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The use of the word "contains" is a bit misleading, there. When we talk about a set $A$ "containing" a subset $B$, what we really mean is that $A$ contains all elements of the set $B$ (that is, for every $x\in B,$ we have $x\in A$). In that sense, we're saying that an empty bag is a bag that contains exactly what an empty bag contains: nothing at all. It is ...

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For this kind of question, it's often best to make a table. Here's the data when put into a table. $$\begin{array}{c|cc|c} &\text{ Is Reasonable } & \text{ Not Reasonable } & \text{ Total} \\ \hline \text{Is Worse} &363 & ??? &487 \\ \text{Is Not Worse} & ??? & ??? & ???\\ \hline \text{Total} &756 & ??? & ... 19 Sure, why not.$$\mathbb{R} = \bigcup_{k \in \mathbb{Z}, \ \alpha \in [0,1)} \{5k+\alpha, 5k+1+\alpha, 5k+2+\alpha, 5k+3+\alpha, 5k+4+\alpha\}$$11 There's nothing deep going on here. Its just that: Its often convenient to identify \mathbb{N} with the least infinite ordinal \omega. Its often convenient to identify each well-orderable cardinal number \kappa with the least ordinal \alpha such that |\alpha| = \kappa. This is called the Von Neumann cardinal assigment. Under these ... 10 The objects in set theory are sets. Only sets in \sf ZFC and its related theories. This means that if you want to interpret a mathematical object in set theory you need to assign it a set. Of course you are free to assign to it any set that you wish, as long as you have Tue axiom of replacement set theory is more or less interpretation agnostic (in the ... 10 Do you know what it really means for one set to be a subset of another set? If the set A is a subset of the set B, then we write A\subseteq B; now, to show that A\subseteq B, we must show that x\in A\to x\in B. The fact that \varnothing\subseteq\varnothing is really not that surprising if you are aware of what a so-called "vacuous truth" is. ... 9 How is a chair defined? Surely something as simple and fundamental as a chair should have a reasonable definition. In the dining room, the table is certainly not a chair. Because you don't eat from a chair. But in the university lobby, sometimes I would sit on the table when talking to someone who's standing up. And maybe I decided to sit on the floor and ... 8 what about the set of all positive reals? Let l be the least element. What can you say about \frac{l}{2}? 8 You can call them equinumerous. 7 Probably the most standard term of the sort you're looking for is "equipotent", though it isn't used particularly often (more often, people will just say two sets "have the same cardinality" or "are in bijection"). People who think about categories a lot sometimes say "isomorphic as sets". 7 You are confusing "subset of" ( \subseteq ) with "element in" ( \in ). This is possibly because you are confusing the set of elements (marbles) with the container (the bag) which holds the set. The bag is not the set, the set is the collection held by the bag. The set is empty, as there are no marbles held by the bag. This empty set ... 5 Suppose we could list the rational elements of [0,1] in order; specifically, suppose that [0,1] \cap \mathbb{Q} = \{ r_n : n \in \mathbb{N} \} with$$r_1 < r_2 < r_3 < \dots$$Then we'd need r_1 = 0 and r_2 > 0. But then \frac{r_2}{2} is a rational in [0,1], so r_2 = r_k for some k \in \mathbb{N}; and moreover k \ne 1 since r_1 ... 4 Your last statement "an empty bag is a bag with nothing but an empty bag inside it." is not correct. Because, A is a subset of B does not mean that B contains the set A. It simply means that A is another bag(container) like B(which is also a container) which contains the items taken from inside the bag B. 3 The function y = \tan(x) is bijective on \left(-\frac{\pi}{2},\frac{\pi}{2}\right). We would like to shift/stretch it so that it's bijective on (a,b). The period should be b-a, so we would at least have y = \tan\left(\frac{\pi}{b-a}x \right). Now translate it. 3 Since the other two answers mostly focus on the first part of your question, I'll take the second one: Does every non empty subset of positive real numbers contain a least element? If so, why? If it is not, then why not? The set of positive integers is subset of positive real numbers and contains a least element, which is 1. This means some non-empty ... 3 The original Peano's axioms (1889, page 1) started from 1 : Axiomata 1 \in N [...]. This "tradition" is followed by many mathematicians, like : Edmund Landau, Foundations of analysis : The arithmetic of whole rational irrational and complex numbers (ed or 1930, page 2) : Axiom 1 : 1 is a natural number. In the mathematical ... 3 Well, usually they do mean "the set," just different mathematicians mean different sets. It's like a language difference. There are lots of cases in math of this sort. For example, "the standard normal distribution" means something different to different mathematicians. Basically, good luck trying to sort this out. Historically, the natural numbers were ... 3 Notice that the sets 2\mathbb Z and 3\mathbb Z (as well as translations thereof) are both periodic and share 6 as a period - that is, you can reduce the problem mod 6 and note that, since your sets are equal mod 6 they are equal in general. This essentially tells us that your visual proof suffices as a rigorous one. To give a more algebraic proof ... 3 Suppose that A\cap B\neq \emptyset. Then there exists x\in A\cap B. If f(X)=(A,\emptyset), X\cap B=\emptyset, so x\not\in X. But X\cap A=A so A\subseteq X and x\in X. This is a contradiction. For the other direction, assume that A\cap B=\emptyset. Let (C,D)\in P(A)\times P(B). Then if X=C\cup D, f(X)=(C,D). 3 Here's one direction of part a). Suppose A\cup B=E, then for any X\subseteq E, X=(A\cap X)\cup (B\cap X). If f(X)=f(Y) then (A\cap X,B\cap X)=(A\cap Y,B\cap Y) which implies A\cap X=A\cap Y and B\cap X=B\cap Y. So,$$X=(A\cap X)\cup(B\cap X)=(A\cap Y)\cup(B\cap Y)=Y.$$Hopefully this will get you thinking along the right track. 3 The usual diagonal argument works fine "in reverse": Suppose g:\mathcal P(X)\to X, and consider$$ A = \{ g(Y)\mid Y\subseteq X, g(Y)\notin Y \} $$Let x=g(A). If x\notin A, then setting Y=A we see that x\in A, which is a contradiction. So x\in A. On the other hand x\in A implies that there is an Y such that g(Y)=x=g(A) and x\notin Y. ... 3 We say "A injects into B", and may write f: A \hookrightarrow B. 2 If x+6=6, then x=0 and it's the only one solution... And 0^2\neq 16 , hence the set A is empty... Where's the problem? The x in the set must satisfy both conditions, that is x+6=6 and x^2=16, but there are no such numbers. 2 Another option:$$\left\{n\in\mathbb{Z}_{\geq2}: m\in\mathbb{Z}\wedge1<m<n\implies m\not\mid n\right\}$$(the set of integers n that are at least 2, such that whenever m is an integer and m is between 1 and n, then m won't divide n) or$$\left\{n\in\mathbb{Z}_{\geq2}: m\mid n\implies |m|=1\vee|m|=n\right\}$$(the set of integers n that ... 2 How about this?$$\{n\in\mathbf{Z} : \forall a,b \in \mathbf{Z},\, n\mid ab \Rightarrow n\mid a \vee n \mid b\}

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Remember that when you're looking for overlapping elements of two groups, you're looking for their intersection. This is a big upside down U in set notation. An apostrophe signifies a "not". So answer a here essentially says, "every element of the men group that is also in the children group". a) $M \cap C$ b) $M' \cap C$ c) $M' \cap C'$ d) $C' \cap H ... 2 Yes, that is correct.$X$is not an element of$Z$, because$Z$contains only one element, and that element is$\{\{1,2,3\}\}$The only way for$X$to be an element of$Z$would be if$X=\{\{1,2,3\}\}$, but since$\{\{1,2,3\}\}$contains one element and$X$contains three elements (assuming that$1\neq 2$and$2\neq 3$and$1\neq 3$), this is not true. 2 Yes it look's correct, basically what you construct is a piece of the space that has the same cardinality as$\mathbb(R)$and that is only covered with countably number of points. On the same theme would be that each line touches the unit circle at at most two points, making those points where a line touches the unit circle countable, but the unit circle ... 2 Hint: Set$X_k^n$to be the set of integer sequences that have period$k$, starting at index$n$. Now observe that$X = \bigcup \limits_{(k, n) \in \mathbb{N}^2} X_k^n$. 2 Suppose$f$were not surjective. Then there is$b \in F$such that$f(a) \not = b$for all$a \in E$. That is, there is$b \in F$such that$f^{-1}(\{ b \})$is empty. What is another set which has empty preimage? Suppose$f^{-1}$were not injective. Then there are$A, B$distinct such that$f^{-1}(A) = f^{-1}(B)$. Equivalently,$\{ x \in E \vert f(x) \in A ...

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