Let $A,B$ sets and $B$ infinite set , $\lvert A\rvert=a , \lvert B\rvert=b$  and $1&lt;a\leq b$, and let $D=\{f \mid f:B \to A\} $. Prove that $\lvert D\rvert=2^b$.  I know that $\lvert D\rvert=\lvert A\rvert^{\lvert B\rvert}$ and there is a subset of $B$ that has the same cardinality as $A$, but how to prove $\lvert D\rvert=2^b$?
122768
Let $2\leq \kappa\leq \gamma$ where $\gamma$ is an infinite cardinal number and $\kappa$ is any cardinal. Prove that $2^\gamma = \kappa^\gamma$.  One direction is obvious but I'm stuck with the other. i.e. showing $\kappa^\gamma\leq 2^\gamma$.  Any hints or solutions will be appreciated.
22520
I came to point where I suppose for case 1 that $A\subseteq B$ and conclusion is trivial. For case 2 I suppose that $A\not\subseteq B$ and try to prove $B\subseteq A$, but that gets me nowhere. Any pointers here are most welcome.
25990
Let $x,y,a,b$ be real numbers.  For $x&lt;y$ and $a&lt;b$, show the cardinality of $[a,b]$ equals the cardinality of $[x,y]$.  I did the above problem, by defining a linear function as a bijection between the intervals. The next part of the problem states,     Extend your result (from above) to open and half-open intervals.    I'm just not sure what exactly I'm supposed to do next, the question seems vague. Am I supposed to show $(a,b)\sim (x,y)$? $[a,b]\sim (x,y)$? $[a,b)\sim (x,y]$?
59402
I am trying to prove: If A is an infinite set and B is at most countable set, prove that A and $A \cup B$ have the same cardinality.  Set definitions here are from baby Rudin.   Attempt: There are 4 possible cases:  1.A is countable and B is finite.  2.A is countable and B is countable.  3.A is uncountable and B is finite.  4.A is uncountable and B is countable.  Then for each case I need to show that $A \cup B \sim A$. I can prove 1-3, but I am unsure how to prove case 4. And in general, it seems that my proof strategy is inefficient and there exist much faster proof then proof by cases.
94485
I'm pretty sure I've asked way too many questions on sets already(still, a big thanks to everyone who helped) and even though I'm starting to see what's going on there are still some issues, for example this question:  Prove that $\bigcup P(X)=X$, where $P(X)$ is the power set for the set $X$  So, what I can do is the following:  $a \in \bigcup P(X) \implies \exists b \subset X$ s.t. $a\in b \implies a \in X$ and for the other direction I can take $b=X$.  However, I'm not sure whether this is correct, and in particular, the bit with $b \subset X, a\in b \implies a \in X$  If I take $X=\{\{1\},\{2\}\}$, then $P(X)=\{\emptyset,\{1\},\{2\},\{\{1\},\{2\}\}\}$, hence $\bigcup P(X)=\{1,2,\{1\},\{2\}\}$ which is not equal to $X$ by ZF$1$  Where's the problem?  Thanks
154
I need to prove the following letting A and B be subsets of a universal set U. $$A\cup B = U \iff A^c\subset B$$ It just isn't straight forward to me.
48867
Proving that $[0,1)$ and $(0,1)$ have the same cardinality (without assuming any previous knowledge) can be done easily using Cantor-Bernstein theorem.  However I'm wondering if someone can build an explicit bijection between these sets.  It's easy to build a bijection between $(0,1)$ and $\mathbb R$, so a bijection from $[0,1)$ to $\mathbb R$ will also fit the bill.
31992
Possible Duplicate:          Let $X$ and $Y$ be sets, and suppose that $|\mathscr{P}(X)| = |\mathscr{P}(Y)|$ (where $\mathscr{P}$ denotes the power set).  Does it follow that $|X|=|Y|$?  Remark: It's obviously true for finite sets, as $2^m=2^n$ implies $m=n$.
105971
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!!
57200
Proof that the following equality is wrong: $$\mathcal{P}(A\Delta B) \setminus \{\emptyset\} = \mathcal{P}(A) \Delta\mathcal{P}(B) $$ I choose elements of sets $A$ and $B$ so that they are separate.  $A = \{1, 2\}$ $ \mathcal{P}(A) = \{\{1\}, \{2\}, \{1, 2\}, \emptyset\}$ $B = \{3, 4\}$ $\mathcal{P}(B) = \{\{3\}, \{4\}, \{3, 4\}, \emptyset\}$ $A\cup B = \{1, 2, 3, 4\}$ $A\cap B = \emptyset$  Now I see that: On the left side: $\mathcal{P}(A\Delta B) = \mathcal{P}((A\cup B) \setminus (A\cap B)) = \mathcal{P}((A\cup B) $ $ \mathcal{P}((A\cup B) =\{\{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{1, 2, 3, 4\},\emptyset \} $ $ \mathcal{P}((A\cup B) \setminus \{\emptyset\} =\{\{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{1, 2, 3, 4\}\} $ On the right side: $\mathcal{P}(A) \Delta\mathcal{P}(B) = (\mathcal{P}(A) \cup \mathcal{P}(B)) \setminus (\mathcal{P}(A) \cap \mathcal{P}(B)) = \{ \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\} \} $ Therefore: $\{\{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}, \{2, 3, 4\}, \{1, 2, 3, 4\}\} \neq \{ \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{3, 4\} \} $ i.e. $\mathcal{P}(A \,\Delta\,B) \setminus \{\emptyset\} \neq \mathcal{P}(A) \Delta\mathcal{P}(B)$. Is my proof valid? Are there any errors? I could take smaller $A$ and $B$ sets but I was scared that it wouldn't be so obvious that way.
102083
I'm well aware of the standard proof based on cardinality arithmetic to show that these two sets have the same cardinality and the question of defining a bijection between the two sets came up. I worked on it a little while and was wondering if there were any gaps, or a simpler method of coming up with such a bijection.   The main idea I had was to take a family of sequences indexed by the real numbers. So that each one of these sequences corresponds to a unique real number. Then I can map each one of these sequences to another number, by applying the binary function from $\mathbb R$ to $\{0,1\}$ and mapping the infinite binary sequence to a real number, use one of the standard bijections.  First let $E$ be a family of sequences indexed by the real numbers. Such that $E_{xi}\neq E_{yj}$ for all $x,y \in \mathbb R$ and $i,j \in \mathbb N$ also with the property that  $\mathbb R= \bigcup_{x \in \mathbb R, i \in \mathbb N} E_{xi}$.   What sort of theorems would I need to prove such a family existed, is it enough to note that the reals can be represented as a uncountable union of countably infinite sets?  Also let $P: 2^{\mathbb N} \rightarrow \mathbb R$ be you favorite bijection.  Now consider an element of $2^{\mathbb R}$ as a function $f: \mathbb R \rightarrow \{0,1\}$ and $g \in \mathbb{ R^R}$ in a similar manner. Then our bijection between the two sets will be the function $\Phi$ which maps $f$ to a function $g$, such that $g(x)=P(F(E_x))$ where $F$ applies $f$ to each element of the sequence $E_x$.   I'm fairly sure this is an onto and one-to-one function. The one-to-one property should follow because the sequences are disjoint, cover $\mathbb R$ and $P$ is a bijection. The onto argument is a little hazier, but it should be possible to recreate the function $f$, by first inverting each element with $P$ and then defining $f$ based on these sequences.
61029
Task is:  Let $A$ be a finite set and let $B$ be a subset of $A$ with $|A|=n$,$|B|=m$ and $0&lt; m &lt; n$.Find a formula for the number of subsets of $A$ that contain $B$ and prove your statement.  I assume this is $2^m$ + something. $2^m$ would count the subsets containing only elements of $B$, but i'm not sure how the sets of the form $(a_1,a_2,b_1)$ or similar would be counted. I'm having troubles with this so help would be appreciated.
23409
Is $2^{|\mathbb{N}|} = |\mathbb{R}|$? If so, how?  I was reading the Wiki page on the , and it says "Moreover, $\mathbb{R}$ has the same number of elements as the power set of $\mathbb{N}$", but I don't see how this is true?  I feel like it has something to do with binary, but I'm not too sure how it works? Do I have to show a map of all reals can be done in binary? I'm just very confused, and any advice would be appreciated!
9688
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?
56925
Prove that if $a^2=a$ for each infinite cardinal $a$ then $b + c = bc$ for any two infinite cardinals $b,c$.   I tried $b+c=(b+c)^2=b^2+2bc+c^2=b+2bc+c$, but then I'm stuck there.
41613
Let  $$A = \{x | x = 2n+1, n\in\mathbb{Z}\}$$ and  $$B = \{x | x = 2m-21, m\in\mathbb{Z}\}.$$  I am trying to prove $A =B.$ I understand that I need to prove  $A\subseteq B$ and $B\subseteq A$; But my difficulty is how to prove it.  Any help would be appreciated!
120485
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.