFan
2,8091422
answered Dec 10 '18 at 0:13
YiFanYiFan
2,8091422
2,8091422
add a comment |
add a comment |
$begingroup$
Draw Venn Diagrams. This should be obvious.
Now just prove it with Element chasing or definitions or whatever you like.
answered Dec 9 '18 at 23:10
fleabloodfleablood
69.3k22685
$endgroup$
add a comment |
$begingroup$
Draw Venn Diagrams. This should be obvious.
Now just prove it with Element chasing or definitions or whatever you like.
answered Dec 9 '18 at 23:10
fleabloodfleablood
69.3k22685
$endgroup$
add a comment |
$begingroup$
Draw Venn Diagrams. This should be obvious.
Now just prove it with Element chasing or definitions or whatever you like.
answered Dec 9 '18 at 23:10
fleabloodfleablood
69.3k22685
$endgroup$
Draw Venn Diagrams. This should be obvious.
Now just prove it with Element chasing or definitions or whatever you like.
answered Dec 9 '18 at 23:10
fleabloodfleablood
69.3k22685
answered Dec 9 '18 at 23:10
fleabloodfleablood
69.3k22685
answered Dec 9 '18 at 23:10
fleabloodfleablood
69.3k22685
answered Dec 9 '18 at 23:10
fleabloodfleablood
69.3k22685
69.3k22685
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%2f3033128%2fhow-can-i-prove-that-for-any-a-b-if-a-subseteq-b-and-b-subseteq-c-then%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
$begingroup$
Rewrite each statement as an implication in memberships.
$endgroup$
– J.G.
Dec 9 '18 at 22:38
$begingroup$
Consider $A={1}, B={1,2}, C={1,2,3}$. We have $Asubseteq Bsubseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is ${2,3}$, so $Ccap A^c={2,3}$ and $Bcap A^c={2}$.
$endgroup$
– Shaun
Dec 9 '18 at 22:39
$begingroup$
$A subset C$ means everything in $A$ is in $C$. $Csetminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $Csetminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C setminus A$ with $A setminus C$? Setminus is not commutative. $A setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $Asetminus C = emptyset$. But $Csetminus A$ is just... $C setminus $A$.
$endgroup$
– fleablood
Dec 9 '18 at 23:09