Prove for every two sets $A$ and $B$ that $A − B$, $B − A$ and $A ∩ B$ are pairwise disjoint.  I've been looking at this problem and keep thinking it isn't true every time I think I make progress.  My main issue is that $A$ and $B$ can be any sets, including themselves. If $A = B$ then $A-B=\{\}$ but also $B-A=\{\}$ so they can't be disjoint.  Obviously, I'm missing something, probably will feel dumb after I figure it out.
121755
Prove for every two sets $A$ and $B$ that $A-B$, ${B-A}$ and $A \cap B$ are pairwise disjoint.     I'm really stuck on this one. I know pairwise disjoint means no two elements in $A$ and $B$ are the same. $A \cap B$ is the empty set, but how would I prove this?
110657
Let A, B be two sets. Prove that $A \subset B \iff A \cup B = B$  I'm thinking of using disjunctive syllogism by showing that $\neg \forall Y(Y \in A).$ However, I'm not sure how the proving steps should proceed such that it leads me to that premise. Edit: Thanks for the input. FYI, I need to prove this using predicate logic.
120134
Prove that $A \subset B$ if and only if $A \setminus B = \emptyset$.  What is the correct and mathematically strict way to prove the above? (slightly different than  )
55029
Let $A,B$ be any two sets. I really think that the statement $|A|\leq|B|$ or $|B|\leq|A|$ is true. Formally:     $$\forall A\forall B[\,|A|\leq|B| \lor\ |B|\leq|A|\,]$$   If this statement is true, what is the proof ?
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Can you verify my proof if it is right?  Let $A$ and $B$ be sets.  (a) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ union $C$ is a subset of $B$ union $C$. (b) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ intersect $C$ is a subset of $B$ intersect $C$.  Edited Solution :  (a)  (=>) Given A is a subset of B, this implies A U B = B. Given a set C, A U B U C = B U C. Since A is a subset of B, this implies A U C is a subset of B U C.      Solution 1 :(&lt;=) Given A U C is a subset of B U C , For A U C , x is a member of A U C . For B U C, x is a member of B U C. Suppose x is a member of A but not a member of B, meaning A is not a subset of B. Therefore , A U C is not a subset of B U C. Hence, A must be a subset of B.      Solution 2 :(&lt;=) Given A U C is a subset of B U C, For A U C, x is a member of A U C and for B U C, x is a member of B U C. Suppose C is a disjoint set. Since A U C is a subset of B U C, it follows that A is a subset of B.  (b) (=>) Since A is a subset of B , it follows that A U B = B. [A intersect C] is a subset of A and  [B intersect C ] is a subset of B by definition of intersection. Let x be member of A and member of C. Since [ A intersect C ] is a subset of A, x must be in A. Let x be a member of B and member of C. Since [ B intersect C ] is a subset of B, x must be in B. Therefore, A intersect C is a subset of B intersect C.  (&lt;=) Since A intersect C is a subset of B intersect C, suppose we let Set C be a subset of B and a subset of A as well. C is a subset of A implies A intersect C is equals to C. C is a subset of B implies B intersect C is equals to C as well. Since A intersect C is equals to B intersect C, A is a subset of B.
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I have the following question:  $\mathbb R^3\setminus \mathbb Q^3$ is a union of disjoint lines. This exercise is in the book: Set theory for the working mathematician of Krzysztof Ciesielski.  someone can give me a help on how to proceed?
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Statement of the problem: Prove that if $A_1 \supseteq A_2 \supseteq A_3 \cdots$ are all finite, nonempty sets of real numbers, then the intersection $\bigcap_{n=1}^{\infty} A_n$ is also finite and nonempty.  My solution: $\bigcap_{n=1}^{\infty} A_n = \emptyset.$ Then at least one $A_n$ is disjoint from the rest - contradiction, since each $A_n \subseteq A_{n-1}.$ Next assume that the intersection has infinite cardinality - contradiction, since $A_n \supseteq \bigcap_{n=1}^{\infty} A_n,$ but we have $A_n$ finite.  This is a homework problem and this is what I will be turning in regardless of any answers I get. I would like to know if there is a direct proof of this claim. I had considered trying to show that $\exists x \in A_n$ for each choice of $n,$ which would show that the intersection is nonempty, but: a) I don't know how to do this, and b) I still would not be able to prove the intersection was finite except by contradiction.  Thanks!  edit: Moreover, we were told not to use induction on this problem. But if $A_1$ is finite, can't we reduce the collection of $A_n&#39;s$ to a subset of the power set of $A_1$ and then induct?
120353
I am working with proofs in discrete math. Help to prove:  For the sets $A$ and $B$, we define the symmetric difference of $A$ and $B$ to be $A \triangle  B = (A-B)\cup(B-A).$  Prove that if $A \triangle B = C \triangle B,$ then $A = C.$
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I just started Set theory And I have the following question. Suppose I have $2$ sets, $A$ and $B$. $A$ is a subset of $B$ (contained inside $B$), then the union of these $2$ sets is $B$ right $?$ If no, could you guys explain to me why, if yes, could you guys provide a simple proof.  Thank you for reading my post.
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I know that this is valid because by thinking about it, you can see it's true, but I'm just not sure how I would go about actually proving it mathematically. I've kind of tried proof by contradiction, since that's what I use most of the time, but it doesn't seem to work.    In my mind, A ∪ B ⊆ C kind of just implies that A ⊆ C and B ⊆ C.
97793
Let there be $B \subset \{10,11,...,99\}$, with $|B| =10$. I need to prove, using the pigeonhole principle, that there are at least two disjoint subsets of $B$ (and none empty) where the sums of their elements are identical. I.e. $\exists B_1 \ne \varnothing, B_2 \ne \varnothing$ such that $B_1 \bigcup B_2 = B$, $B_1 \bigcap B_2 =\varnothing$ and $$\sum_{a \in B_1} a = \sum_{b \in B_2} b$$
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Hi I just want to find out if the following is a acceptable proof for the proposition:   "Consider metric space $(X,d)$. If $A \subset X$ is connected and $A \subset B \subset \bar{A}$ then $B$ is connected."  Proof:  Assume that $A \subset X$ is connected and $B$ is not connected. There exists disjoint open sets (in $(B,d)$) $V_{1}$ and $V_{2}$ such that $B = V_{1} \cup V_{2}$. Since $A \subset B$, it follows that $A \cap V_{1}$ and $A \cap V_{2}$ are open in $(A,d)$. Also then $A = (A \cap V_{1}) \cup (A \cap V_{2})$. Thus since $A $ is assumed connected, either $A \cap V_{1} = \emptyset$,  $A \cap V_{2} = \emptyset$, $A \cap V_{1} = A$ or $A \cap V_{2} = A$.   W.L.O.G assume that $A \cap V_{1} = \emptyset$, then there exists an $x \in B \subset \bar{A}$ and $\epsilon &gt; 0$ such that $B(x,\epsilon) \cap A = \emptyset$. This contradicts $x \in \bar{A}$. Assume now W.L.O.G that $A \cap V_{1} = A$, then $A \subset V_{1}$ and $A \cap V_{2} = \emptyset$. If we take $x \in V_{2} \subset B \subset \bar{A}$ then again we find $\epsilon &gt; 0$ such that $B(x,\epsilon) \cap A =\emptyset$. This contradicts $x \in \bar{A}$. Therefore there cannot exist disjoint open sets $V_{1}$ and $V_{2}$ in $B$ such that $B = V_{1} \cup V_{2}$. Hence, $(B,d)$ cannot be disconnected and is therefore connected.   $\square$
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I haven't been able to figure out if the following reasoning is correct, so I'd like to have the opinion of other people on that.  The goal is to prove that the empty set is unique.  In order to do that, let $E$ be a set, $A$ be an empty set and  $B$ be an empty set. I want to prove that $A = B$.  First, I can try proving that $A \subset B$.  I know that $\forall x \in E, x \notin A$. Now I can consider this proposition : $x \in A \Rightarrow x \in B$  Since $x \notin A$, I can safely say that that the proposition above is true.  Hence, I can conclude that $A \subset B$  Am I right in my conclusion?
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Question:  Show that if $ \ A \cup B = A$ and $ \ A\cap  B = A$ then $ \ A = B$  My attempt:  Proof by contradiction:  Assume $ \ A \cup B = A$ and $ \ A\cap  B = A$ and $ \ A \neq B$   Case 1: $ \exists \ x \in A, x\notin B$  If $ x \in A \implies x\in A \cap B \implies x \in A \ and \  x\in B  \implies x\in B$, a contradiction.   Case 2: $ \exists \ x \in B, x\notin A$  If $ \ x \in B \implies x\in A \cup B \implies x\in A$, since $ \ A \cup B = A$. Contradiction.   Is this approach correct? Could someone please show me how to do a direct proof?
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I know how to prove when the two equivalence classes are not disjoint, i.e. $[a]=[b]$. I see that the proof works for proving a equivalence class is disjoint, but I don't get it. Can someone explain it to me?
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By definition:     If$ \ A$ and $B\ $are sets and every element of$ \ A\ $ is also an element of $ \ B \ $then we can say $ \ A\ $is a subset of $ \ B, \ $and denote this by $ \ A\subset B\ $or $ \ A\subseteq B.\ $Or, equivalently, we can say that$ \ B$ is a super set of $\ A \ $(if every element of A    is also an element of B), which is denoted by $ \ B\supseteq\ A.\ $If $ \ A \ $ is $\neq\ B \ $(there exists at least one element of $ \ B \ $which is not an element of $ \ A), \ $then $ \ A \ $is a proper subset of $ \ B \ $, which is denoted$ \ A\subsetneq\ B. \ $ Or, equivalently, $ \ B$ is a proper super set of $\ A \  $(if $A\subset B,$ but $A\neq B$) which is denoted by $ \ B\supsetneq\ A.\ $   In the case where where $ \ A \ $ is $\neq\ B, \ $but every element of $A$ is also an element of $B,$ are $ \ A\subset B,\ A\subseteq B,\ B\supseteq A,\ A\subsetneq B,\ B\supsetneq A \ $all equivalent? In the case where where $ \ A=B \ $are $\ A\subset B,\ B\subset A,\ A\subseteq B,\ B\supseteq A,\ A\supseteq B,\ B\subseteq A\ $all equivalent? It's kind of strange to think that if two sets are equal to one another, that they are also both supersets of each other. It seems more "natural," to me at least, that the whole superset "concept," where $ \ A\supseteq B\ $or $ \ B\supseteq A,\ $should be reserved for when $ \ B\neq A\ $(but every element of $B$ is in $A$) or $ \ A\neq B \ $(but every element of $A$ is in $B$). Of course in both of these cases we would use the proper superset notation.
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Why is the cardinality of set S is 3^n, where S is the collection of all ordered pairs of disjoint subset of {1,2,3,...,n} where n is non negative. if n = 2, then subset A and B are disjoint and can only contain elements of 1,2, or the empty set, is this correct? I can't get 9 ordered pair. Can someone explain how do I get 9 ordered pair?