I understand why this is the case comparatively , if a set is looked at as being a 'box', then the empty set is considered to be an empty box. So the set $\{1,2,3\}=A$ is considered to be a 'box' containing elements $1,2,3$, and $\{1,2,3,\emptyset \}=B$ is the set containing the elements  $1,2,3$ and an 'empty box'?   What I struggle with is how this intuition would make sense given the empty set is a subset of every other set.   Also, I know that $\{\emptyset \} \neq \emptyset$, so would it be the case that for the given set $A$, $\emptyset \subset A$, but that $\emptyset  \notin A$, as the set containing no elements is not in the set $A$, but $\emptyset \in B$ is true, right?  In the midst of forming this question I thought of a way of looking at this question, but I am not sure whether it is correct:  So continuing with the comparison of looking at a set being a 'box', $\emptyset \subset A$ simply means that the the set $A$ has some 'free space' in the box, and so the set could be visualised by $A=\{1,2,3,\qquad\}$?, and every set\'box' has some 'empty space', whereby the 'empty box' represents simply  the collection of that void.  Does this analogy capture the true picture of Why $\{1,2,3\} \neq \{1,2,3,\emptyset \}$ and $\emptyset \subset A$?  Just a very quick question,  Is the power set of $C=\{1,\emptyset\}$,  $$\mathcal{P}(C)=\{\{1\},\{ \emptyset \},\{1, \emptyset \},\emptyset \}?$$
122243
If the empty set is a subset of every set, why it isn't written with the elements of a set? like so $\{1,2,3,\emptyset\}$   Or why isn't $\{\emptyset,\{a\}\}=\{\{a\}\}$? I know one has two elements and the other has 1 but since the empty set is a subset of both, then why it isn't being mentioned explicitly in the definition of the set?
101960
This is a homework question of mine that Ive answered and I'd really like some feedback on my solution.       Prove or find a counter example to the claim that for all sets $A,B,C$, $$\text{if  }A\cap B=B\cap C=A\cap C=\emptyset \text{ then } A\cap B\cap C\neq \emptyset$$.   Since I can only either prove it is true, OR find a counter example I have gone with the latter because I think this statement is not true. My counter example is as follows:  Let $A = \{1,2,3 \}$ , $B = \{4,5,6 \}$ and $C = \{7,8,9 \}$. It is then clear that $A ∩ B =  B ∩ C =  A ∩ C = \emptyset$ .  If we let $x ∈ A\cap B\cap C$ then $x$ must be in $A$ and $B$ and $C$. Since this cannot be possible because there are no common elements then this implies that $A∩B∩C$ has to be equal to a null set i.e $A∩B∩C = \emptyset$ and thereby showing that (at least for this scenario) if $A \cap  B =  B \cap  C =  A \cap  C = \emptyset$ then $A\cap B\cap C = \emptyset$.  Is this okay? Any feedback will be greatly appreciated thank you!
56405
I have came across this in a textbook:     $\{2\}\nsubseteq\{\{2\}\}$ but $\{2\}\in\{\{2\}\}$   I understand that $\{2\}$ is an element (member) of the other set but considering $\{2\}$ is a set itself, why is it not a subset?
42501
Let $A$ be a set of size fifteen. Let $\mathcal{P}(A)$ denote the power set of $A$, that is the set of all the subsets of $A$. How many elements does $\mathcal{P}(A)$ contain?   This is the same as being $0$ or $1$, binary sets of size $1\to 14$?  Does this mean the answer is  ${15\choose 0} + {15\choose 1} + {15\choose 2} +\dots +{15\choose 13}+{15\choose 14}+{15\choose 15}$?  Or is there something I am not considering?
107055
Prove or disprove:  The set: $\{ \emptyset\}^{\Bbb N}$ has only one element.  The set $\emptyset^{\Bbb N}$ is empty.    I'm pretty sure both are true, for 1. all the natural numbers go to the empty set and that is the only element. Where as 2. all the natural numbers go to nothing so that's why it's empty. (Same as $ {\Bbb N}^{\emptyset} $ nothing goes to something is nothing (I think)). The problem is I have no idea how to show this so advice would be appreciated.  Edit: My defintion of a function is: $f:B\to A$ if  $(a_1,b),(a_2,b)\in f \ \Rightarrow \ a_1=a_2$ So for 2. I can simply say that: $(1,\emptyset),(2,\emptyset)\in f \Rightarrow a_1\neq a_2$  ?
112855
I am having difficulty understanding  :     Every non-empty set $\rm A$ contains an element $\rm B$ which is disjoint from $\rm A.$   So from my understanding if I have a set like:   $$\{1, 2, 3, 4, 5\}$$  Then one of $[1,2,3,4,5]$ is not an element in $\rm A?$ huh?
1148
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.
15781
Show that: Let A be a set and let $P(A)$ be the set of all subsets of $A$. Then there is no surjection $f: A→P(A)$.  Here is what I thought:  if $A=\{a,b\}$ then it has only two elements where $P(A)=\{∅,\{a\},\{b\},\{a,b\}\}$ has 4 elements. Therefore $f:A→P(A)$ cannot be surjective. But I have some problems:  1) How is it possible that any $f$ function to take $\{a\}$ from set $A$ to $\{a,b\}$? Maybe because I am thinking mainly about functions with real values like $f(x)=2x$, I find it a little bit strange that a function to take an element of a set to another set which has more elements. Is it possible?  edit: Now I thought that if $f(x)$ is $\sqrt{x}$, then $f(4)=±2$ which means it took an element from a set to a set which has 2 elements. But still I find it kind of strange to denote $f(\{a\})=\{a,b,c,...\}$  2) How can I construct a explicit proof for this question?  Regards
84176
Let $A$ to be a nonempty set and $B= \emptyset$; then $ A \times B$ is a set. And let $F$ be a function $A \to B$. Then $F \subseteq A \times B$. By the axiom of specification, $F$ must exists (if I didn't mess up something).   But the book I'm reading, Elements of Set Theory by Enderton, says that no function could have a nonempty domain and an empty range, and no more detail is given.  So my confusion arises. His statement against my proof. The only axiom as far as I know to prove such a set does not exist is the axiom of regularity. But I can't give such a proof. So I need help. I hope someone could clearly explain why such a function doesn't exist, and why this doesn't contradict the axiom of specification.    Let me explain my confusion in more detail:  First of all, I know that $A \times \emptyset$ is empty. But $ \emptyset \times A$ is also an empty set, yet there is no problem with functions with an empty domain.  The book I'm reading defines a function as:      "A function is a relation such that for each $x$ in $\operatorname{dom} F$ there is only one $y$ such that $x \mathop F y$."   and a relation as:     "A relation is a set of ordered pairs."   Now, let me define the function $F$ as: $$F =  \{\langle x,y \rangle \mid x \in A \text{ and } y \in B \text{ and ... other conditions} \}$$  so it's equal to: $$F = \{ \langle x,y \rangle \in A \times B \mid \text{... other conditions} \}.$$  Doesn't that mean that $F$ meets the conditions of the axiom of specification, and therefore exists?    What I would specifically like to ask is:   How do you define a function $F$ (as precisely as possible)? Why does the argument above not imply that a function $F: A \to B$ always exists? Does $F \subseteq A \times B$ still hold when $B = \emptyset$?   (The answer to #2 cannot simply be that $A \times B = \emptyset$, because $ B \times A = \emptyset$ as well, but a function $B \to A$ does exist.)
35828
In terms of purely set theory, the axiom of choice says that for any set $A$, its power set (with empty set removed) has a choice function, i.e. there exists a function $f\colon \mathcal{P}^*(A)\rightarrow A$ such that for any subset $S$ of $A$, $f(S)\in S.$ Is this correct?  My question then is about proving this fact, so that we do not need to put it as an axiom. Now as per the research done on this single object- Axiom of Choice, I believe here that there should be some falsity in my argument. I do not find the mistake.  For any $S\in \mathcal{P}^*(A)$, since $S\neq \emptyset$, $\exists s\in S$. Define $f(S)=s$. Then $f$ is a choice function.  This was showing me that the axiom of choice is proved, but then why it had been put as an axiom? For example, in  book, the author asserts that      It is a metatheorem of mathematical logic that it is impossible to specify the function that assigns to each non-empty subset of $\mathbb{R}$, an element of itself.     There are several notes and books on axiom of choice, but here I am trying to understand through doing some argument for some problem, where problem actually arises.
81044
Let $A$, $B$ be sets and denote by $A\leq^* B$ an assertion that if $A$ is non-empty, then there's a surjection from $B$ onto $A$. This might be understood as a dual definition of cardinality. (Typically we use injections to capture the notion of A not being bigger than B. The $\leq^*$ notation is to be read as B not being smaller than A.)  Is this relation trichotomous without using the axiom of choice? Meaning for every two sets $A$ and $B$, it holds $A\leq^* B$ or $B\leq^* A$. (I would be surprised). Would trichotomy imply some interesting extra-ZF axiom?  More broadly, is there anything to be said about $\leq^*$? For example how does it relate to the usual comparisons of cardinality by injections?  I suspect there's no clear answer, since it is somehow reminiscent of Partition Principle ($A\leq^* B \implies A\leq B$) and Weak Partition Principle ($A\leq^* B \implies B\not&lt; A $) which both are independent of ZF.
119909
I am only a first semester undergraduate (of mathematics) so I would appreciate the answers to not be too complicated. Also English is not my first language, so my explanations might be a little off.    While trying to prove that the Powerset $\mathcal{P}(\mathbb{N})$ is uncountable I ran into something that I wasn't able to understand.  So, I define the set $$ \mathbb{N}_n=\{\{a_1,a_2,\ldots,a_n\}:a_i\in \mathbb{N} \; \forall i \text{ and } a_i \neq a_j \; \forall i \neq j\} \text{ for } n \in \mathbb{N} $$ basically the set goes like this $$ \mathbb{N}_0=\{ \emptyset \} \\ \mathbb{N}_1=\{ \{1\},\{2\},\{3\},\ldots \} \\ \mathbb{N}_2=\{ \{1,2\},\{1,3\},\{2,3\},\{1,4\},\ldots \} \\ \vdots $$ and we can conclude that $$ \text{if } \mathbb{N}_i \cap \mathbb{N}_j = \emptyset \text{ then } i \neq j \\ \bigcup\limits_{i \in \mathbb{N} \cup \{0\}} \mathbb{N}_i = \mathcal{P}(\mathbb{N}) $$ Okay, so there are a countable number of sets $\mathbb{N}_i$. Also the set $\mathcal{P}(\mathbb{N})$ is uncountable. Also as briefly mentioned in yesterdays lecture (at my university), a countably infinite union of countable sets is a countable set. From this I have concluded that:  $ \text{There exists an } i \in \mathbb{N} \text{ such that the set } \mathbb{N}_i \text{ is uncountable.} $  Now, there is the possibility of a mistake that I have made in a previous statement. But in the case that I haven't, this last statement seems very counter intuitive to me. I have also tried proving / disproving it but failed.  My question is where have I either made a mistake, or if I haven't how do I prove or disprove my final statement.
118585
Theorem: If $X$ is a set, then $X$ is not equivalent to its power set.  Proof: suppose for a contradiction that $f:X\to P(X)$ is a bijection. Define $B:=\{x \in X, x\not\in f(x)\}$. Because $f$ is surjective (onto), there is a $b$ in $X$ with $B=f(b)$. Now $b$ in $B$ would imply $b$ in $B=f(b)$. By definition of $B$, this would mean that $b$ is not an element of $f(b)=B$. Thus we infer $b$ is not an element of $B$. But then $b$ is not an element of $B=f(b)$, which by definition of $B$ forces $b$ in $B$, a contradiction.  I understood this proof until he reached the part that says "Now $b$ in $B$ would imply $b$ in $B=f(b)$..."  This seems problematic to me because the assumption $b$ in $B$ seems unwarranted. Since $B$ is a subset of $X$ it makes sense if we had $b$ in $B$ and then arrived at $b$ in $X$ but the reverse does not seem like a valid step and I'm left wondering if the resulting contradiction that arised came simply from the assumption $b$ is in $B$ and therefore only means that $b$ is not not an element of $B$, And that this has no bearing on the bijection part of the argument. In fact I dont even see how this connects to the part of the argument that used the onto property of bijectiveness.
89973
From what I have been told, everything in mathematics has a definition and everything is based on the rules of logic. For example, whether or not .  My question is what the definition of a set is?   I have noticed that many other definitions start with a set and then something. A  with an operation, an equivalence relation is a set, a , even the  of other sets containing the empty set.  I understand that there is a whole area of mathematics (and philosophy?) that deals with . I have looked at a book about this and I understand next to nothing.  From what little I can get, it seems a sets are "anything" that satisfies the axioms of set theory. It isn't enough to just say that a set is any collection of elements because of various paradoxes. So is it, for example, a right definition to say that a set is anything that satisfies the ZFC list of axioms?
29057
Some explanations:  A set S is  if there exists an injective function $f$ from $S$ to the natural numbers ($f:S \rightarrow \mathbb{N}$).  $\{1,2,3,4\}, \mathbb{N},\mathbb{Z}, \mathbb{Q}$ are all countable.  $\mathbb{R}$ is not countable.  The  $\mathcal P(A) $ is defined as a set of all possible subsets of A, including the empty set and the whole set.  $\mathcal P (\{\})=\{\{\}\}, \mathcal P (\mathcal P(\{\}))=\{\{\}, \{\{\}\}\} $  $\mathcal P(\{1,2\})=\{\{\}, \{1\},\{2\},\{1,2\}\}$  My question is:  Is $\mathcal P(\mathbb{N})$ countable? How would an injective function $f:S \rightarrow \mathbb{N}$ look like?
29946
I've heard of some other paradoxes involving sets (i.e., &quot;the set of all sets that do not contain themselves&quot;) and I understand how paradoxes arise from them.  But this one I do not understand. Why is &quot;the set of all sets&quot; a paradox?  It seems like it would be fine, to me.  There is nothing paradoxical about a set containing itself. Is it something that arises from the &quot;rules of sets&quot; that are involved in more rigorous set theory?
66884
In naive set theory, I believe ∅ = {∅} = {{∅}} is correct, but just wanted to make sure that I understood this correctly.   ∅ is an empty set, so having an empty set as an element of a set that contains nothing else is pretty much the same thing as e.g. marking the number 2 as 2.000000, right? Doesn't the symbol ∅ intrinsically imply that this is an empty set which contains it self?