# Tag Info

0

Your example is not true, because $g_n = \chi_{[n,n+1]}$ isn't Cauchy measure so this sequence isn't convergence to 0 in measure.

1

The numbers for which such a factorization exists have been studied by Maier and Tenenbaum, On the set of divisors of an integer, Invent. Math. 76:1 (1984), pp. 121-128. Erdos conjectured that almost all positive integers $n$ have a factorization $n=ab$ with $a\le b\le2a$ ("almost all" means all but a set of density zero), and Maier and Tenenbaum proved a ...

1

$35 \cdot 36$ is a counterexample. $35 \leq 36 \leq 70$ and $30 \leq 42 \leq 60$. As this counterexample suggests, such numbers cannot have very many prime factors, otherwise they can be rearranged into very nearly identical factorizations.

1

Hint: define $f: {\mathbb R}^2 \to {\mathbb R}$ such that $f(x,y) = 0$ unless $x^2 < y < 2 x^2$. Note that the intersection of any line through the origin with the exceptional set $A = \{(x,y): x^2 < y < 2 x^2\}$ misses some interval around the origin, so what $f$ does in $A$ does not affect the Gâteaux derivative at the origin.

2

A sequence can be seen as an ordered list and is typically considered countable, especially in real analysis/calculus, but there are uncountable lists. I would bet that you professor means countable list (ordered by the natural numbers), but one can have "sequences" that are longer than the natural numbers, or even uncountable (I put quotes since sequences ...

4

From what I understand, your question comes down to why a set being ordered does not imply it being countable. The slightly subtle notion here is the difference between cardinal and ordinal numbers. In general, there are two standard ways to compare the sizes of two sets. You can construct a bijection (as you've probably seen), or you can construct an ...

1

The interval $(0,1)$ is an ordered, uncountable set in $\mathbb{R}$. It is perhaps slightly tighter to restate his definition as a map from the natural numbers $\{1,2,3,\dots\}$ to the set in question, $\mathbb{R}$ in this case. So while there are uncountable ordered sets in $\mathbb{R}$, they are not sequences since there are not enough natural numbers to ...

1

Consider a particle which is at position $x(t) = \frac{t^4}{4!}$ at time $t$. Its acceleration is $\ddot{x}(t) = \frac{t^2}{2!}$, which means that at $t=0$ it is stationary and it does not (at the moment) accelerate. Shouldn't it stay in place forever? No, the thing is that it will accelerate in a moment (precisely for any positive time), and the reason is, ...

4

More simply, sticking with the same $f$ you may consider $$g(x) = \begin{cases} 0, & \text{if x=0} \\ 1, & \text{if x\in ]0,1]} \\ \end{cases}$$

1

Consider $R=S[a,b]$ for any commutative ring $S$ and indeterminates $a$ and $b$. Alternatively, notice $ax+by=1$ in $R$ implies $b$ is invertible in $R/(a)$, and invertibility doesn't follow from simply not being a zero divisor (as the $b\in S[b]$ example is an illustration of).

0

Let's start from a finite dimensional case. In $\mathbb R^2$ the convex hull of $e_1,e_2$ is a segment, each point of which has $l^1$-distance from $0$ exactly $1$. In $\mathbb R^3$, the convex hull of the basis $e_1,e_2,e_3$ is a triangle, each point of which, has $l^1$-distance from $0$ exactly $1$. And so on... the same argument holds for the space $l^1$, ...

2

Because being lazy is an art-form, the simplest examples I can think of: a)$A=\pmatrix{0&1&0\\0&0&0\\0&0&0}$. $B = \pmatrix{0&0&0\\1&0&0\\0&0&0}$ b) $A$ as above and $B = 2A$ c) $A$ as above. $B = 2A$ and $C = 3A$. d) $A$ as above and $B = A$.

1

Hints: Just generate two random matrices and check whether $AB=BA$ is true (if it is, start again). It unlikely to happen by chance. Take one of them as $2$ times the identity matrix. Try $$A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$$ and $$B=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & ... 0 a) A=\left( \begin{array}{ccc} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & 6 & 7 \\ \end{array} \right) B=\left( \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{array} \right) A\cdot B =\left( \begin{array}{ccc} 30 & 36 & 42 \\ 54 & 66 & 78 \\ 78 & 96 & 114 \\ ... 1 Note that by definition A is dense in \operatorname{cl}(A), so it suffices to prove the following: if f_1, f_2: X \to Y are continuous between spaces X and Hausdorff Y, and if for a dense subset D of X we have that \forall x\in D: f_1(x) = f_2(x), we have equality for all of X. This follows from the following facts: \Delta_Y = \{(y,y): ... 2 Consider the sets U = [ 0 , \omega+1 ) \subseteq \omega_1 and V = [ 0 , \frac{1}{2} ). Clearly U and V are open in \omega_1, [0,1), respectively, and so U \times V is open in \omega_1 \times [0,1) with the product topology. However this set is not open in the dictionary order topology. Note that \langle \omega , 0 \rangle \in U \times V. ... 0 According to the definition of Arora, trigonometric functions are not good counter-examples since they are not functions from \mathbb{N} to \mathbb{N}. However, the function TUC : \mathbb{N} \rightarrow \mathbb{N} defined by TUC(n) = n if M_n(n) \neq n and TUC(n) = 2n + 1 otherwise, is not time-constructible: Suppose there exists M that ... 0 There is no 1 (multiplicative identity) so it is not a ring. x sin(x) is either 0 (when x=0) or is a transcendental number due to Lindemann–Weierstrass theorem when x <> 0 0 Two fantastic books that every mathematician should know are Counterexamples in Analysis by Olmsted Counterexamples in Topology by Steen and Seebach, already mentioned in the comments. 1 Let A=\begin{bmatrix} 1 & 0\\ 0 & 2\end{bmatrix} and B=\begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix}. Clearly the first equality always holds, however \begin{bmatrix} 1 \\ 1\end{bmatrix} is an eigenvector of B, but not of A. 1 The following example (which I am fairly certain is not a corkscrew-like construction) is essentially taken from A. Mysior, A regular space which is not completely regular, Proc. Amer. Math. Soc. 81 (1981), pp.652-653, MR601748, AMS link Let X = (\;\mathbb{R} \times [ 0 , 2 )\;) \cup \{ \langle 0 , -1 \rangle \}, and topologise X as follows: all ... 1 I think you want a counterexample with more conditions that you wrote, but to help you with your question I give this counterexample f(x)=0 if x\neq 0 and f(0)=1. We have that f'(x)=0 for x\neq 0 and thus \lim_{x\to 0} f'(x)=0 and f'(0) doesn't exist since f is not continuous at 0. 1 For your function f(x), if x\ne 0 then f'(x)=-\frac{1}{x}\cos(1/x)+\sin(1/x). If 1/x is of the form 2n\pi, where n is a non-zero integer, then f'(x)=-2n\pi. So there are x near 0 at which |f'(x)| is very large. By looking at 1/x of the form (2n+1/2)\pi, we can find x arbitrarily close to 0 such that f'(x)=0. Thus \lim_{x\to ... 3 Consider a computer system which has a data type of integers, called \def\Int{\mathtt{Int}}\Int, and a data type of errors, called \def\Err{\mathtt{Err}}\Err. Now consider a function, which might run successfully and return an \Int, or unsuccessfully and return an \Err. Such a function returns values from the coproduct type \Int + \Err. ... 3 Consider g_2(x)+a\,x for appropriate values of a. 1 As you said, you think of \mathbb R as a line and \mathbb R\times\mathbb R as a plane. Well, we think of \mathbb R \sqcup \mathbb R as two disjoint lines. Of course already thinking of \mathbb R as a line implies a topology on \mathbb R at least, as well as thinking of \mathbb R\times\mathbb R as a plane implies a topology, after all \mathbb ... 2 \mathrm{Humans} = \mathrm{Men} \sqcup \mathrm{Women} \mathrm{Integers} = \mathrm{odd} \sqcup \mathrm{even} Do you really seriously ask if there is any use of the disjoint union of sets? It arises everywhere. 1 Your solution to (3) appears correct to me (a tiny bit more later). For (4) you just need to show that every nonempty open set contains a rational number. The nonempty open sets in this space are just unions of the sets you described (i.e., sets of the form (a,b) or (a,b) \cap \mathbb{Q} for a<b). So it really suffices to show that all of these ... 6 Look at the following diagram (created using http://Presheaf.com. I didn't know how to do the \Bbb-letters there) Since both paths simply identify (\Bbb N×\Bbb Q×\{0\})\cup(\Bbb N×\{0\}×\Bbb Q)∪(\{0\}×\Bbb Q×\Bbb Q) to a point, we have induced set maps s and \tilde s, where the later is a bijection. Since the lower left arrow is a quotient map, ... 4 There appears to be a detailed proof due to Kathleen Lewis in section 1.5 of Parametrized Homotopy Theory by J.P. May and J. Sigurdsson. 0 Take any infinite-dimensional Banach space and choose its Hamel basis \left(e_i\right)_{i\in I}, say, with \left\|e_i\right\|=1. Define f\colon X \to X by$$ f(e_i) = \alpha_i e_i $$where {\alpha_i} is any unbounded collection of (nonzero) numbers. Clearly f is a bijection, and f is unbounded, hence discontinuous. Now, the above relies ... 0 We can build a bijective linear function which is not continuous even from a Banach space E in itself (clearly B can't be finite dimensional). Using AC (we need the Hamel basis) it is easy to build a discontinuous linear functional$$\varphi : B \to \mathbb{R}$$Let u\in B such that \varphi(u)=1. Then define$$S : B \to B S(x) = x - ...

1

This requires some form of the Axiom of Choice. Take two Banach spaces that are not isomorphic, but both have Hamel bases of the same cardinality (which will be the case e.g. if they are both infinite-dimensional and have cardinality $\bf c$) and define an operator using a one-to-one correspondence between Hamel bases.

1

This is incorrect, the quotient map from $\mathbb{R}$ to $\mathbb{R}/\mathbb{Q}$ is an open map. Indeed, $\mathbb{Q}$ is dense in $\mathbb{R}$ so one can easily see that, given any (non-empty) open set $U \subset \mathbb{R}$, the image of $U$ under the quotient map is the whole of $\mathbb{R}/\mathbb{Q}$ (which is of course open in the quotient space). You ...

2

A typical situation occurs in the numerical solution of differential equations and in numerical differentiation by finite differences. In the numerical integration, decreasing the step length and increasing the number of steps increases the precision of the solution up to a certain point where the accumulated noise of floating point truncations over the ...

1

We know that we are looking for a non $\sigma$-finite example. And it's pretty obvious that the non-uniqueness will follow from the non-uniqueness of the solution to the equation $\infty+x=\infty$ (since $\lambda$ must be defined pointwise as $\lambda(M)=\mu(M)-\nu(M)$, which whenever this difference can be defined - which is whenever we don't have ...

2

Another one I came across was a function composition problem. Let $g(x)=x^2.$ Find $(g\circ g)(x)$. Well it should be $$(g\circ g)(x)=g(g(x))=g(x^2)=(x^2)^2=x^{2\cdot{2}}=x^4$$ But of course I should have caught that my students would do the "natural thing" and say $$(g\circ g)(x)=g(x)\cdot g(x)=x^2\cdot x^2=x^{2+2}=x^4$$ I blame myself for not catching ...

1

If $T : X\rightarrow Y$ is linear bijection between Banach spaces $X$ and $Y$, then the following are equivalent $T$ is continuous. $T^{-1}$ is continuous. $T$ is an open map. $T^{-1}$ is an open map. $T$ has a closed graph. $T^{-1}$ has a closed graph. You may want to read the previous post in light of this.

4

Of course not. Take $\Omega=\mathbb R$, $\mu$ the Lebesgue measure, and $f(t)=1$.

13

Suppose $A$ were a ring; then in particular, we have $2\sin 1=n\sin n$ for some $n$. Now, use the duplication formulas for $\sin$ to write $\sin n$ as $(\sin 1)\cdot P_n(\cos 1)$, where $P_n(x)$ is a polynomial of degree $n$ (this can be shown straightforwardly via induction; these polynomials are known as the Chebyshev polynomials of the second kind). ...

0

Tread very careful here. First of all conjectures can say two different things: a) There is something with a described property. or b) Everything has a described property. Depending on which kind of conjectures you want to disproof the method differs: If the conjecture is of the kind: Everything has a described property. Then it is ...

2

No. If $\sigma(ab)=\{ 0\}$, then $\sigma(ba)\subset\{ 0\}$ since, as you mentioned it, $\sigma(ba)\setminus\{ 0\}=\sigma(ab)\setminus\{0\}=\emptyset$. But $\sigma(ba)$ is nonempty, so you must have $\sigma(ba)=\{ 0\}$. What is possible is that $\sigma(ab)$ contains $0$ and $\sigma(ba)$ does not. For example, take $A=\mathcal L(\ell^2)$, the algebra of all ...

0

David Williams, Probability with martingales (Cambridge Mathematical Textbooks).

0

"minimal counterexample method of proof is to assume to opposite of an argument is true and then finding a counterexample for the opposite and then concluding the validity of the original argument" This is false. The minimal counterexample method is to assume the opposite of a claim is true, and then to examine a minimal (in a sense dependent on context) ...

4

Theorem: If $F$ is a field and $x \in F$ with $x \neq 0$, then either: there exists $y \in F$ such that $y + y = x$, or $x + x = 0$.

7

The multiplicative structure in $\mathbb{Z}$ is determined by the additive structure. We start with the axiom of the multiplicative identity: $1 \times n = n$ for all $n \in \mathbb{Z}$. The axiom of distributivity implies $$2 \times n = (1 + 1) \times n = (1 \times n) + (1 \times n) = n + n = 2n$$ This determines multiplication by positive integers. I'll ...

1

If for each $n\ge 3$ there are at most $7$ indices $i$ with $x_i=n$, then we can bound the left hand sum by $$\frac1{x_1^4}+\ldots \frac1{x_{96}^4}<7\cdot\sum_{n=3}^{14} \frac1{n^4}\approx 0.138<\frac16$$ This is because we may assume wlog. that $x_1\le x_2\le\ldots\le x_{96}$. Then by the assumption of no more than seven equal numbers we get ...

6

$\mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$ is not noetherian. For more examples from field extensions, see math.SE/19426 and your own question math.SE/694440. For example, if $K/F$ is a field extension which is not finitely generated, then $K \otimes_F K$ is not noetherian. From this one also gets more examples by localization, for example $\mathbb{Z}_p ... 0 Answering so that this isn't unanswered. In the comments, Alex Ravsky suggested the following example which was just what I needed. Consider any infinite group$G$equipped with the cofinite topology. Then the multiplication map$G \times G \rightarrow G$defined by$(x,y) \mapsto xy$is separately continuous everywhere, but is not jointly continuous at any ... 2 No such example exists. Assume that there were a countable$\gamma$-cover that is not an$\omega$-cover. Then there is some finite set$a_1, \ldots, a_n$not included in any member of the cover. By the property of a$\gamma$-cover, there are only finitely many (say$N_1$) sets in the cover not containing$a_1$, and$N_2$set not containing$a_2\$, etc. Then ...

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