# Tag Info

1

Let me take a sidestep. Trying to understand $\sigma$ algebra through Vitali sets and Lebesgue measure, etc. is, in my opinion, the wrong approach. Let me offer a much, much simpler example. A probability space is a measure space $(X, \sigma, \mu)$ where $\mu(X)=1$. What's the simplest thing you can model with probability theory? Well, the flip of a fair ...

0

The idea is that there is an injection $f$ from $A\cup\{x\}$ into $A$. But the range of the injection misses $x$ (because it's not in $A$, otherwise the claim is trivial). Using the fact that $f(x)\in A$, deduce that for every $n\neq m$, $f^n(x)\neq f^m(x)$. From this conclude the wanted conclusion.

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What are the 'privileged subsets of X'? Borel sets for instance? Is it that because all Borel sets form the smallest sigma algebra (smallest meaning it has the least elements among all sigma algebras containing all Borel sets), and Borel sets have the nice property that they are made of open intervals, which means we can assign size to those sets that is ...

1

The statement $f^{-1}(Y \setminus Z)=f^{-1}(Y) \setminus f^{-1}(Z)$ is correct, but your proof is problematic. For example, as already mentioned in a comment, $f$ need not be invertible, and then equations like $x=f^{-1}(y)$ make no sense. But you can proceed as follows (using the definition $f^{-1}(W)=\{x \in A \ : \ f(x) \in W\}$): \begin{aligned} x ... 0 Prove or disprove(A\times B) \setminus (C\times D) = (A\setminus C)\times (B\setminus D)$$Counterexample Let A = \{1,2\}, B = \{2,1\}, C = \{2,8\}, and D = \{1,10\}. Then A\times B = \{(1,2),(1,1),(2,2),(2,1)\} and C \times D = \{(2,1), (2,10), (8,1), (8,10)\}. Taking the set difference of A \times B and C \times D renders the set ... 0 Let me try to do a sum-up and give my own thinking. As suggested here, in the case of vector spaces A+B=\{a+b:a\in A,b\in B\}. That is, A+B is the set of all sums of two vectors, one in A and one in B. If A,B are open, then let a+b\in A+B. I can find V_A\subseteq A,V_B\subseteq B such that they are open and contain a and b respectively. In ... 0 Context is important, and notations differ between different authors. + for union is not very common these days, but you will find it in some older books and papers. An example I just happened to pull off my shelf is Natanson, "Theory of Functions of a Real Variable". 0 When A and B are vector subspace, there is a nice illustrative example: Let$$ A =\{(x,0)\mid x\in\Bbb R\} \qquad \text{and}\qquad B=\{(0,y)\mid y\in\Bbb R\}.$$Geometric interpretation: A is the set of points on the the "x-axis" and B is the set of points on the "y-axis" in the plane. Then$$ A\cup B =\{(x,y)\mid x =0 \text{ or } y =0\} \qquad ...

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If they are sets without any structure (so no addition operation on elements) then it could mean disjoint union, i.e., a synonym for U but only to be used when it has already been established that the sets have no elements in common.

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Usually $A+B$ means $\{a+b\,:\,a\in A,b\in B\}$.

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Since $f$ is strictly decreasing on $(0,\infty)$, we have that for any $x_1,x_2 \in (0,\infty)$ such that $x_1 < x_2$, $f(x_1)>f(x_2)$, which in turn means that $f(f(x_1)) < f(f(x_2))$. Hence, in fact $f \circ f$ is strictly increasing.

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Here is one answer, which may answer the question as I understand it (but perhaps I do not understand the question correctly). Start with a measure $\mu$ defined on a $\sigma$-algebra $\mathcal A$. Then define the outer measure $\mu^*$ associated with $\mu$, in the sense of Caratheodory. Once you have this outer measure, you can define the class $\mathcal ... 0 Its probably worth point out that this is a special case of a more general result. Given a set$X$, assert that the wimpy powerset of$X$is denoted$\mathcal{W}(X)$and is defined as follows. $$\mathcal{W}(X) = \{S \subseteq X : |S|<|X|\}$$ The basic result about wimpy powersets is: Proposition 0. Given an infinite set$X$, we have: ... 2 Other than zero, which corresponds to the empty set, the resulting fractions from that procedure are those of the form $$\frac{j}{2^k}$$ for some natural numbers$j,k$and$j$odd. Therefore you can list them out in this order: $$\frac 1{2^1}$$ $$\frac 1{2^2},\frac 3{2^2}$$ $$\frac 1{2^3},\frac 3{2^3},\frac 5{2^3},\frac 7{2^3}$$ and so on: I'm sure you ... 1 Essentially you're proving the contrapositive: if$T\subsetneq S$then$T$is not inductive. Suppose$T\subsetneq S$. Then$S\setminus T\ne \emptyset$, so it has a least element$x$. For any$s\in S$, if$s<x$then$s\in T$by definition of$x$; however,$x\notin T$. Therefore, $$\exists s\in S\,((\forall t\in S\,(t<s\to t\in T) \land s\notin T),$$ ... 1 This answer makes sense when$F$is a subset of${\cal P}({\cal P}(A))$, so that the elements of$F$are "boolean functions", subsets of${\cal P}(A)$, or alternatively functions$\{0,1\}^A\to\{0,1\}$. You stated that$F$is a set of subsets of$A$, however, which is incompatible with this assumption. My only guess in this case is that the projections are ... 7 It is not wrong for you to worry, and the same problem occurs elsewhere: We might define$\Bbb Z$from$\Bbb N_0$as a set of equivalence classes of$\Bbb N_0^2$where$(a,b)\sim (c,d)\iff a+d=b+c$. But then$\Bbb N\not\subset \Bbb Z$. We might define$\Bbb Q$as a set of equivalence classes of$\Bbb Z\times \Bbb N$where$(a,b)\sim (c,d)\iff ad=bc$. But ... 0$x \ne (x,0)$but$f: R \times 0 = \{(x,0)\} \rightarrow R$via$f((x,0)) = x$is a isoomorphism so$R \cong R \times 0 \subset R^2$And actually, in a sense, (x,0) is the same thing as x. It's fundamental that we can take two (or more) real numbers put them in an ordered pair and that doing so and creating a third thing, the ordered pair, we aren't ... 11 No matter how you define$\mathbb C$, the crucial point is that$\mathbb C$contains a copy of$\mathbb R$, which we still denote by$\mathbb R$, such that$\mathbb C = \mathbb R + i\mathbb R$. In mathematics, it matters less what things are than how they behave. If two things behave exactly the same, they're taken to be the same thing, even if the ... 0 From the second definition you gave, the real numbers are just the complex numbers with$b=0$. 1 Fix a set$P$, we want to show that if there is a surjection$f\colon X\to Y$, and$g\colon P\to Y$, then there is a function$h\colon P\to X$such that$g=fh$. Since$f$is surjective,$\{f^{-1}(y)\mid y\in Y\}$is a non-empty family of sets. Use the axiom of choice to choose$x_y\in X$such that$f(x_y)=y$for all$y\in Y$, and use this information to ... 1 The role of the axiomatic method in the development of mathematics has been crucial from the beginning; See : Euclid's Elements. But it has been crucial also in natural science; see : Newton's Philosophiae Naturalis Principia Mathematica, and also in use by some phliosophers; see : Spinoza's Ethics. Regarding Set theory, the early discovery of the ... 1$\displaystyle f(x) = \tan (x{\pi/2})$2 $$f : (0,1) \to (0,\infty) , \, x \mapsto -\ln(1 - x)$$ 2 Consider first the bijection$f\colon x\mapsto \frac 1 {1-x}\colon (0,1) \to (1,+\infty)$, which is continuous and order-preserving. Now$f-1$is the bijection you want: $$x\mapsto \frac x {1-x}\colon (0,1) \to (0,+\infty)$$ It too is continuous and order-preserving. 2 Here is a continuous one:$x \mapsto \frac{1}{x}-1$0 The answer to your actual question is probably that your professor mentioned it about another part. That the image of$f$is as you mention nothing controversial at all. Then the rest depends on the exact definitions that are used. Given the axiom of choice you can phrase definition in one of many equivalent ways, but when dropping the axiom of choice these ... 3 Yes. You need the axiom of choice in order to prove that every infinite set has a countably infinite subset. And choice is used to prove there is an injective function in the first place. Once you have that injection, enumerating the image is choice free. The axiom of choice is needed here because$X$is any arbitrary set. In order to prove that there is an ... 0 Your proof of$U \subseteq \bigcap_\alpha \pi_\alpha^{-1}\bigl[\pi_\alpha[U]\bigr]$is fine. For the other direction: Suppose that$x \in \bigcap_\alpha \pi_\alpha^{-1}\bigl[\pi_\alpha[U]\bigr]$, then for each$\alpha$we have$x \in \pi_\alpha^{-1}\bigl[\pi_\alpha[U]\bigr]$, that is, by definition of the preimage,$\pi_\alpha(x) \in \pi_\alpha[U]$. Now, ... 1 Unless I am extremely confused, the assumption$A_i \subset A_{i-1}$implies that, for example,$A_4$is a subset of$A_3$. Then clearly, the principle of induction can be used to show that$\forall i, A_i \subset A_1,$then$A=\cup_i A_i = A_1.$Since by assumption$A_1 \subsetneq B$, we have$A\subsetneq B$. If the assumption was actually that each ... 3 Hint: A bijection maps the first element to something. How many possibilities are there? Once you choose where the first element maps to, how many possibilities are there to map the second element to? The third? The fourth? The fifth? How many choices total are there? 3 " i.e. there exists at least one element in B that is not in Ai for every Ai." Not quite. The element of$B$that is not in$A_i$could be an entirely different b that is not in$A_j$. In fact you don't even need infinite sets. Let$A_1 = \{1,3,5\}; A_2 = \{2, 4, 6\}; B = \{1,2,3,4,5,6\}$.$A = A_1 \cup A_2$,$A_i \subsetneq B$but$A = B$===== With ... 8 No. Take$A_i = \{ 1,2,...,i \} \subset B = \mathbb{N}$. However,$A = \cup_i A_i = B$. 1 Every set other than the empty set has a subset to which it is not equipotent. So having a subset "smaller" than the original set is a property equivalent to being non-empty (in standard versions of set theory, anyhow). The property of being infinite is characterised by having at least one proper subset which is equipotent with the original set. 0 The statement should be read "if there exists a proper subset to which it is equipotent". As you have shown, that does not mean that every proper subset qualifies. 1 Here's a more precise statement of the theorem: A set$X$is infinite, if and only if there exists a proper subset$Y\subset X$, such that$X$and$Y$are equipotent. So,$X$is not expected to be equipotent to any proper subset of itself, but if$X$is infinite, then it is equipotent to some proper subset of itself. In your example, if ... 0 The fact you found one subset which is not equipotent with$\Bbb N$does not mean there are no others. For example$\{2n\mid n\in\Bbb N\}$is equipotent with$\Bbb N$. 1 We don't assume that$ m( A \cup B ) = m(A) + m(B) $for every pair of sets$A, B$, but only for when they're disjoint. Refer to a simple example of a measure like Lebesgue measure on the real line, i.e. "length". We expect to be able to add the lengths of two segments when they don't touch (or only touch in small ways), but if I have two segments that ... 1 If$I_{1}$and$I_{2}$are sets (such as intervals of real numbers), define$I_{1} + I_{2}$to be their disjoint union and$I_{1} \times I_{2}$to be their Cartesian product. Since the Cartesian product distributes over disjoint unions, $$(I_{1} + I_{2}) \times (I_{1} + I_{2}) = (I_{1} \times I_{1}) + (I_{1} \times I_{2}) + (I_{2} \times I_{1}) + (I_{2} ... 3 You would have that {\frak m}A = 2\,{\frak m}A if the union A = A \cup A were disjoint. This is not the case, unless A = \varnothing (and here 0 = 2 \cdot 0 is ok). 0 You haven't justified how one uses (2) and (3) to get (4), yet. From (4), you can fairly directly conclude that$$\mathcal P(A\cap B)\subseteq\mathcal P(A)\cap\mathcal P(B),\tag{$\star$}$$but (1) isn't sufficient to give you the reverse inclusion on its own. It's important in a proof not to just say things like "we can get" or "it follows that." ... 1 Let C\in P(A)\cap P(B) then C\in P(A) and C\in P(B) implying C\subseteq A and C\subseteq B implying C\subseteq A\cap B implying C\in P(A\cap B). Thus, P(A)\cap P(B)\subseteq P(A\cap B). Now let D\in P(A\cap B) then D\subseteq A\cap B implying D\subseteq A and D\subseteq B implying D\in P(A) and D\in P(B) implying D\in P(A)\cap ... 5 \newcommand{\Symb}{\clubsuit}You've got correct answers, but in case it helps to consider the analogous question with digits replaced by \Symb's and the association with numerical limits removed: In the sequence$$ \begin{array}{r@{}ccccc} x_{1} = & \Symb & & & & \\ x_{2} = & \Symb & \Symb & & & \\ x_{3} = ... 5 Note that$\Bbb{N^N}$is a subset of$\mathcal P(\Bbb{N\times N})$. Now use the fact that$\mathcal P(\Bbb{N\times N})$and$\mathcal P(\Bbb N)$have the same cardinality. 0 You are on the right track. I'll assume that your definition of cardinality is that two sets have the same cardinality if there exists an injection both ways (or that you know the Cantor-Bernstein-Schröder theorem). You probably know that cardinal subtraction isn't well defined, but if it was it would be easy to see that$|\mathbb R| - |\mathbb Z| = ...

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You establish the following facts: The diagonal of the set equals 1 The set approaches 1 (from below). But you seem to assume, incorrectly, that: a. The diagonal of a set must be an element of the set. b. The diagonal of a set must equal the last element of the set. c. Every set has a "last element". Your example is actually a very important ...

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Your problem comes from the fact that "belonging to a set" is not a continuous function! For all $n \ge 1$, call $x_n = 0.999 \dots 9$ with $n$ nines (i.e. $x_n = 1-10^{-n}$), and call $S= \{ 0.9 , 0.99, \dots \} = \{ x_n : n \ge 1\}$. You say $x_1 \in S$, $x_2 \in S$ , $x_3 \in S$ and so on, so we can conclude that $$1= \lim_{n \to \infty} x_n = ... 21 Your mistake is that there is no "last" element of your set. Just like there is no last natural number. If x is in your set, then it has only finitely many 9's, so it is some \sum_{n=1}^k\frac9{10^n}. Therefore \sum_{n=1}^{k+1}\frac9{10^n} comes "after" x in the set. Therefore there is no last element, and in particular 1 is not an element of ... 0 A \subseteq \bigcap_{n=1}^{\infty} A_n? No. Suppose we have a measure space (S, \Sigma, \mu) and measurable sets A_1, A_2, ... '\lim A_n = A' is an abbreviation for$$\limsup A_n = \liminf A_n = A$$Definitions:$$\limsup A_n = \bigcap_{m \ge 1} \bigcup_{n \ge m} A_n\liminf A_n = \bigcup_{m \ge 1} \bigcap_{n \ge m} A_n$$If \omega \in ... 2 Here's an approach you can take. For every x\in(0,1], show that there is a unique sequence a_1,a_2,a_3,\dots such that every a_n\in\{0,1\}, there are infinitely-many a_n=1, and$$x=\sum_{n=0}^\infty\frac{a_n}{2^n}. Use this to show that the set $S$ of all such sequences together with the sequence of all $0$s, has the same cardinality as $[0,1].$ ...

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