Order of Set Operations
up vote
0
down vote
favorite
A particular question states: Show if $A ⊆ B^c$ then $A ∩ B = ∅$.
Being very new to set theory, I attempted to start some proof, which appears below, where $S =$ universe of discourse:
$$
A cap B = Acap(Ssetminus B^c) = (Acap S)setminus B^c = Asetminus B^c = emptyset.
$$
This may or may not be correct; I have no solution personally offered to this question. But I am wondering if it was permitted to "slide" the parentheses from $Acap (Ssetminus B^c) $ to $Acap B = (A cap S)setminus B^c $.
proof-verification elementary-set-theory
add a comment |
up vote
0
down vote
favorite
A particular question states: Show if $A ⊆ B^c$ then $A ∩ B = ∅$.
Being very new to set theory, I attempted to start some proof, which appears below, where $S =$ universe of discourse:
$$
A cap B = Acap(Ssetminus B^c) = (Acap S)setminus B^c = Asetminus B^c = emptyset.
$$
This may or may not be correct; I have no solution personally offered to this question. But I am wondering if it was permitted to "slide" the parentheses from $Acap (Ssetminus B^c) $ to $Acap B = (A cap S)setminus B^c $.
proof-verification elementary-set-theory
This is not true unless $A=varnothing$?
– Rushabh Mehta
Nov 13 at 18:23
Are you sure you read the question right? If $Asubseteq B$ then $Acap B=A$.
– Anguepa
Nov 13 at 18:37
My apologies. I've re-modified the question.
– Julia Kim
Nov 13 at 19:05
$A cap S = A.$
– Will M.
Nov 13 at 19:06
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A particular question states: Show if $A ⊆ B^c$ then $A ∩ B = ∅$.
Being very new to set theory, I attempted to start some proof, which appears below, where $S =$ universe of discourse:
$$
A cap B = Acap(Ssetminus B^c) = (Acap S)setminus B^c = Asetminus B^c = emptyset.
$$
This may or may not be correct; I have no solution personally offered to this question. But I am wondering if it was permitted to "slide" the parentheses from $Acap (Ssetminus B^c) $ to $Acap B = (A cap S)setminus B^c $.
proof-verification elementary-set-theory
A particular question states: Show if $A ⊆ B^c$ then $A ∩ B = ∅$.
Being very new to set theory, I attempted to start some proof, which appears below, where $S =$ universe of discourse:
$$
A cap B = Acap(Ssetminus B^c) = (Acap S)setminus B^c = Asetminus B^c = emptyset.
$$
This may or may not be correct; I have no solution personally offered to this question. But I am wondering if it was permitted to "slide" the parentheses from $Acap (Ssetminus B^c) $ to $Acap B = (A cap S)setminus B^c $.
proof-verification elementary-set-theory
proof-verification elementary-set-theory
edited Nov 13 at 19:03
asked Nov 13 at 18:21
Julia Kim
414
414
This is not true unless $A=varnothing$?
– Rushabh Mehta
Nov 13 at 18:23
Are you sure you read the question right? If $Asubseteq B$ then $Acap B=A$.
– Anguepa
Nov 13 at 18:37
My apologies. I've re-modified the question.
– Julia Kim
Nov 13 at 19:05
$A cap S = A.$
– Will M.
Nov 13 at 19:06
add a comment |
This is not true unless $A=varnothing$?
– Rushabh Mehta
Nov 13 at 18:23
Are you sure you read the question right? If $Asubseteq B$ then $Acap B=A$.
– Anguepa
Nov 13 at 18:37
My apologies. I've re-modified the question.
– Julia Kim
Nov 13 at 19:05
$A cap S = A.$
– Will M.
Nov 13 at 19:06
This is not true unless $A=varnothing$?
– Rushabh Mehta
Nov 13 at 18:23
This is not true unless $A=varnothing$?
– Rushabh Mehta
Nov 13 at 18:23
Are you sure you read the question right? If $Asubseteq B$ then $Acap B=A$.
– Anguepa
Nov 13 at 18:37
Are you sure you read the question right? If $Asubseteq B$ then $Acap B=A$.
– Anguepa
Nov 13 at 18:37
My apologies. I've re-modified the question.
– Julia Kim
Nov 13 at 19:05
My apologies. I've re-modified the question.
– Julia Kim
Nov 13 at 19:05
$A cap S = A.$
– Will M.
Nov 13 at 19:06
$A cap S = A.$
– Will M.
Nov 13 at 19:06
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Usually if you need to ask if a step is permitted, then the answer is "no, unless..."—what comes next is either:
- ...it is part of the definition of something involved; or
- ...it invokes a result that has already been proved.
In this case, you want to know whether you can apply the rule $A cap (B setminus C) = (A cap B) setminus C$. It turns out that this rule is true in general, but if you wanted to use it then you'd need to prove it, or you'd need to cite somewhere that it is proved. I suspect that this isn't a result that you can use without justification, so you'd need to prove this too.
Be careful though! For example, it is not true in general that $A cup (B setminus C) = (A cup B) setminus C$, even though this looks very similar.
Coming back to the problem at hand: when you're new to set theory, the temptation in answering questions like this is to rearrange equations until you get the answer. I would advise against this, as it often leads to mistakes and increases the risk of invoking illegal 'rules'. Instead, to prove two sets are equal, you should prove that they have the same elements.
So for your question: assume that $A subseteq B^c$. To prove that $A cap B = varnothing$ you need to show that $A cap B$ and $varnothing$ have the same elements. Since $varnothing$ has no elements, this amounts to assuming that there is some $x in A cap B$ and then deriving a contradiction. This, in turn, is almost immediate from the assumption that $A subseteq B^c$.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Usually if you need to ask if a step is permitted, then the answer is "no, unless..."—what comes next is either:
- ...it is part of the definition of something involved; or
- ...it invokes a result that has already been proved.
In this case, you want to know whether you can apply the rule $A cap (B setminus C) = (A cap B) setminus C$. It turns out that this rule is true in general, but if you wanted to use it then you'd need to prove it, or you'd need to cite somewhere that it is proved. I suspect that this isn't a result that you can use without justification, so you'd need to prove this too.
Be careful though! For example, it is not true in general that $A cup (B setminus C) = (A cup B) setminus C$, even though this looks very similar.
Coming back to the problem at hand: when you're new to set theory, the temptation in answering questions like this is to rearrange equations until you get the answer. I would advise against this, as it often leads to mistakes and increases the risk of invoking illegal 'rules'. Instead, to prove two sets are equal, you should prove that they have the same elements.
So for your question: assume that $A subseteq B^c$. To prove that $A cap B = varnothing$ you need to show that $A cap B$ and $varnothing$ have the same elements. Since $varnothing$ has no elements, this amounts to assuming that there is some $x in A cap B$ and then deriving a contradiction. This, in turn, is almost immediate from the assumption that $A subseteq B^c$.
add a comment |
up vote
0
down vote
Usually if you need to ask if a step is permitted, then the answer is "no, unless..."—what comes next is either:
- ...it is part of the definition of something involved; or
- ...it invokes a result that has already been proved.
In this case, you want to know whether you can apply the rule $A cap (B setminus C) = (A cap B) setminus C$. It turns out that this rule is true in general, but if you wanted to use it then you'd need to prove it, or you'd need to cite somewhere that it is proved. I suspect that this isn't a result that you can use without justification, so you'd need to prove this too.
Be careful though! For example, it is not true in general that $A cup (B setminus C) = (A cup B) setminus C$, even though this looks very similar.
Coming back to the problem at hand: when you're new to set theory, the temptation in answering questions like this is to rearrange equations until you get the answer. I would advise against this, as it often leads to mistakes and increases the risk of invoking illegal 'rules'. Instead, to prove two sets are equal, you should prove that they have the same elements.
So for your question: assume that $A subseteq B^c$. To prove that $A cap B = varnothing$ you need to show that $A cap B$ and $varnothing$ have the same elements. Since $varnothing$ has no elements, this amounts to assuming that there is some $x in A cap B$ and then deriving a contradiction. This, in turn, is almost immediate from the assumption that $A subseteq B^c$.
add a comment |
up vote
0
down vote
up vote
0
down vote
Usually if you need to ask if a step is permitted, then the answer is "no, unless..."—what comes next is either:
- ...it is part of the definition of something involved; or
- ...it invokes a result that has already been proved.
In this case, you want to know whether you can apply the rule $A cap (B setminus C) = (A cap B) setminus C$. It turns out that this rule is true in general, but if you wanted to use it then you'd need to prove it, or you'd need to cite somewhere that it is proved. I suspect that this isn't a result that you can use without justification, so you'd need to prove this too.
Be careful though! For example, it is not true in general that $A cup (B setminus C) = (A cup B) setminus C$, even though this looks very similar.
Coming back to the problem at hand: when you're new to set theory, the temptation in answering questions like this is to rearrange equations until you get the answer. I would advise against this, as it often leads to mistakes and increases the risk of invoking illegal 'rules'. Instead, to prove two sets are equal, you should prove that they have the same elements.
So for your question: assume that $A subseteq B^c$. To prove that $A cap B = varnothing$ you need to show that $A cap B$ and $varnothing$ have the same elements. Since $varnothing$ has no elements, this amounts to assuming that there is some $x in A cap B$ and then deriving a contradiction. This, in turn, is almost immediate from the assumption that $A subseteq B^c$.
Usually if you need to ask if a step is permitted, then the answer is "no, unless..."—what comes next is either:
- ...it is part of the definition of something involved; or
- ...it invokes a result that has already been proved.
In this case, you want to know whether you can apply the rule $A cap (B setminus C) = (A cap B) setminus C$. It turns out that this rule is true in general, but if you wanted to use it then you'd need to prove it, or you'd need to cite somewhere that it is proved. I suspect that this isn't a result that you can use without justification, so you'd need to prove this too.
Be careful though! For example, it is not true in general that $A cup (B setminus C) = (A cup B) setminus C$, even though this looks very similar.
Coming back to the problem at hand: when you're new to set theory, the temptation in answering questions like this is to rearrange equations until you get the answer. I would advise against this, as it often leads to mistakes and increases the risk of invoking illegal 'rules'. Instead, to prove two sets are equal, you should prove that they have the same elements.
So for your question: assume that $A subseteq B^c$. To prove that $A cap B = varnothing$ you need to show that $A cap B$ and $varnothing$ have the same elements. Since $varnothing$ has no elements, this amounts to assuming that there is some $x in A cap B$ and then deriving a contradiction. This, in turn, is almost immediate from the assumption that $A subseteq B^c$.
answered Nov 13 at 19:47
Clive Newstead
49.3k472132
49.3k472132
add a comment |
add a comment |
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%2f2997087%2forder-of-set-operations%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
This is not true unless $A=varnothing$?
– Rushabh Mehta
Nov 13 at 18:23
Are you sure you read the question right? If $Asubseteq B$ then $Acap B=A$.
– Anguepa
Nov 13 at 18:37
My apologies. I've re-modified the question.
– Julia Kim
Nov 13 at 19:05
$A cap S = A.$
– Will M.
Nov 13 at 19:06