I know that if for any set $U$, the power set of $U$ forms a monoid(a commutative one) under set union, because it is closed under the operation, is associative and it has an identity element $\emptyset$.  Following the argument that this is true for any set, it seems reasonable to think about the class of sets forming a monoid with $\emptyset$ as its identity, but, is it correct to say so? Or it must be a set even though there is a monoid for any set?
121228
My question if it possible for us to have, for example, a monoid(any algebraic structure really) with a class as the collection of the elements, instead of a set.  An example would be the class of sets and set union as addition. Clearly, it is closed under the operation, is associative and it has an identity element, so it is a monoid. Is it possible to say so?
54372
I am presented with the following task:     Show that if $U$ is the collection of all units in a ring $\langle R, +, \cdot\rangle$ with unity, then $\langle U, \cdot\rangle$ is a group.   I am still not confident with proofs in Abstract Algebra, and I'd like to receive some verification/slaughter/constructive critique on my reasoning here. I do not know if there are any mistakes, or how serious they are.     We know that there is an identity element, since the unity itself is a unit, and therefore in the group. It follows directly from the definition of a unit that there exists an inverse for every element in the group, and associativity follows directly from the Ring-axioms. As for closure, we need to show that the product of two units is also a unit. Let $a, b \in U$ be two elements with the property that $a\cdot b \ne e$, where $e$ is the identity element in $\langle U, \cdot \rangle$. We know that there exists elements $a', b' \in U$ with the property that $a' \cdot a = e = a' \cdot a$ and $b' \cdot b = e = b \cdot b'$. Let $a', b'$ be two such elements with the property that $a' \cdot b' \ne e$. We may now compute $(a\cdot b) \cdot (a' \cdot b')$, and from the Ring-axioms we have associativity under multiplication, so we may write the expression as $a \cdot (b \cdot a' \cdot b') = a \cdot (b \cdot b' \cdot a') = a \cdot (e \cdot a') = a \cdot a' = e$. Thus we have shown that the product of any two elements is a unity, which leads to closure. All the group-axioms are met and $\langle U, \cdot \rangle$ is a group.    I have to apologize if you find the argument needlessly long; I've been having some trouble with assuming things I am not supposed to assume (like what I'm going to prove), so I wanted to stay clear of that.
82763
In ZF, the Power Set Axiom (POW) says that given any set $A$, there exists a set $\mathcal{P}(A)$ such that $$ \forall a(a\in\mathcal{P}(A)\leftrightarrow a\subseteq A)\tag{1} $$     Questions:         On many occasions, I've seen POW stated like this: "For any set $A$ there is a set consisting of all subsets of $A$."      Would a more precise formulation be: "For any set $A$ which is known to exist there exists a set containing all subsets of $A$ which are known to exist."?   An instance of the Axiom Schema of Separation (SEP) essentially gives the existence of a subset of a given set. So, why doesn't POW imply SEP? Is it because POW doesn't give the existence of the individual subsets but only the set containing them all?   Conversely, does SEP and the remaining ZF axioms imply POW?
56262
I have a non-empty set $R$ with a binary operation $*$. If the pair has an identity element $e\in R$ and  $$(a*b)*(c*d)=(a*c)*(b*d)$$  holds for all $a,b,c,d \in R$, how do I then prove that this is associative?   I understand the concept that $x*(y * z) = (x * y)*z$, but I am unsure how to apply it here.
33233
$\{\emptyset\}$ 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 $\{\emptyset\}$".
41088
We have learned that for universal set $U$ and  for set $A$ belonging to $U$, $A^c=U-A$ but what about for the empty set?
77527
(Reference : Fraleigh, A first course in abstract algebra)     Prove that a nonempty set $G$, together with an associative binary operation * on $G$ such that $a*x=b$ and $y*a=b$ have solutions in G, $\forall a,b\in G$, is a group.   Because * is an associatie operation, we have to prove that: there exists $e\in G$ such that $e*a=a=a*e$ for any $a\in G$, and that for any $a\in G$ there exists $\alpha \in G$ such that $a*\alpha=e=\alpha *a$.  This is very confusing, first I tried to prove the identity exists. If we consider a=b, then $a*x=a=y*a$, and because of the hypothesis $x,y\in G$ always exists, now we have to show that $x=y$, but I just can't do this with out inverses, so I tried to prove that the inverses exists, however I have no idea how the identity looks like, I'm stuck.  And I tried to get unstuck, so I thought particular cases, first if $G$ had only one element, then what would happend if it had just two elements... but I remembered that we don't know if G is finite or not, and I think this makes things a little bit more complicated.  What can I do?
71778
I'm only in high school, so excuse my lack of familiarity with most of these terms!  A monoid is defined as "an algebraic structure with a single associative binary operation and identity element."   A binary operation, to my understanding, is something like addition, subtraction, multiplication, division i.e. it involves 2 members of a set, a single operation, and the resulting third member within that set. And an identity element is a special type of element of a set, with respect to a binary operation on that set, which leaves other elements unchanged when combined with things. Examples are "$0$" as an additive identity and "$1$" as a multiplicative identity.   How does this definition correspond to a category with a single object? What are some examples of monoids?
32316
So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity element, other than 0. Because if 0 is the identity element, then this group won't have inverses.  The set explicitly excludes -1, which I found to be its identity element, which makes going about proving that this is a group mighty difficult.
7146
In Munkres’ topology, there are two instances where we would have to represent the empty set as a union of sets.  First, we have a lemma that says if $X$ is a set; and $\mathcal{B}$ is a basis for a topology $\tau$ on $X$, then $\tau$ is equal to the collection of all unions of elements of $\mathcal{B}$.  Later on, we define a subbasis $S$ of a topology on a set $X$ to be a collection of subsets of $X$ whose union equals $X$, and the topology generated is the collection of all unions of finite intersections of elements of $S$.  So there are a few questions I wanted to ask about these just to make sure that I'm understanding them correctly. First in the case of the subbasis, when Munkres says finite intersections, he's allowing for a set to be intersected with itself right? There's nothing that says the sets have to be different? This is important because we will need the elements of S to be in the collection of finite intersections so that we can get the full set X.  Secondly, and more importantly, when dealing with the empty set, how do we say that this set is a union of anything? If I have a collection of sets D and I take the empty collection $\mathcal{A}$, then can I say the union of the empty collection is the empty set? I think this is ok to do, and Munkres mentions this earlier in the book, so is he allowing this to occur when he says all unions of elements from a collection of sets? I suppose its not that hard to understand but I guess I'm not used to thinking this way.  Lastly, in the proof of the lemma I mentioned first, Munkre's proves that every open set is a union of basis elements in the following way. Let $U \in \tau$, choose for each $x \in U$ an element $B_x$ of the basis such that $x \in B_x \in U$. Then $U = \bigcup_{x \in U} B_x$, so $U$ equals a union of elements of the basis.  I'm having trouble interpreting this statement if we start with $U=\emptyset \in \tau$. Then there is no $x \in U$, so how do we interpret the rest of the statement?  Thank you
29860
If a monoid is a category with a single object, is there a "monoid-theoretical" concept that the categorical product translates to? As an analogue, in a poset the product translates to the notion of meet (and the coproduct to its dual, the join).  Just by unwinding the definition one can see that this boils down to the existence of two monoid elements $p_1$ and $p_2$ such that for every $x$ and $y$ in the monoid there's a unique element $\langle x,y \rangle$ that factors as $x = p_1\langle x,y \rangle$ and $y = p_2\langle x,y \rangle$; the question is whether the concept is a known and/or useful one.  Incidentally, if the monoid in question is a group then the existence of a product makes the whole thing collapse into the trivial group; this might or not hint that the concept I'm looking for is "too boring" to have a name. In fact it suffices for one of the projections to be right cancellable, or for the pair $\langle x,x\rangle$ to be left cancellable for some $x$, to make the whole thing collapse.
86166
Prove or Disprove: For any family of sets $\{A_n\}_{n\in\mathbb N}$  $$\bigcup_{n=1}^\infty\mathcal P \left({A_n}\right)\subseteq \mathcal P \left({\bigcup_{n=1}^\infty A_n}\right)$$  How do I approach proving this? I know how to unpack the definition of powersets ($\mathcal P \left({A}\right) = \{x | x \subseteq A\}$) but I'm not sure what else I can do. I've done powerset proofs before but none involving indexed family of sets.
87396
I am not very well versed on set theory or syntax, but I thought I knew the basics.  However, in a book about databases I am reading now, the author uses $2^x$ to signify "a set of $x$."  For example, $2^{\text{dogs}}$ is a set of $\text {dogs}$, etc.    The author never really explained this or why he does it, I just picked up the meaning from context.    I am not sure why the exponent operator is used, nor am I sure what the number $2$ has to do with it.  The sets being represented are NOT powers of $2$ (in size)... they come in all sizes.  Is this a valid notation?  I have not seen it anywhere before...
87311
A semiring $\Pi$ on a set $X$ is a non-empty family of subsets of $X$ with the following properties.     1) $P \cap Q \in \Pi$ whenever $P\in \Pi$ and $Q\in\Pi$.      2) $P - Q$ is a finite disjoint union of members of $\Pi$ whenever $P\in \Pi$ and $Q\in\Pi$.   Let $\lambda: \Pi\rightarrow [0, \infty]$ be a non-negative extended real-valued set function such that $\lambda(\emptyset) = 0$. Suppose $\lambda$ is additive, i.e. if $P, Q_1, Q_2$ are members of $\Pi$ and $P = Q_1\cup Q_2$ is a disjoint union, then $\lambda(P) = \lambda(Q_1) + \lambda(Q_2)$.  Then $\lambda$ is finitely additive? In other words, if $P, Q_1, \cdots, Q_n$ are members of $\Pi$ and $P = \bigcup_{i=1}^n Q_i$ is a disjoint union, then $\lambda(P) = \sum_{i=1}^n \lambda(Q_i)$?  I came across this problem when trying to solve this question:   I also found this is an exercise in Halmos's Measure Theory, Ch. II, sec. 7. It has some hints, but I was unable to solve it.
47573
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?
43890
$G$ is the set of all subsets of a set $A$, under the operation of $\triangle\;$: Symmetric Difference of sets.  $A$ has at least two different elements.  I need to check if this is a group, and if it does to show if the group is abelian and/or finite.  Associativity - easy from Symmetric Difference. Identity element - empty group. Inverse element - each element is inverse to itself. Am I right?  abelian? finite?
63065
It has come to my attention that the pairing axiom and the separation axiom schema are rarely listed since they follows from the replacement axioms. I see how this works for the pairing axiom, since we can take some set, say $A=\{\emptyset,\{\emptyset\}\}$, which exists by the power set axiom and the existence of the emptyset, and define a nominating formula on $A$ by $$ \phi(u,v)\colon (u=\emptyset\land v=a)\lor(u\neq\emptyset\land v=b) $$ to see that $\{a,b\}$ is a set for any sets $a$ and $b$. However, I don't see it for the separation schema. Could someone enlighten me on how they follow? Thanks.