A simple textbook reference would probably suffice here; I’m just not sure where to look, as I haven’t done much set theory.  Let $C$ be a class that is not a set. Is it true/provable/whatever that for any set $S$, there is an injection from $S$ to $C$? (I’m not even totally sure what the right way to phrase the question is, but hopefully the intuitive idea is clear.)
123552
Some proper classes are "too big" to be a set in the sense that they have a subclass that can be put in bijection with $\alpha$ for every cardinal $\alpha$. It is implied in  post that every proper class is "too big" to be a set in this sense, however I have been unable to prove it. It's true if every two proper classes are in bijection, but it's consistent with ZFC for there to be a pair of non-bijective classes.  So, is the following true in ZFC:     For all proper classes, $C$, and $\alpha\in\mathbf{Card}$, $\exists S\subset C$ such that $|S|=\alpha$?   If not, is there something reasonable similar that preserves the intuition about classes that are "too big to be sets"?
28071
I know that the usual ordering of $\mathbb R$ is not a well-ordering but is there an uncountable $S\subset \mathbb R$ such that S is well-ordered by $&lt;_\mathbb R$?   Intuitively I'd say there is no such set but intuitively I'd also say there is no well-ordered uncountable set at all, which is obviously wrong. I still struggle to grasp the idea of an uncountable, well-ordered set.
115977
I'm making my way through Thomas W Hungerfords's seminal text "Abstract Algebra 2nd Edition w/ Sets, Logics and Categories" where he makes the statement that the Continuum Hypothesis (There does not exist a set with a cardinality less than the reals and no set strictly greater than the natural numbers.) is neither true or false.  This is utterly baffling to me, If it's possible to construct a set between $\mathbb{N}$ and $\mathbb{R}$ then this statement is demonstrably false, but if not then the statement is true.  This seems to be a straitforward deduction, but many with a more advanced understanding of the topic matter believe CH to be neither.  How can this be?
50561
I've been trying to understand in a more formal way what a set actually is, but I have some questions. According to the axiom of regularity for every non-empty set A there exists an element in the set that's disjoint from A. That would mean that such element is also a set, right?   I read here:  , that in axiomatic set theory everything is a set, I understand that natural numbers are constructed from the empty set, integers are constructed from the naturals, rationals from the integers, and reals from the rationals. I can see how every element in such sets is also a set. But, for example, in the set of all the letters of the alphabet, or the sample space of an experiment when the possible results are not numbers, or the set of my classmates; it's not clear to me how their elements are also sets. So, are they really sets? Is every element in a set also a set?  Thank you
93389
Given an arbitrary set $X$, can one always find an element outside $X$? I think this question boils down to whether there exists a universal set that contains everything, including itself. But wiki says that assuming the existence of a universal set leads to . So given a set $X$, can one explicitly construct an element that does not belong to $X$? For example, is $\{X\}$ such an element?
69987
I know what a set is. I have no idea what a class is.   As best as I can make out, every set is also a class, but a class can be "larger" than any set (a so-called "proper class").  This obviously makes no sense whatsoever, since sets are of unlimited size. It's not like you get up to 25 elements and go "oh, hey, sets are limited to a maximum of 20 elements, so we'll have to call this 25-element thing something else".  Does anybody know what the actual difference between a set and a class is?  Several people have pointed out that you can write paradoxes which involve sets. I get that. But where do classes come into the picture?
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?
43365
I'm not sure how to approach the proof for this. I know this statement is vacuously true, because the empty set does not contain any accumulation points. But I'm having a hard time writing a proof for this. I have looked at other answers on here, however we have not learned what a "ball" is in lecture.
118176
Looking at the top answer on this page,    A user said,      There are models of ZF such that $A,B$ are sets for which exists   surjections from $A$ onto $B$ and vice versa, however there is no   bijection between the sets.   Can anyone give an example of this?  If A is the set of real numbers and B is the set of natural numbers , then there will be a surjection from A to B, but no surjection from B to A.  Is this what surjection even means.  Here is a surjection from real to naturals  1.5 → 1  1.7 → 1  2.3 → 2  2.6 → 2  f(A) = int(A) → B  But there is no surjection from the naturals to the reals (Cantor's diagonal proof)  So, how is it even possible to have "There exists surjections from $A$ onto $B$ and vice versa, and there is no bijection between the sets"? Did he, maybe, mean there is no way to find the formula for the bijection?
25123
I am trying to understand   How can a set contain itself? Can you show example of set which is not a set of all sets and it contains itself.
59118
Let $S$ be an infinite subset of $\mathbb{R}$ with the property that the existence of $S$ can be proved within ZFC (and in particular the definition of $S$ does not invoke the negation of the continuum hypothesis), but the cardinality of $S$ is undecidable within ZFC.  Are there some concrete examples of such sets?  I.e. something more constructive than $S:=$ some subset of cardinality $\aleph_1$?  This question results from musing about intuitively obvious equivalents of $\neg CH$ in order to find evidence against the validity of $CH$. My (undoubtedly too optimistic) thinking was that if some such example of $S$ is concrete enough, it would be perhaps intuitively obvious whether it should be countable or continuum, despite this being undecidable formally.  My best guess for the moment is something along the lines of  $$S=\left\{\sum_n\frac{1}{a_n}:\; (a_n)\;\text{is a strictly increasing sequence of primes}\right\}$$  but surely the cardinality can be determined (though not quite sure how).
37416
I have been using the following definition of infinity (from Dedekind): Set $S$ is infinite iff there exists a proper subset $S'$ of $S$ and bijection $f: S\rightarrow S'$. Occasionally, I have seen &quot;injection&quot; used instead of &quot;bijection.&quot; Exactly why is this wrong? Can anyone give me a counterexample demonstrating why these are not equivalent definitions?  Follow-up 8 years later: See my formal proof of the equivalence using a form of natural deduction and basic set theory at:
52817
DEFINITION 1:  A set $S$ is finite  if there is a bijection from the set $[n]=\left\{ 1, ..., n \right\} \text{for some } n\in N$   to $S$.  A set is infinite if its not finite.   Claim : $\nexists g:[n]\to A$ an injection, where $A\subsetneq [n]\;\forall  n\in N$  Proof of claim : Let $P(n) $ be the statement  $\nexists g:[n]\to A$ an injection, where $A\subsetneq [n]\;\forall  n\in N$  $P(1)$ true as $\nexists g:[1]\to A $ a function, where $A\subsetneq[1] (=\phi)$  Assume $P(n)$ to be true.  Consider for $P(n+1)$ : Let $A\subsetneq [n+1]$  if $\exists g:[n+1]\to A$ an injection then  Case $1.$ $n+1\in rng(g)$  Case $1.1$ if $g(n+1)=n+1$, then $g_{[n]}$, the restriction of $g$ on $[n]$ is injection from $[n]$ to $A\setminus\{n+1\}\subsetneq [n]$, which is a contradiction to induction hypothesis.  Case $1.2$ if $g(n+1)\neq n+1$, suppose $g(i)=n+1$ and $g(n+1)=j$ then we can always interchange the value of $g$ at $i$ and $n+1$ to go back to case $1.1$ and conclude contradiction.  Case $2.$ $n+1 \notin rng(g)$ then the injection $g_{[n]}:[n]\to A\setminus\{g(n+1)\}\subsetneq [n]$ again leads to contradiction.  Back to main proof:  Since every finite set of cardinality $m$ is in bijective correspondence to $[m]$, we conclude by our above claim that there can not exist an injection from a finite set to its proper subset.  Now, on contrary, suppose $N$ is finite  $\Rightarrow \exists f:N\to[n]$ for some $n\in N$  But $[n]\subsetneq N$ leading to contradiction. Hence, $N$ can't be finite. So $N$ is infinite.  Is this proof correct?
105317
Theorem:  No set is equinumerous with its power set.  Proof:  Let $A$ be a set. We want to show that if $f: A \to \mathcal{P}A$ (a random function) then $f$ is not surjective.  We define the set $D=\{ x \in A: x \notin f(x)\}$ and obviously $D \in \mathcal{P}A$.  We assume that there is an $a \in A$ such that $f(a)=D$.  Then we have:  $$a \notin D \leftrightarrow a \notin f(a) \leftrightarrow a \in D, \text{ contradiction}$$  Therefore for each $x \in A, f(x) \notin D$, i.e. $f$ is not surjective.  Could you explain me the proof from the point where we assume that there is an $a \in A$ such that $f(a)=D$?  We have show that $D \in \mathcal{P}A$ and we want to show that $D \notin f(A)$ in order to show that $\mathcal{P}A$ isn't the image of $f$. Why do we do it like that? Why do we assume an $a \in A$ such that $f(a)=D$?
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
120110
My question is, does there exist a set of subsets of the naturals, $S$, such that $S$ is uncountable, where for any two subsets $A,B \in S$ we have either $A \subset B $ or $B \subset A $?  Initially my thoughts were that we could construct a chain $A_{1} \subset A_{2} \subset \cdots $ and find a bijection by picking out one element which is 'new' to each set by axiom of choice and so we have bijected to a subset of naturals, however I'm not convinced this fully works.  Furthermore, I was thinking there may be a nice elegant solution by looking for such a set in either $ \mathbb{Q}^2 $ or $ \mathbb{Z}^2 $ for example, but any attempt seems to fail due to the 'ordering' part of the question.
101325
Can a set be neither open nor closed? An example would do. I cant think of any.   Thanks in advance!