# Tag Info

2

Consider the tree of order $8$ with degree sequence $(4,4,1,1,1,1,1,1).$ If we color it with colors $1$ and $2,$ the sum of the colors is $12.$ Using three colors, we can color the leaves with color $1$ and the vertices of degree $4$ with colors $2$ and $3,$ so that the sum of the colors is $11.$

3

The tree that is Figure 3.2 in this paper is an example. The chromatic number is $2$, but the lowest sum using only two colors is $21$. A sum of $19$ can be achieved with $3$ colors, using color $3$ for the root, and color $1$ for all leaves.

5

$1-2\left|x-\frac12\right|$ has the three-cycle $2/7,4/7,6/7$. This example comes from the paper "The Sharkovsky Theorem: A Natural Direct Proof", by Keith Burns and Boris Hasselblatt. A preprint is freely available online, and it's a great read.

5

Just pull some function values out of a hat -- for example $$f(0) = 42 \qquad f(42)=117 \qquad f(117)=0$$ and then do Lagrange interpolation (or for that matter linear interpolation, whatever floats your boat) between those points. Your attempt with a first-degree polynomial failed because a first-degree polynomial iterated three times is still a first-...

8

In this answer (and with more detail in this answer) I gave an example, due to Sundaresan, of a compact Hausdorff space $X$ such that if $Y$ and $Z$ are the results of adding one and two isolated points, respectively, to $X$, then $X\sim Z\not\sim Y$, where $\sim$ denotes homeomorphism. Thus, $X\sqcup X\sim X\sqcup Z\sim Y\sqcup Y$, even though $X\not\sim Y$....

0

The function $$f(x) = \left ( \frac{2+\cos x}{3} \right) ^{x^4}$$ has this property. This follows from the fact that $\int_0^{2\pi} [(2+\cos x)/3]^n\, dx \sim 1/\sqrt n$ as $n\to \infty$ (by Laplace's method).

0

The fractional part function $\{x\} = x - \lfloor x \rfloor$ is discontinuous but satisfies the IVP. An strictly increasing function can only have jump discontinuities and so must be continuous if it satisfies the IVP.

3

a solution is $$g(x) = \sum_{n=-\infty, n \ne 0}^\infty e^{-n^4 (x-n)^2}$$ it is real analytic since it is complex analytic it is $L^1$ since $$\|g\|_{L^1} = \sum_{n=-\infty}^\infty \int_{-\infty}^\infty e^{-n^4 (x-n)^2} dx = \sum_{n=-\infty, n \ne 0}^\infty \frac{1}{n^2}\int_{-\infty}^\infty e^{-x^2} dx = 2\frac{\pi^2}{6}\sqrt{2\pi}$$ and clearly $g(x) \... 7 Spaces in which countable intersections of open sets are open are called$P$-spaces. (Warning: the same term is also used with a completely different meaning.) The co-countable topology on an uncountable set is an example of a non-discrete$T_1P$-space. In general we can start with any space$\langle X,\tau\rangle$and let$\tau'$be the collection of$G_\...

1

Consider $D_4$, the group of symmetries of the square. You can show that $Z(D_4)=\{e,a^2\}$, where $a$ is an element of order $4$. Here is a proof sketch: Let $D_4 = <a,b>$, where $o(a)=4$, $o(b)=2$, and $ab=ba^{-1}$. It follows from this that $a^n b = b a^{-n}$ for any $n$. We have $xa=ax$ and $xb=bx$ if and only if $x^n=x^{-n}$, which is the case ...

0

This is along the lines of a nonmeasurable set, and Feynman may have rejected it on physical grounds, but theorem: There exists a nowhere dense set with positive measure, a fat cantor set

1

Presumably you wanted an example of a discontinuous function satisfying IVP. If $f$ is differentiable then the Mean Value Theorem shows that $f'$ has IVP, although $f'$ need not be continuous.

-1

An example is the identity. Hint (for $f$ increasing): $f$ must be surjective in $[a,b]$ for all $[a,b]\subset I$

1

no, not necessary. "$f,g$ are homotopic relative to $\{1\}$" is equivalent to "$[f]=[g]$ in $\pi_1(X)$". "$f$ is homotopic to $g$ relative to $\emptyset$" is equivalent to "$[f]$ and $[g]$ lay in the same conjugacy class". it is easy to construct an example for arbitrary $X$ with nonabelian $\pi_1$.

0

Take a lattice which is not totally ordered and has good number of elements. Most of the times you can delete an element from the lattice and the subset remaining would be a poset and not a lattice. (Easy to see from the Hasse diagram)

3

I was quite surprised to find that "proexample" is actually a word. Google turns up, for example (or should that be "for proexample"?) Corcoran, J. 2005. Counterexamples and proexamples. Bulletin of Symbolic Logic 11(2005) 460. Are you that John Corcoran? If so, you seem to be the main published user of this word, so you are the one who should be ...

7

HINT: You can add $\Bbb Q\cap[0,1]$ to any measurable subset of $[0,1]$ without changing its measure.

-1

Example: A group of people is allowed to buy liquor (with assumed utility for all) if at least one of them carries an ID. (Carrying an ID is an earlier, individual, uncoordinated choice that is assumed not to hurt or benefit your utility otherwise.) In the ESS everybody will carry their ID. NB: Your statement that (1) defines a NE is not correct. (1) is not ...

0

A strategy is an ESS if it is not profitable for a small fraction of the population to deviate from the societal norm. Think of folks developing software in a research lab. Folks can either use the .NET platform on Windows or the Python platform on Linux. If most folks adopt Linux, then it doesn't make sense for a minority to play Windows. The folks adopting ...

3

I don't know if you'd call this a formal definition, but a counterexample to $$\forall(x \in X, y \in Y)P(x,y)$$ is a basically a substitution $(x:= x_0, y:= y_0)$ together with a proof that $$(x:= x_0, y:= y_0) P(x,y) \rightarrow \bot.$$ By the way, we can think of a substitution like $(x:= x_0,y:= y_0)$ as being a bit like an ordered tuple, in this case $(... 20 First a joke: I don't know what a counterexample is, but I can recognize one when I see one. In a first order context, something like the following begins to capture the notion. Let$T$be a theory over the language$L$. and let$\phi$be the sentence$\forall x_1\dots\forall x_n\psi(x_1,\dots,x_n)$. Then a counterexample to$\phi$in the context$T$is ... 5 You can think of our knowledge of math as divided into (1) knowledge about mathematics itself—theorems of algebra, analysis, number theory, topology, etc.—and (2) knowledge about how to do mathematics—paradigmatic cases, heuristics for approaching a problem, counterexamples, etc. Counterexamples are part of our knowledge of how to do ... 0 To use proof by counterexample, the original statement has to be of the form "For all x, statement$A_x$is true". (1) Thus, showing that for some x, the statement$A_x$is false, proves that the original statement is false. A formal definition of a counterexample could be "an example that proves a statement false" or something. This is a rigorous way of ... 10 A counterexample is a special case of a general claim: a case that shows the claim to be false. This is really just a matter of common sense and everyday logic that applies far beyond mathematics. It isn't usually regarded as something that needs formal definition. Once you have a counterexample, you know that the general claim isn't true, and that's the end ... 1 Let's ask when, given a module$M$with a proper nonzero submodule$L$, the set$(M\setminus L)\cup\{0\}$is a submodule. Let$x\in M\setminus L$and let$y\in L$,$y\ne0$. Then also$x+y\in M\setminus L$and$-x\in M\setminus L$; on the other hand $$(x+y)+(-x)=y\notin (M\setminus L)\cup\{0\}$$ Therefore the set is not even an additive subgroup. So your ... 4 Consider$X = \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Then both$(1,1)$and$(-1,0)$are in$(X \setminus TX) \cup \{0\}$, but their sum isn't. In fact let$R$be any ring that has a module$M$containing a nonzero torsion element$m$. Then in$R \times M$, both$(1, m)$and$(-1,0)$are torsion-free, but their sum isn't. In other words, there is no ... 0 I'd stay with your original layout, as it answers the question asked, not emphasising a true statement; just change the labels: Proposition: the sum of two integers is odd Counterexample: 1 + 1 = 2 Thus the sum of two integers may be even 0 If a sentence$\varphi$is false, then$\neg \varphi$is a theorem, and needs to be proven as such. So I'd write: Proposition. There exist integers$a$and$b$such that$a+b$isn't odd. Proof. Define$a=1$and$b=1$. We will show that$a+b$isn't odd. Observe that: $$a+b=1+1 =2.$$ But since$2$isn't odd, hence$a+b$isn't odd, as required. 0 In general terms you assume that$\forall x \in V : P(x)$. But you've found, say, an$n \in V$for which$\neg P(n)$, (i.e. your counterexample), therefore you may conclude that$\exists n \in V : \neg P(n)$. Using de Morgan this is equivalent to$\neg(\forall n \in V: \neg \neg P(n))$, and therefore$\neg \forall n \in V: P(n)$. But this conclusion and the ... 2 If you want to be formal, one way would be to negate the assertion. The original statement claimed: The sum of any two integers is odd. More formally, For all integers$x,y$, their sum$x+y$is odd. The negation of this is: There exist integers$x,y$whose sum is not odd. Or if you prefer, There exist integers$x,y$whose sum is even. ... 4 It looks to me like the biggest step you might be missing is how to formulate the statement you want to prove, which is the complement of the (false) assertion you were given in the exercise. You certainly wouldn't write something like this in a paper (instead you would just express that you are showing a counterexample), but it might be helpful to your ... 3 I would suggest trying to present your example in a different manner though mostly correct. I also recommend presenting it in a manner of using exact language to avoid ambiguity and variables so as to mathematically represent the statement and proceed to use a Proof by contradiction where a Counterexample would be used. So to present your problem: ... 27 In these cases on my homework when I type it up, I just change the statement of the theorem to match what is true. Given that problem on my homework I would write$\textbf{Theorem:}$It is not necessarily the case that the sum of two integers is odd.$\textbf{Proof:}$Observe that$1+1=2$. Since$1$is an integer and$2$is not odd, we have proved the ... 5 The "theorem" you are proving, in the specific case you give, is not "the sum of two odd integers is always odd." A "theorem", by definition, is a true (or at least provable) statement. So the "theorem" cannot be "the sum of two odd integers is always odd," because your argument shows that statement to be false. In fact what you are proving is "the sum of ... 3 To be a bit more formal: If I am choosing to disprove proposition$P$is false, by exhibiting a counterexample, I would be proving. "The proposition$P$is false." And the proof would start from "If$P$, then for all integer (or whatever)$n$,$<$some statement about$n$that is implied by$P>$". Therefore, in particular, for$n=$my ... 14 Prove or find a counterexample: the sum of two integers is odd. The above statement is not true and a counterexample can be easily found. It suffices to check the first few positive integers to discover that$1+1=2$. I think the idea is that since it's not a Proof you shouldn't precede the statement of the counterexample with the word "Proof" like ... 3 It would be as follows: This theorem or proposition is false because the following example satisfy the hypothesis of the theorem but it doesn't satisfy the conclusion. Example ....... 30 I would say something along the lines of: The proposed result is false. Here is a counterexample... 6 Cite the counterexample. Since$1 + 1 = 2$, the sum of two arbitrary integers is not always odd. 4 I think you should add some text to help the reader. Here's my interpretation of an acceptable presentation: Suppose that for any two integers$a$and$b$,$a+b$is odd. Letting$a=b=1$, it follows that$a+b=1+1=2$is odd, a contradiction. Edit: As pointed out in the comments, your presentation is not acceptable, since you have written "Theorem: the ... 0 That is correct.$1/x$on$(0,1)$is easier. 2 Try$f_n(x):=x^n$on$[0,1]$endowed with the Borel$\sigma$-algebra and Lebesgue measure. For any$\varepsilon$, we have the uniform convergence of the sequence$\left(f_n\right)_{n\geqslant 1}$on$[0,1]\setminus [1-\varepsilon,1]$, hence the almost uniform convergence. However, we do not have the uniform convergence on$[0,1]$and even on$[0,1]\...

0

Consider a $$\{ (x,y,z)| x^2+y^2=1,\ z\geq 0\}$$ We have a complete manifold $M_1$ by making a suitable $cap$ around $z=0$ And around $(x-2+\epsilon )^2+y^2=1$ we do same thing so that we have $M_2$ Here they are intersect along $x= \frac{2-\epsilon}{2}$ So by cutting along the intersection and deleting we attach them And by small perturbation we have ...

2

$A = \bigcup_{i=1}^{\infty} \left(\frac{1}{2i+1}, \frac{1}{2i}\right]$, $B = [-1, 0]$ $A$ in answer 1. I think these would work. *Edited : $A$ was originally an open set, whose complement in $\mathbb{R}$ would be closed and thus locally compact. Thanks to clark's critique, the example was modified.

1

If $N > 1$ and $\Omega = B(0,1)$ then $u(x) = \log \log \left( 1 + \frac{1}{||x||} \right) \in W^{1,N}(\Omega)$, the function is continuous on $B(0,1) \setminus \{ 0 \}$ and unbounded at the origin and so does not equal a.e to any continuous function defined on $B(0,1)$. If $N = 1$ then any Sobolev function has a (locally) absolutely continuous ...

2

No. $a=12, b=20, n=30...$ Or $a=p^2 q,\; b=p^2r, \;n=pqr$ with $p,q,r$ primes with $p<q<r.$

8

A counterexample would be $a=6$, $b=6$, and $n=9$. Now $0<a,b< n$ and $ab=36=4\cdot 9\equiv 0\pmod 9$, but $a\nmid n$ and $b\nmid n$.

1

A space $X$ is anticompact if and only if the only compact subsets of $X$ are the finite subsets. Let $X$ be $T_1$, anticompact, and not discrete. Since $X$ is not discrete, it has a non-empty subset $A$ that is not open. $X$ is $T_1$, so the relative topology on each finite subset of $X$ is discrete, and therefore $A\cap K$ is open in $K$ for each compact $... 3 Take$X$to be an uncountable set with the cocountable topology (a set is open if and only if its complement is countable). You can verify that the compact subsets of$X$are finite sets with the discrete topology. Hence, take any$A\subseteq X$that is not open. Then, for any compact$K\subseteq X$,$A\cap K$is open in$K$. 4 One example is what is called a fusion system. A fusion system is a category where the objects are the subgroups of some fixed$p$-group$S$and where the morphisms is a subset of the set of injective homomorphisms between the subgroups which contains all those induced by conjugation by some element from$S$. Further, it is required that any morphism$\...

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