Mixing
How to prove if $ A subset B $ then $ A cup C subset B cup C $
$begingroup$
I don't really know how to approach this question:
Let $ U $ be a universe. Use an element argument to prove the following
statement.
For all sets $ A, B $ and $ C $ in $ P(U) $, if $ A subset B $ then $ A cup C subset B cup C $
My working:
Logical form: $ ( A subset B ) Rightarrow ( A cup C subset B cup C) $
Let $ x in ( A subset B ) $
I don' really know how to do from here. Can anyone here help me understand? Thanks a lot.
discrete-mathematics
edited Jan 21 at 5:18
user549397
1,5061418
asked Jan 21 at 4:35
J.SJ.S
585
$endgroup$
add a comment |
$begingroup$
I don't really know how to approach this question:
Let $ U $ be a universe. Use an element argument to prove the following
statement.
For all sets $ A, B $ and $ C $ in $ P(U) $, if $ A subset B $ then $ A cup C subset B cup C $
My working:
Logical form: $ ( A subset B ) Rightarrow ( A cup C subset B cup C) $
Let $ x in ( A subset B ) $
I don' really know how to do from here. Can anyone here help me understand? Thanks a lot.
discrete-mathematics
edited Jan 21 at 5:18
user549397
1,5061418
asked Jan 21 at 4:35
J.SJ.S
585
$endgroup$
1
$begingroup$
Well let $x in A cup C$, then what happens if $x in A$? Or if $x in C$?
$endgroup$
– Fede Poncio
Jan 21 at 4:39
add a comment |
$begingroup$
I don't really know how to approach this question:
Let $ U $ be a universe. Use an element argument to prove the following
statement.
For all sets $ A, B $ and $ C $ in $ P(U) $, if $ A subset B $ then $ A cup C subset B cup C $
My working:
Logical form: $ ( A subset B ) Rightarrow ( A cup C subset B cup C) $
Let $ x in ( A subset B ) $
I don' really know how to do from here. Can anyone here help me understand? Thanks a lot.
discrete-mathematics
edited Jan 21 at 5:18
user549397
1,5061418
asked Jan 21 at 4:35
J.SJ.S
585
$endgroup$
I don't really know how to approach this question:
Let $ U $ be a universe. Use an element argument to prove the following
statement.
For all sets $ A, B $ and $ C $ in $ P(U) $, if $ A subset B $ then $ A cup C subset B cup C $
My working:
Logical form: $ ( A subset B ) Rightarrow ( A cup C subset B cup C) $
Let $ x in ( A subset B ) $
I don' really know how to do from here. Can anyone here help me understand? Thanks a lot.
discrete-mathematics
discrete-mathematics
edited Jan 21 at 5:18
user549397
1,5061418
asked Jan 21 at 4:35
J.SJ.S
585
edited Jan 21 at 5:18
user549397
1,5061418
asked Jan 21 at 4:35
J.SJ.S
585
edited Jan 21 at 5:18
user549397
1,5061418
edited Jan 21 at 5:18
user549397
1,5061418
edited Jan 21 at 5:18
user549397
1,5061418
1,5061418
asked Jan 21 at 4:35
J.SJ.S
585
asked Jan 21 at 4:35
J.SJ.S
585
asked Jan 21 at 4:35
J.SJ.S
585
585
1
$begingroup$
Well let $x in A cup C$, then what happens if $x in A$? Or if $x in C$?
$endgroup$
– Fede Poncio
Jan 21 at 4:39
add a comment |
1
$begingroup$
Well let $x in A cup C$, then what happens if $x in A$? Or if $x in C$?
$endgroup$
– Fede Poncio
Jan 21 at 4:39
1
1
$begingroup$
Well let $x in A cup C$, then what happens if $x in A$? Or if $x in C$?
$endgroup$
– Fede Poncio
Jan 21 at 4:39
$begingroup$
Well let $x in A cup C$, then what happens if $x in A$? Or if $x in C$?
$endgroup$
– Fede Poncio
Jan 21 at 4:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Sometimes a figure is worth 1000 words:
answered Jan 21 at 4:40
David G. StorkDavid G. Stork
11k41432
$endgroup$
1
$begingroup$
Oooppss.... I'll fix the figure and repost. (Thanks.) . But the concept should nevertheless be clear.
$endgroup$
– David G. Stork
Jan 21 at 5:36
1
$begingroup$
Now that the figure is corrected, I've deleted my previous comment to avoid any future confusion.
$endgroup$
– Gregory J. Puleo
Jan 21 at 5:39
$begingroup$
Thank you, @David G. Stork. This helps me in visualising and understand better.
$endgroup$
– J.S
Jan 21 at 15:05
add a comment |
$begingroup$
Take an arbitrary element $x in A cup C$. Then $x in A vee x in C$. If $x in A$, then $x in B$, so $x in B cup C$ by the properties of unions. If $x in C$, it still follows that $x in B cup C$. By proving for an arbitrary element $x in A cup C$ that $x in B cup C$, we prove that $A cup C subseteq B cup C$.
However, it is not necessarily true that $A cup C subset B cup C$ (note the proper subset vs. subset). Suppose that $U = mathbb{N}$, $A = {1,2}$, $B = {1,2,3}$, and $C = {3,4,5}$. Then clearly, $A cup C not subset B cup C$.
Note: I see you used $P(U)$, not $U$, but a similar intuition can be applied to create a counterexample for power sets.
answered Jan 21 at 5:17
HyperionHyperion
636110
$endgroup$
$begingroup$
Thanks alot, I understand better. Just to confirm on whyA∪C⊄B∪C. Is it becauseA∪C = {1,2,3,4,5}andB∪C = {1,2,3,4,5}Where bothA∪CandB∪Chave exactly the same element. However,A subset B, means A needs to have lesser element than B. ThusA∪C⊄B∪C?
$endgroup$
– J.S
Jan 21 at 15:03
$begingroup$
@J.S Effectively, yes. For $A$ to be a proper subset of $B$, there must be some element in $B$ that is not in $A$.
$endgroup$
– Hyperion
Jan 21 at 16:23
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%2f3081516%2fhow-to-prove-if-a-subset-b-then-a-cup-c-subset-b-cup-c%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
$begingroup$
Sometimes a figure is worth 1000 words:
answered Jan 21 at 4:40
David G. StorkDavid G. Stork
11k41432
$endgroup$
1
$begingroup$
Oooppss.... I'll fix the figure and repost. (Thanks.) . But the concept should nevertheless be clear.
$endgroup$
– David G. Stork
Jan 21 at 5:36
1
$begingroup$
Now that the figure is corrected, I've deleted my previous comment to avoid any future confusion.
$endgroup$
– Gregory J. Puleo
Jan 21 at 5:39
$begingroup$
Thank you, @David G. Stork. This helps me in visualising and understand better.
$endgroup$
– J.S
Jan 21 at 15:05
add a comment |
$begingroup$
Sometimes a figure is worth 1000 words:
answered Jan 21 at 4:40
David G. StorkDavid G. Stork
11k41432
$endgroup$
1
$begingroup$
Oooppss.... I'll fix the figure and repost. (Thanks.) . But the concept should nevertheless be clear.
$endgroup$
– David G. Stork
Jan 21 at 5:36
1
$begingroup$
Now that the figure is corrected, I've deleted my previous comment to avoid any future confusion.
$endgroup$
– Gregory J. Puleo
Jan 21 at 5:39
$begingroup$
Thank you, @David G. Stork. This helps me in visualising and understand better.
$endgroup$
– J.S
Jan 21 at 15:05
add a comment |
$begingroup$
Sometimes a figure is worth 1000 words:
answered Jan 21 at 4:40
David G. StorkDavid G. Stork
11k41432
$endgroup$
Sometimes a figure is worth 1000 words:
answered Jan 21 at 4:40
David G. StorkDavid G. Stork
11k41432
edited Jan 21 at 5:37
answered Jan 21 at 4:40
David G. StorkDavid G. Stork
11k41432
answered Jan 21 at 4:40
David G. StorkDavid G. Stork
11k41432
answered Jan 21 at 4:40
David G. StorkDavid G. Stork
11k41432
11k41432
1
$begingroup$
Oooppss.... I'll fix the figure and repost. (Thanks.) . But the concept should nevertheless be clear.
$endgroup$
– David G. Stork
Jan 21 at 5:36
1
$begingroup$
Now that the figure is corrected, I've deleted my previous comment to avoid any future confusion.
$endgroup$
– Gregory J. Puleo
Jan 21 at 5:39
$begingroup$
Thank you, @David G. Stork. This helps me in visualising and understand better.
$endgroup$
– J.S
Jan 21 at 15:05
add a comment |
1
$begingroup$
Oooppss.... I'll fix the figure and repost. (Thanks.) . But the concept should nevertheless be clear.
$endgroup$
– David G. Stork
Jan 21 at 5:36
1
$begingroup$
Now that the figure is corrected, I've deleted my previous comment to avoid any future confusion.
$endgroup$
– Gregory J. Puleo
Jan 21 at 5:39
$begingroup$
Thank you, @David G. Stork. This helps me in visualising and understand better.
$endgroup$
– J.S
Jan 21 at 15:05
1
1
$begingroup$
Oooppss.... I'll fix the figure and repost. (Thanks.) . But the concept should nevertheless be clear.
$endgroup$
– David G. Stork
Jan 21 at 5:36
$begingroup$
Oooppss.... I'll fix the figure and repost. (Thanks.) . But the concept should nevertheless be clear.
$endgroup$
– David G. Stork
Jan 21 at 5:36
1
1
$begingroup$
Now that the figure is corrected, I've deleted my previous comment to avoid any future confusion.
$endgroup$
– Gregory J. Puleo
Jan 21 at 5:39
$begingroup$
Now that the figure is corrected, I've deleted my previous comment to avoid any future confusion.
$endgroup$
– Gregory J. Puleo
Jan 21 at 5:39
$begingroup$
Thank you, @David G. Stork. This helps me in visualising and understand better.
$endgroup$
– J.S
Jan 21 at 15:05
$begingroup$
Thank you, @David G. Stork. This helps me in visualising and understand better.
$endgroup$
– J.S
Jan 21 at 15:05
add a comment |
$begingroup$
Take an arbitrary element $x in A cup C$. Then $x in A vee x in C$. If $x in A$, then $x in B$, so $x in B cup C$ by the properties of unions. If $x in C$, it still follows that $x in B cup C$. By proving for an arbitrary element $x in A cup C$ that $x in B cup C$, we prove that $A cup C subseteq B cup C$.
However, it is not necessarily true that $A cup C subset B cup C$ (note the proper subset vs. subset). Suppose that $U = mathbb{N}$, $A = {1,2}$, $B = {1,2,3}$, and $C = {3,4,5}$. Then clearly, $A cup C not subset B cup C$.
Note: I see you used $P(U)$, not $U$, but a similar intuition can be applied to create a counterexample for power sets.
answered Jan 21 at 5:17
HyperionHyperion
636110
$endgroup$
$begingroup$
Thanks alot, I understand better. Just to confirm on whyA∪C⊄B∪C. Is it becauseA∪C = {1,2,3,4,5}andB∪C = {1,2,3,4,5}Where bothA∪CandB∪Chave exactly the same element. However,A subset B, means A needs to have lesser element than B. ThusA∪C⊄B∪C?
$endgroup$
– J.S
Jan 21 at 15:03
$begingroup$
@J.S Effectively, yes. For $A$ to be a proper subset of $B$, there must be some element in $B$ that is not in $A$.
$endgroup$
– Hyperion
Jan 21 at 16:23
add a comment |
$begingroup$
Take an arbitrary element $x in A cup C$. Then $x in A vee x in C$. If $x in A$, then $x in B$, so $x in B cup C$ by the properties of unions. If $x in C$, it still follows that $x in B cup C$. By proving for an arbitrary element $x in A cup C$ that $x in B cup C$, we prove that $A cup C subseteq B cup C$.
However, it is not necessarily true that $A cup C subset B cup C$ (note the proper subset vs. subset). Suppose that $U = mathbb{N}$, $A = {1,2}$, $B = {1,2,3}$, and $C = {3,4,5}$. Then clearly, $A cup C not subset B cup C$.
Note: I see you used $P(U)$, not $U$, but a similar intuition can be applied to create a counterexample for power sets.
answered Jan 21 at 5:17
HyperionHyperion
636110
$endgroup$
$begingroup$
Thanks alot, I understand better. Just to confirm on whyA∪C⊄B∪C. Is it becauseA∪C = {1,2,3,4,5}andB∪C = {1,2,3,4,5}Where bothA∪CandB∪Chave exactly the same element. However,A subset B, means A needs to have lesser element than B. ThusA∪C⊄B∪C?
$endgroup$
– J.S
Jan 21 at 15:03
$begingroup$
@J.S Effectively, yes. For $A$ to be a proper subset of $B$, there must be some element in $B$ that is not in $A$.
$endgroup$
– Hyperion
Jan 21 at 16:23
add a comment |
$begingroup$
Take an arbitrary element $x in A cup C$. Then $x in A vee x in C$. If $x in A$, then $x in B$, so $x in B cup C$ by the properties of unions. If $x in C$, it still follows that $x in B cup C$. By proving for an arbitrary element $x in A cup C$ that $x in B cup C$, we prove that $A cup C subseteq B cup C$.
However, it is not necessarily true that $A cup C subset B cup C$ (note the proper subset vs. subset). Suppose that $U = mathbb{N}$, $A = {1,2}$, $B = {1,2,3}$, and $C = {3,4,5}$. Then clearly, $A cup C not subset B cup C$.
Note: I see you used $P(U)$, not $U$, but a similar intuition can be applied to create a counterexample for power sets.
answered Jan 21 at 5:17
HyperionHyperion
636110
$endgroup$
Take an arbitrary element $x in A cup C$. Then $x in A vee x in C$. If $x in A$, then $x in B$, so $x in B cup C$ by the properties of unions. If $x in C$, it still follows that $x in B cup C$. By proving for an arbitrary element $x in A cup C$ that $x in B cup C$, we prove that $A cup C subseteq B cup C$.
However, it is not necessarily true that $A cup C subset B cup C$ (note the proper subset vs. subset). Suppose that $U = mathbb{N}$, $A = {1,2}$, $B = {1,2,3}$, and $C = {3,4,5}$. Then clearly, $A cup C not subset B cup C$.
Note: I see you used $P(U)$, not $U$, but a similar intuition can be applied to create a counterexample for power sets.
answered Jan 21 at 5:17
HyperionHyperion
636110
edited Jan 21 at 6:36
answered Jan 21 at 5:17
HyperionHyperion
636110
answered Jan 21 at 5:17
HyperionHyperion
636110
answered Jan 21 at 5:17
HyperionHyperion
636110
636110
$begingroup$
Thanks alot, I understand better. Just to confirm on whyA∪C⊄B∪C. Is it becauseA∪C = {1,2,3,4,5}andB∪C = {1,2,3,4,5}Where bothA∪CandB∪Chave exactly the same element. However,A subset B, means A needs to have lesser element than B. ThusA∪C⊄B∪C?
$endgroup$
– J.S
Jan 21 at 15:03
$begingroup$
@J.S Effectively, yes. For $A$ to be a proper subset of $B$, there must be some element in $B$ that is not in $A$.
$endgroup$
– Hyperion
Jan 21 at 16:23
add a comment |
$begingroup$
Thanks alot, I understand better. Just to confirm on whyA∪C⊄B∪C. Is it becauseA∪C = {1,2,3,4,5}andB∪C = {1,2,3,4,5}Where bothA∪CandB∪Chave exactly the same element. However,A subset B, means A needs to have lesser element than B. ThusA∪C⊄B∪C?
$endgroup$
– J.S
Jan 21 at 15:03
$begingroup$
@J.S Effectively, yes. For $A$ to be a proper subset of $B$, there must be some element in $B$ that is not in $A$.
$endgroup$
– Hyperion
Jan 21 at 16:23
$begingroup$
Thanks alot, I understand better. Just to confirm on why
A∪C⊄B∪C. Is it because A∪C = {1,2,3,4,5} and B∪C = {1,2,3,4,5} Where both A∪C and B∪C have exactly the same element. However, A subset B, means A needs to have lesser element than B. Thus A∪C⊄B∪C ?$endgroup$
– J.S
Jan 21 at 15:03
$begingroup$
Thanks alot, I understand better. Just to confirm on why
A∪C⊄B∪C. Is it because A∪C = {1,2,3,4,5} and B∪C = {1,2,3,4,5} Where both A∪C and B∪C have exactly the same element. However, A subset B, means A needs to have lesser element than B. Thus A∪C⊄B∪C ?$endgroup$
– J.S
Jan 21 at 15:03
$begingroup$
@J.S Effectively, yes. For $A$ to be a proper subset of $B$, there must be some element in $B$ that is not in $A$.
$endgroup$
– Hyperion
Jan 21 at 16:23
$begingroup$
@J.S Effectively, yes. For $A$ to be a proper subset of $B$, there must be some element in $B$ that is not in $A$.
$endgroup$
– Hyperion
Jan 21 at 16:23
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%2f3081516%2fhow-to-prove-if-a-subset-b-then-a-cup-c-subset-b-cup-c%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
$begingroup$
Well let $x in A cup C$, then what happens if $x in A$? Or if $x in C$?
$endgroup$
– Fede Poncio
Jan 21 at 4:39