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 want to proof that:  $A \subseteq B  \Leftrightarrow  A \cup B = B$  At the moment I have no idea on how to start. Please give me a hint.  Thx in advance!
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What does it mean when you have a set like A = {1, {2}, 3, 4}? Is {2} a subset within A?  Say I have the following:  A = {1, {2}, 3, 4}, B = {1, 2, {3}}, C = {1, 2}  How many elements are in $A \cap B \cap C$? 1 or 2?  What if I want to find $A \times B$, is there 6 elements or 12?
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Given two sets$$A=\{1,3,5\}\quad,\quad B=\{1,3,8\}.$$ Then I compute the $A\setminus B=\{5\}.$   But my book$^\dagger$said: ... . Also, observe that $x\notin A\setminus B$ does not mean that $x\notin A\lor x\in B$. Why?  I don't know how to explain the Why, please give me some example.    $^\dagger$, second edition, page $4$.  EDIT: My intuition tells me that it does mean that. I guess there is some special case about null-set or about the restriction of the universal set.
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if $$A=\left\{\varnothing \right\}$$ we know that $A$ is a Singleton Set whose single element is a Null set. That is $A$  is a set containing another set which is Null set.  Now are these two statements True?  $1.$ $\varnothing \in A$  $2.$ $\varnothing \subset A$
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There are two sets $A$ and $B$ which are bound and are not empty. Now we'll define another set as: $A + B = \{a + b | a \in A, b \in B\}.$  I need to prove that $\sup(A + B) = \sup A + \sup B.$   I don't know if the way in which I proved it is enough or is missing something. This is my proof:  I know that for every $a \in A, b \in B$: $\sup A \geq a$ and $\sup B \geq b.$  Therefore $\sup A + sup B \geq a + b.$ I'll take a $c \in A+B$ and by $A+B$ definition: $c = a + b$. Then, it is safe to say that $c = a + b \leq \sup A + \sup B$ which gives: $c \leq \sup A + \sup B.$  I know that $c \leq \sup A + \sup B$ is true for every general $c \in A+B$, so the supremum of $A + B$ is: $\sup(A + B) = \sup A + \sup B.$  Is this a way to prove it or is it not enough? Thanks.
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$\{\emptyset\}$ is a set containing the empty set. Is $\{\emptyset\}$ a subset of $\{\{\emptyset\}\}$?  My hypothesis is yes by looking at the form of "the superset $\{\{\emptyset\}\}$" which contains "the subset $\{\emptyset\}$".
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I Have a question. Given only the definition of equality of two sets, how can we prove that there is one and only one empty set. I mean by equality of two sets the following: $$A=B \iff  \forall x (x \in A \iff x \in B)$$
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I have been learning set theory from scratch again and I have a question about the distributive property for sets. It seems the only proof I found of it actually USES the very same distributive property while proving it. This seems circular to me, could someone explain why it is allowed?  This is what I am trying to prove. $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$  I got this proof from a quora answer.  Let $x \in A \cup (B \cap C)$. If $x \in A \cup (B \cap C)$ then $x$ is either in $A$ or in ($B \cap C$).  $x \in A$ or $x \in (B \cap C)$  $x \in A$ or ($x \in B$ and $x \in C$)  HERE!!! How does one get from the statement above to the one below without actually using the distributive property itself?  ($x \in A$ or $x \in B$) and ($x \in A$ or $x \in C$)  $x \in (A \cup B)$ and $x \in (A \cup C)$  $x \in (A ∪ B) \cap (A ∪ C)$  $x \in A \cup (B \cap C) \implies x \in (A \cup B) \cap (A \cup C)$.  Thanks a lot!!
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Consider the following putative theorem.   Theorem? Suppose A, B, and C are sets and $A \subseteq B \cup C$. Then either $A \subseteq B$ or $A \subseteq C$.   What's wrong with the following proof?   Proof. Let $x$ be an arbitrary element of A. Since $A \subseteq B \cup C$, it follows that either $x \in B$ or $x \in C$.   Case 1. $x \in B$. Since x was an arbitrary element of A, it follows that $\forall x \in A(x \in B)$, which means that $A \subseteq B$.  Case 2. $x \in C$. Similarly, since x was an arbitrary element of A, we can conclude that $A \subseteq C$.  Thus, either $A \subseteq B$ or $A \subseteq C$.   I have proved that the theorem is incorrect but I can't understand why the above proof is not correct.
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Note: Actually there's no error in the book and the manual. I actually misread it. The answer is of a different question : True or False: {0} ⊂ {0}  This question is from Discrete Math Book by Rosen.     {{∅}} ⊂ {∅,{∅}}   Answer in the manual is:     This  is false. For one set  to  be  a  proper  subset  of  another,  the  two sets  cannot  be  equal.   How is {{∅}} is equal to the set {∅,{∅}} ? I know the two sets have different cardinal numbers. So they cannot be equal.
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Statement: Let $A$, $B$, and $C$ be sets. Then $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$.   Proof: Let $(x,y) \in ( A \setminus B) \cap (A \setminus C)$, where $ x \in A$ and $y \in (B \cap C$ by definition of intersection. So, $(x,y) \in (A \setminus B \cap C)$ which show that $(A \setminus C) \cap (A \setminus C) \subseteq A \setminus (A \cup C)$. \ Now let $(x,y) \in A \setminus(B \cup C)$, where $ x \in A $. So, $ (x,y) \in (A \setminus B) \cap (A \setminus C)$ by definition of intersection, which shows that $A \setminus (B \cup C) \subseteq (A \setminus B) \cap (A \setminus C)$. Therefore, by the Axiom Set of Equality we know that $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$ .   Am I on the right track to proving that this is a true statement or is there a better way to show this?
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If $(X,\mathcal{T})$ is a connected space, and $Y$ a connected subset, and $X\setminus Y=A\cup B$ for separated sets $A$ and $B$, then why is $A\cup Y$ connected as well?  Thank you kindly.
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Is this true?  If $A$ and $B$ are sets, then $A=\emptyset $ if and only if $B = A \bigtriangleup B$.  If $A=\emptyset$ then $B=\emptyset$ too?  Could someone help me please?
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Sorry but I don't think I can know, since it's a definition. Please tell me. I don't think that $0=\emptyset\,$ since I distinguish between empty set and the value $0$. Do all sets, even the empty set, have infitinite emptiness e.g. do all sets including the empty set contain infinitely many empty sets?
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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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I've been trying to challenge myself but with no luck. Maybe one of you will have a better idea. How can it be proven that there can't be a set which contains its own powerset, using only russell's paradox? I've managed to prove it with other principles or axioms, such as that a set can not belong to itself, or that a powerset cannot belong to itself.
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More generally, if a set $S$ has cardinality $\mathfrak{m}$, how many of its subsets have cardinality $\mathfrak{n}$? Clearly there are at least $2^\mathfrak{n}$ such subsets. I don't see how many more though.  Thanks