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Based on your remark that you allready proved existence I preassume that you have proved that $A:=\cup\mathcal F$ satisfies the conditions (a) and b). Let it be that the set $A'$ also satisfies these conditions. So: (a') $\mathcal{F}\subseteq\wp\left(A'\right)$ (b') $\forall B\left[\mathcal{F}\subseteq\wp\left(B\right)\implies A'\subseteq B\right]$ Then ...

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HINT: Let $A=\bigcup\mathscr{F}$, and suppose that $B$ also has the desired properties. Then on the one hand (b) applied to $A$ tells you that $A\subseteq B$, and on the other hand (b) applied to ... ?

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$\{0,1\}^\Bbb N$ is the set of sequences of $0$s and $1$s. For example, $001101010\dots$. $\{0,1,2\}^\Bbb N$ is the set of sequences of $0$s, $1$s, and $2$s. For example, $110210100\dots$. Define $f:\{0,1,2\}^\Bbb N\to\{0,1\}^\Bbb N$ by the following. For any sequence $S\in\{0,1,2\}^\Bbb N$, let $f(S)$ be $S$ with all $1$s replaced by $10$s, and all $2$s ...

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Consider the inclusion map $i:\{0,1\}^{\mathbb N}\to \{0,1,2\}^{\mathbb N}$, i.e. for a sequence $x:=\{x_n\}\in\{0,1\}^{\mathbb N}$, $i(x)_n = x_n$ for all $n$. Clearly $i$ is injective, so $\#\{0,1\}^{\mathbb N}\leqslant \#\{0,1,2\}^{\mathbb N}$. Now consider the map $f:\{0,1,2\}^{\mathbb N}\to\{0,1\}^{\mathbb N}$ where $$f(x) = ... 1 Since you already know \{0,1\}^{\mathbb{N}} \cong \mathbb{N}^{\mathbb{N}}, there is an injective map \{0,1,2\}^{\mathbb{N}} \to \mathbb{N}^{\mathbb{N}} \cong \{0,1\}^{\mathbb{N}}. There is also an injective map in the other direction, just the inclusion. Hence, you may apply the Schröder-Bernstein theorem. 1 Why not just use the Schröder-Bernstein theorem? Let we build an injective map from \{0,1,2\}^\mathbb{N} to \{0,1\}^\mathbb{N}. Given A\in\{0,1,2\}^\mathbb{N}, represented by \{a_n\}_{n\in\mathbb{N}}, with a_n\in\{0,1,2\}, we may consider the binary representation of:$$ r_A = \sum_{n\geq 0}\frac{a_n}{10^{n+1}}.$$In the opposite direction, given ... 2 Hint: It is much easier to show that$$\{0,1,2,3\}^{\mathbb N}\sim \{0,1\}^{\mathbb N}$$and then show that there there are 1-1 functions:$$ \{0,1\}^{\mathbb N}\to \{0,1,2\}^{\mathbb N}\to \{0,1,2,3\}^{\mathbb N}$$2 The claim that the diagonal argument proves is: There is no surjective (i.e., onto) function \mathbb N\to\mathbb R. (This implies that there can be no bijection \mathbb N\to\mathbb R either, because bijections are surjective functions, and thus that \mathbb R is, by definition, not countable). The presentation of the argument is often phrased as ... 0 Suppose, for example, that the first number on the original list is \frac 49=0.\dot 4 that 0.4 is the second number on the original list, and that 0.4001 is the third number on the list. With your truncation the list begins 0.4; 0.40; 0.400 - so your method can introduce duplicates into the list. You haven't controlled for this effect, and as you ... 0 Subsets of P(P(\phi))=\{\phi,\{\phi\}\} are \{\phi\},\{\{\phi\}\},P(P(\phi)) itself and empty set i.e. \phi again. (Remember that empty set and the whole set are always subsets of a given set.) Hence P(P(P(\phi)))=\{\phi,\{\phi\},\{\{\phi\}\},P(P(\phi))=\{\phi,\{\phi\}\}\}. 3 Probably the clustering of empty set symbols and nested set braces is the most confusing here. The empty set has only one subset: \emptyset itself, hence P(\emptyset) is a one-element set with \emptyset as only element. For any one-element set X=\{a\} we have P(X)=\{\emptyset,X\} as the empty set and the full set itself are the only subsets of ... 1 We can solve this problem rather easily by recalling that if A is a set with cardinality k, then the set P(A) has cardinality 2^k. First, the cardinality of \phi is 0. Applying the above principle, the cardinality of P(\phi) is 2^0=1. Then the cardinality of P(P(\phi)) is 2^1=2. Finally, the cardinality of P[P(P(\phi))] is 2^2=4. -3 think a=\phi ,b=\{\phi \}, then P[\{a,b\}]=\{\phi,\{a\},\{b\},\{a,b\}\} 2 Yes, De Morgan laws hold for arbitrary unions and intersections, as can easily be verified directly. The proof in the finite case is essentially the same as for the infinite case. 0 Okay: G={(y,x): (x,y) ∈ F) what does it mean? F, G, R, S are all sets. R is going to have a bunch of elements, say {r, s, t}. S is going to have a bunch of elements, say {a, b} Then the notation R x S is a set whose elements are all ordered pairs where the first term of the pair is in R and the second term is in S. We write this as RxS = {(x, y)| x in ... 0 Of course (as Wojowu points out in the comments) V_\omega is an example of such a set (as indeed is any V_\lambda for \lambda a limit ordinal), but we can also prove in ZFC that the set$$\{\{\}, \{\{\}\}, \{\{\{\}\}\}, . . .\}$$of finite Zermelo ordinals exists, which I think is what you were specifically looking for. We can prove this set exists ... 1 I know it's old, but here's how I solved it using logical equivalences. For people who coming from the future, I was required to solve it like this. We know that: A=\{x|x\in A\}, which reads: there is an x in the set of A. And we know that the complement of A is A^c=\{x| x\notin A\}, which reads: there is an x which is not in A. We start with (A^c)^c ... 0 i think it's because of the fact (\cup_{i=1, i\neq j}^{k} A_i)\cap A_j=\phi  0 You're way over thinking this, and I know this question is old, but for anybody else who sees this in the future, this is what the answer should look like. 1 "not all alpha's are men" \Leftrightarrow"there is an alpha who is not a man". i.e.$$ \exists a \in A \text{ such that } a \not\in M $$2 You could write A\backslash M\ne\emptyset. Meaning that when you take all the men out of the alphas, there are alphas remaining. 0 {a < x < b} is not empty so there is a number x_0 in the set so a< x_0 < b. But is {a < x < x_0} Empty? Nope for the same reason you gave. So there is a number x_1 such that a < x_1 < x_0 < b. Repeat inductively to get an infinite number of these x_i all of which are in {a < x < b} ===== or.. you have ... 1 If your injective function has domain \Bbb R you have proved even more. You have proved that the codomain is uncountably infinite (which it is here), but you were not asked for that. You have found one number in (a,b), which can be the image of 1. Now can you use the number you found to find another? That one can be the image of 2 and so on. 0 If x is not in A then f(x) is not defined. so g(f(x)) is not defined so x is not in the domain of gf. If a is in A then g(f(a)) exists. So x is in the domain of A. So domain of gf is A. You haven't actually claimed that B is the codomain of f. (The codomain could just be a proper subset of B). Nor that the codomain of g is all of C. In general, f ... 1 sounds ok, but i would recommend switching the first and the second sentence: Since g is surjective, for every c \in C, there's b \in B such that g(b) = c. Since f is surjective, then for this b \in B there exists a \in A such that f(a)= b. Then g(f(a)) = g(b) = c. 1 "Length" is an inappropriate word here, partly because it's confusing and potentially ambiguous. One can say that the set of all numbers between 0 and 1 has finite "length", but it has infinitely many members. It is infinite in the sense usually used when talking about sets, i.e. it has infinitely many members. It is also bounded, both in the sense ... 0 By definition a set B is a subset of A iff every element of B in A. So, the largest subset of a finite set A has exactly as many elements as A, but no more. 1 The proof is very intuitive (as you probably are feeling). But it can be written elaborately as follows, if you wish. Your claim: For any finite set F, there exists an infinite subset I. Try to prove: Let F be a finite set defined as F = \{f_1, f_2, \ldots , f_n\}, where n = 1, 2, \ldots Let I be an infinite set defined as I = \{i_1, i_2, \ldots, ... 0 I assume that S^R is the set of functions R \rightarrow S, a "function" is defined as a binary relation, and we work in ZFC. Assuming that we know that R\times S exists, it's powerset P(R\times S) also exists by the powerset axiom. Then, using axiom schema of separation on the set P(R\times S) and a predicate "f is a function from R to S", ... 0 Essential here is the definition of function. A function F:R\rightarrow S is actually a subset of R\times S that satisfies the following condition:$$\text{ for every }x\in R\text{ there is exactly one }y\in S\text{ such that } \langle x,y\rangle\in F$$Now let R and S both be sets with more than one element. 1) Can you find a subset of R\times ... 1 If we have a neighbourhood U of x\in M such that U\cap M is finite, consider F := (M\cap U)\setminus \{x\}. If F = \varnothing, we're done, otherwise note that F is closed, since it's finite (and \mathbb{C} is Hausdorff, hence T_1), and thus V := U \setminus F is also a neighbourhood of x, and by construction we have V\cap M = \{x\}. ... 0 A binary relation on X is any subset of R \times R.And R^{-1}=\{(x,y) : (y,x)\in R\}.Now (x,y) \in (R \cap R^{-1} iff ((x,y) \in R and (y,x)\in R). From this it should be obvious that if R is transitive then R\cap R^{-1} is also transitive. 0 Given a relation R on a set X (that is, a subset of the Cartesian product X \times X of orders pairs of things in X), the relation R^{-1} is the inverse relation formed by taking the ordered pairs in R and reversing them; that is,$$R^{-1} = \{(y, x): (x, y) \in R\}.$$This should remind you of inverse functions, where 'input' and 'output' are ... 4 It's not a big deal or a problem, but it does require the Axiom of Choice. Here's the point. We want to base our math on set theory. In elementary situations sets are just things that have elements, and the way they work is just the way things with elements obviously work. That's the level at which you'd say what's the big deal, we just select one of these ... 4 The claim that any surjective function has a right inverse is indeed (against standard background assumptions) equivalent to Axiom of Choice. There are whole books on the Axiom of Choice, and on the costs and benefits of accepting it (it has surprising consequences). The Wikipedia entry is not a bad place to start finding out more about AC. And for a more ... 4 I don't know how you got to Therefore f(a) = b , g(a) = b , h(a) =b , so f is surjective. by the way you don't need A,B,C and D because you know by definition of f,g,h that they go from R to R..... so A,B,C,D would simply be R As you said we know per definition \forall y \in R\; \exists x\in R : f(g(h(x)))=y Then simply substitude g(h(x)) by x' ... 8 Let y\in\mathbb{R}. As f\circ g\circ h is surjective, there exists x\in\mathbb{R} such that (f\circ g\circ h)(x) = y. So, if we let x' = (g\circ h)(x), we have f(x') = y, which proves that f is surjective. 5 \mathbb{R}=(f\circ g\circ h) (\mathbb{R})\subseteq{f(\mathbb{R}})\subseteq \mathbb{R}, so f(\mathbb{R})=\mathbb{R}. even f\colon E \to F, g\colon D \to E and h: C \to D, then since \mathbb{F}=(f\circ g\circ h) (\mathbb{C})\subseteq{f(E})\subseteq \mathbb{F}, so f(\mathbb{E})=F. 1 In your language, the empty set can not be "nothing at all" since it is really a mathematical object: a set with no elements, as you understood thanks to the box example. The fact it does not contain anything does not affect the fact it is a set. 1 It does not matter whether singletons are included or not, because any formed pairs will have two elements and a singleton will be either the empty set or a set with one element. The sets formed are distinct: any set formed has exactly two elements, so for two sets to be the same the elements themselves have to be the same, which means the sets have to be ... 5 Think of a box. Empty set is like an empty box. But a box with another box inside it is no longer empty. 4 \{\emptyset \} has one element, while \emptyset has 0 elements. 1 Wrong. The powerset of empty set is not an empty set; it is a single-element set, and its only element is an empty set:$$P(\{\}) = \{\,\{\}\,\}$$This relates also to the cardinalities:$$|P(A)| = 2^{|A|}$$so if A=\{\} and |A|=0 then |P(A)|=2^0=1, hence the powerset has an item — is not empty. 0 Hint: if the empty set were the power set of some set A, what would be the subsets of A? How many would there be? 4 The question is asking you to prove that a relation, \sim, is an equivalence relation. A binary relation is technically a subset of the cartesian product of a set on itself. Although the words may seem foreign, you've been using relations your whole life. Think of it as a way of "comparing" things. For example, 5<7, we say that 5 is "less than" ... -1 It's far easier than step 1. Step 2 : same definition than step 1, but with: \forall x∈[0,1), \space f(x)=x instead of \forall x∈(0,1), \space f(x)=x. Step 3 : \forall x, \space f(x)=1-x. 3 (Work in progress.) Say a subset A of \mathbb{R} is difference-unique if for all z \not = 0, z \in \mathbb{R} there is at most one pair a_z, b_z \in A with a_z - b_z \in A. Say A is nice if additionally for all z \not = 0, z \in \mathbb{R} there is exactly one a_z, b_z \in A with a_z - b_z = z. (That is, your property.) It is immediately ... 2 You can order by size every finite subset of positive rationals. If you could conclude from every finite subset on the infinite set, then ordering of all positive rationals by size would be possible. Of course this is not possible. For instance there is no smallest positive rational. But this shows that you can not conclude from every finite subset on the ... 1 Hint: For m>n>0, we have$$\frac1m<\frac1n.$$In particular, for m\in\Bbb N, we have$$\frac{1}{9m^2}\leq\frac19.$$Can you take it from here? 1 For one thing:$$\frac1{9(N+1)^2}\space<\space\frac1{9N^2}$$so:$$\frac1{9(N+1)^2}+ \frac2{3}\space<\space\frac1{9N^2}+ \frac2{3}\space<\space1

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