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## New answers tagged examples-counterexamples

1

If you know that the $\sigma$-algebra of Lebesgue measurable sets is a proper subset of $\mathcal{P}(\Bbb R)$, then this is easy to prove. The $\sigma$-algebra of Lebesgue measurable sets is complete, which means all subsets of sets of measure $0$ are measurable. Let $A$ be any non-empty subset of measure $0$, e.g., $A = \{1,2,3 \}$. Then if every ...

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Take your favorite measurable subset $A$ of $(0, 1)$, and favorite non-measurable subset $B$ of $(2, 3)$. Then $A\cup B$ is non-measurable, but is a superset of $A$. Note that the example you give in your question, while in the right direction, doesn't quite work - if we take $V$ to be a Vitali subset of $(0, 1)$, and $A=(0, 1)\setminus \{x\}$ for some ...

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Recall that every set is a superset of the empty set.

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This is not because a fuction takes the values $0$ and $1$ that we can declare it is measurable; her, the partial functions of $f$ are not. Fix $y$. The set of points $x$ such that the first partial function $x\mapsto f(x,y)$ is less than $1$ equals $E\setminus \lbrace y\rbrace$, which is not measurable. Thus $f$ is not separately measurable. J. Yeh in ...

0

Answer is no. Let $G$ be a graph such that $\bar{G}$ is the graph in the picture. It is clear from the picture that $\chi(\bar{G})=2$. We will properly color $\bar{G}$ using the colors red and blue. Let $V_{red}=\{a,b,c\},V_{blue}=\{1,2,3\}$ be the set of red and blue vertices respectively. The matching which you describe in formula (1) in your question can ...

0

A try for an example. Put $\displaystyle \sqrt{1+x}=\sum_{n\geq 0} u_n x^n$. We have for $n\geq 3$ $\displaystyle u_n=(-1)^{n-1}\frac{1.3\cdots2n-3}{2^n n!}$, hence $\displaystyle |u_n|\geq \frac{1.2. \cdots (2n-4)}{2^n n!}\geq \frac{1}{2n(n-1)}$. Now put $a_n=b_n=2^nu_n$, these sequences are unbounded. We have $\displaystyle\sum_{n\geq ... 2 Both statements are false, actually. For (1): You've already shown that every such series converges; it's not hard to show that any two series which differ at some point (e.g.$a_n\not=b_n$for some$n$) converge to different reals. Thus, you're really just counting the number of sequences of possible$a_i$s. Using this, can you construct a bijection ... 3 Your question is quite interesting, and deserves more attention than it has been getting. Whether or not a (time homogeneous) Markov process$(X_t)_{t\geq 0}$with state space$E$is strong Markov depends on a precise definition. I hope that my explanation below substracts from, rather than adds to, your confusion. The standard definition of the strong ... 1 Let$R_0$be any ring and let$R=R_0[x_1,x_2,x_3,\dots]$be a polynomial ring over$R_0$in infinitely many variables. Then$R/I\cong R$, where$I=(x_1)$(since$R/I=R_0[x_2,x_3,\dots]$, and you can just shift the variables over by one to get an isomorphism with$R$). There are many other similar examples. For instance, you could take a product ... 1 Somewhere between a joke and an essay, is Impure Math. http://www.snowman-jim.org/science/humor/impure-math.html 0 The condition $$\tag{1} f(x) - \frac{C}{k} \le f(x_k)$$ is actually very strong as there is no conditions on the sequence$x_k \to x$. Indeed you can find a counterexample even for a function as simple as a linear function: Let$f(x) = x$on$(0,1)$and$x_k = \frac 12 - \frac{1}{\sqrt k}$. Then$x_k \to x=\frac 12$and$(1)is the same as $$\frac 12 - ... 0 An example is a random variable X having a student-t distribution with \nu = 2 degrees of freedom Its E[X] = 0 for \nu > 1. Its E[X^2] - Var[X] = \infty for 1 < \nu \le 2 3 Let U be a random variable which is uniform on (0,1]. Let X=Y=U^{-1/2}. Then E[X]=E[Y]=2 but E[XY]=+\infty. You can make lots of examples of functions like this, which are not integrable but their square root is integrable, because it diverges "more slowly". On an infinite measure space you can also have the opposite phenomenon with tailing: ... 6 Any branch of \sqrt{1+z} is bounded in the unit disk, but the derivative \frac{1}{2\sqrt{1+z}} is not. 10 (z+1)\log(z+1) is bounded on the open unit disk |z|\lt1. Its derivative, 1+\log(z+1) is not. 3 Does this work?$$\sum_n \frac{x^{2^n}}{2^n}$$0 The function f is onto: Case 1 Let y \in [3,5) be of the form: 3 + 2^{1-n} for some n \in \mathbb{N}. Then x = 1 + 2^{1-n} is such that f(x) = y. In particular, if n is such that 3 + 2^{1-n} \in [3,5) then n \geq 0. This in turn implies that 1 \leq 1 + 2^{1-n} \leq 2. Case 2 Let y \in [3,5) not be of the form: 3 + 2^{1-n} ... 1 Let 1_{\Bbb Q} be the indicator function (characteristic function) of \Bbb Q, and let f(x)=\lceil |x|\rceil 1_{\Bbb Q} for each x\in\Bbb R. If K\subseteq\Bbb R is compact, then K is bounded, and f[K] is finite, but f[\Bbb R]=\Bbb Z. 2 It's just the plain old liar paradox. Typically nowadays the liar paradox is presented as a sentence that speaks only about its own truth: This sentence is false. but the name is also applied to a sentence that claims that a whole class of sentences which includes itself are all false. The classical example of that was when Epimenides (who was Cretan) ... 1 Your statement has to be false. As you said, if the statement were true, this would imply that the statement is false which is a contradiction. If the statement false, no contradiction occurs. More problematic are sentences like “This sentence is false.” This can be neither true nor false as always a contradiction will follow. For more information on this, ... 2 A \to \lnot A is not a contradiction. If we assume f as truth value for A, then \lnot A is f, and thus the conditional becomes : f \to t which is t. Thus, we have the conclusion that the "correct" answer is : A must be false. In fact : (A \to \lnot A) \to \lnot A is a tautology, as you can easily check via truth table. 1 You know that$$\lim_{n\to\infty}n^2-n=\infty$$By definition this means for all N>0 there exists M(N)>0 such that n>M implies n^2-n>N. But then let N>\log_2(c). Then$$2^{n^2-n}>2^N>2^{\log_2(c)}=c.$$2 This is true, even under slightly weaker assumptions, but is fairly technical to prove. In fact, we have Theorem If \gamma is a rectifiable Jordan curve and f is holomorphic on the interior G of \gamma and continuous on \bar G = G \cup \gamma, then$$ \int_\gamma f(z)\,dz = 0. $$The proof is apparently due to Denjoy and appeared in Compt. ... 1 Just wish to contribute that in general for these problems it is very convenient to think of graphs on the Cartesian Plane. Graphs that do not pass the vertical line test (no two y values for one x value) are not functions. and Graphs that do not pass the horizontal line test (similarly, no two x values for one y value) are not injective. Then it becomes ... 1 Hint: Try A=[-1,1], f(x)=x, g(x)=-x. Then f(x)+g(x) = 0 is not bijection and f(x) g(x)=-x^2 also is not bijection since f(1) g(1)=f(-1) g(-1). 0 Given any c>0 and assuming it is known that e^{-x} satisfies lim_{x\rightarrow \infty}e^{-x}=0, there is by definition of the limit a real number R>0 sucht that |e^{-x}| < \frac{c}{2} for every x>R. This shows, directly, that no real number c>0 with f(x)\ge c can exist. 1 Note that if y \in [0,1[, then there is some \delta > 0 such that |x-y| < \delta implies |f(x) - f(y)| < f(y)/2, implying that f(x) > f(y)/2; hence f(x) > 2^{-1}\inf f[0,1[ for all y \in [0,1[. A problem is that \inf f[0,1[ can be =0, as shown by the example that Thomas provided; hence you know the statement under ... 4 Take any closed subset E of \mathbb R with empty interior, positive measure or not. Then E + \mathbb Q \ne \mathbb R. Proof: Baire. 0 Partial result: Suppose \gamma encloses a convex region G. For convenience, assume 0\in G. Then for 0<r<1, r\gamma is a closed contour in G. Since G is simply connected, \int_{r\gamma} f(z)\,dz = 0 by Cauchy. As r\to 1^-, the uniform continuity of f on \overline G shows$$\int_{r\gamma} f(z)\,dz \to \int_{\gamma} f(z)\,dz.... 2 No, and it follows from a dimension argument. Let C be the cantor set. Then your claim is that C + \mathbb{Q} = \mathbb{R}, i.e. that every real is a sum of an element of C and a rational. We know that \dim_{H}(C) = \frac{\log 2}{\log 3}. But \begin{align*} C + \mathbb{Q} & = \{c + q : c \in C, q \in \mathbb{Q} \} \\ & = \bigcup_{q \in ... 3 Let T:=\inf\{t:X_t=0\}, and suppose that X_0=x\not=0. Fix t>0. Then \Bbb P^x[X_t=0]=0 because X_t\sim\mathscr N(x,t). On the other hand, if X were a strong Markov process, then on the event \{T<t\} we would have X_t=0 because the post-T process \{X_{T+s}:s\ge 0\} would start in state 0, and so stay in state 0 for all time by ... 1 Without loss of generality you can assume that [a, b] = [-1, 1]. The "Markov brothers' inequality" then states that \sup_{x\in [-1, 1]}|p'(x)| \le n^2 \sup_{x\in [-1, 1]}|p(x)| $$for all polynomials p of degree \le n, therefore$$ |p'(x)| \le n^2 $$for all p \in K and all x \in [-1, 1]. It follows that K is equicontinuous. 1 This is not possible. If E is a vector space and W_1,W_2 vector subspaces of E then you have$$(W_1\cap W_2)^\circ = W_1^\circ +W_2^\circ.$$This is true even if E is of infinite dimension. However, Axiom of Choice is required in that case to work with basis. 0 The only way you can make your example work is to take g(y)=0 for all y\geq0 so that g\circ f is the zero map \Bbb R\to\Bbb R. You can however define g(y) to be whatever you like for y<0; in particular you can very well make g not linear (nor injective, surjective, or most any nice property you would like to avoid). The reason that you ... 8 It's not very exciting but in general if f:V \rightarrow W is nonlinear but invertible then f\circ f^{-1} = Id_V is linear. 2 A function g like you wish can be given, but it is not "interesting", as far as I'm concerned. Note that a nonzero linear function \mathbf R \to \mathbf R is one-to-one, as f is not one-to-one, g has to be zero on the image of f. But on (-\infty,0) you can take g to be what you like. That is for example, define g \colon \mathbf R \to \mathbf ... 1 In M = \mathbb R^2, as suggested in the comment, one can take X = \{xy = 0\} and Y = \{xy = 1\}. They are infinitely closed to each other when (x, y) \to \infty. One can also take an example in \mathbb R, letting X = \mathbb N and Y = \{ n + \frac 1{2n} : n\in \mathbb N\}. On the other hand, one cannot find such an example in a discrete ... 1 [I have edited my answer. I don't think this is the best possible response but it is somewhat reasonable.] Many motivate the Lebesgue integral over the Riemann integral by stating (loosely) that the former has "better limit theorems." That is a bit misguided I think. They have exactly the same limit theorems (from one point of view) except that the ... 1 Maybe its to late for You but this might be helpful for someone else A topological space K is said to be extremally disconnected if \bar U is open for every open U ⊆ K; equivalently, in K the following separation axiom is satisfied: If U, V ⊆ K are open and U ∩ V = \emptyset then \bar U ∩ \bar V = ∅. This separation is extremally (sic!) ... 1 Consider \mathcal(S) = \{\emptyset, \Bbb N\} \cup \{X\subseteq \Bbb N\mid \lvert X\rvert\ge 2\}. 1 Hint: Consider X=\{1,2,\ldots,2n\} for some n \in \mathbb{N} and$$\mathcal{F} := \{A \subseteq X; \sharp A \, \text{is even}\}.$$(\sharp A denotes the cardinality of the set A.) 1 Counterexample: Let a_n = 1/n, n = 1,2, \dots. For each n let g be an isosceles triangular spike over a_n of base length b_n < 1/2^n and height h_n = 1-1/n. This can be done so that the bases are disjoint. Define g = 0 everywhere else. Then g is continuous and bounded on (0,1], hence g is Riemann integrable on [0,1]. Define$$f(x) = ... 1 A try for a counter-example. Put forx\in ]0,1]$$$f(x)=\int_{1/x}^{+\infty}\frac{\sin(u)}{1+u^2}du$$ and of course$f(0)=0$. 1) We integrate by parts: $$f(x)=\frac{x^2}{1+x^2}\cos(\frac{1}{x})-\int_{1/x}^{+\infty}\frac{2u\cos(u)}{(1+u^2)^2}du$$ We have$\displaystyle |\frac{x^2}{1+x^2}\cos(\frac{1}{x})|\leq \frac{x^2}{1+x^2}$, and: ... 0 I think I have a much simpler answer... Consider$(\mathbb{R},d)$and$(\mathbb{R},d_{\text{disc}})$where$d$is the standard metric and$d_{\text{disc}}$is the discrete metric. Then, consider the identity mappings $$\iota: (\mathbb{R},d) \to (\mathbb{R},d_{\text{disc}})$$ and $$\iota^{-1}: (\mathbb{R},d_{\text{disc}}) \to (\mathbb{R},d).$$ Notice that ... 5 Our first step to better understand when L'Hospital's Rule is applicable, is to consider its formal statement. It is a theorem, therefore it is always correct; we just have to make sure that all needed assumptions/hypotheses do hold for the limit we are trying to evaluate. Let's take a look! Theorem (L'Hospital's Rule for right-hand limits): Let$f$,$g$... 1 Here's a similar example to Ittay Weiss's, but more nefarious, since in Ittay's example the limit was easily seen to be equal to 1: $$\lim_{x \rightarrow 0}\frac{e^{-\frac{1}{x^2}}}{x}$$ Both the numerator and denominator go to zero. But applying l'Hospital gives: $$\lim_{x \rightarrow 0}\frac{2e^{-\frac{1}{x^2}}}{x^3}$$ And applying l'Hospital again: ... 5 Here is an example of why you should be careful:$ \begin{align*} \lim_{x \to \infty} \frac{x+\sin(x)}{x} &= \lim_{x \to \infty} \frac{\frac{d}{d x}(x+\sin(x))}{\frac{d}{dx}(x)}\\ &= \lim_{x \to \infty} \frac{1+\cos(x)}{1}\\ \end{align*} $This last limit does not exist, so you might conclude that the original limit does not either. In fact,$ ...

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L'Hopital's rule fails if $$\lim_{x\to x_0} \frac f g \text{ exists but} \lim_{x\to x_0}\frac {f'}{g'} \text{doesn't}$$ e.g. $$\lim_{x\to \infty} \frac {x+\sin x}x.$$ I wonder if we could extend generalization of Stolz–Cesàro theorem to continuous case - i.e. if $$\liminf \frac {f'}{g'}\le \liminf \frac f g\le \limsup \frac f g\le \limsup \frac ... 0 There are situations where all conditions are met, but the usage of L'Hôpital leads to circular reasoning. For example,$$\lim_{x \to 0}\frac{\sin x}x = \{\text{L'Hôpital}\}=\lim_{x \to 0}\frac{\cos x}1=\cos 0=1 Here, you have used the fact that $\frac d{dx}\sin x = \cos x$, which is proven by using $\lim_{x \to 0}\frac{\sin x}x$, which can be found is ...

2

Eular characteristic of a non-orientable surface is $2-g$ where $g$ denotes the genus of the surface. So if we consider $g=3$ i.r if we consider a hexagon and identify its edges with the relation $a^2b^2c^2$ , the resultant surface we will get as our required surface.

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