# Tag Info

## New answers tagged examples-counterexamples

1

Let $G$ be a graph with chromatic number $2$ and list chromatic number $3,$ e.g. $G=K_{3,3},$ or just take $C_6$ and add an edge joining two diametrically opposite vertices. Take the graph $3G$ (the union of three vertex-disjoint copies of $G$); add a new vertex $v$ and edges joining $v$ to every vertex of $3G.$ You can easily show that the resulting graph ...

1

The set of torsion elements of an abelian group is certainly a subgroup. For the example you're looking for in non abelian groups, consider a free group $F$ on two elements, which has no nontrivial torsion elements; then consider any finite nontrivial group $G$; then $F\times G$ will give you the example.

3

Real numbers with the discrete metric. I.e. $d(x,y) = 1$ if $x \neq y$.

0

This I think would be the simplest example. Let $(Y,\tau_d)$ be an uncountable discrete space. Consider the space $X=Y\cup \{x\}$ with topology $\tau=\tau_d \cup \{X\}$. $(X,\tau)$ is clearly compact and, since all points in $Y\subset X$ are isolated, not separable. The space I described is $T_0$. @Léo 's answer gives an example of a compact Hausdorff ...

4

If you think of $p \vee q$ as equivalent to $\neg p \implies q$ and $\neg p \vee r$ as equivalent to $p \implies r$, this is used all the time, whenever you have a result that has two cases.

11

In fact, this happens all the time for a nonabelian group $G$. If $G$ is not abelian, then its commutator subgroup $[G,G]$ is nontrivial. Take any $x\in[G,G]$ different from $1$, and let $H$ be the subgroup generated by $x$. Then $H$ is abelian but maps to $0$ in the quotient $G/[G,G]$!

17

No. Take $G=S_3$. The abelianization of $G$ is $C_2$. But $C_3$ is a subgroup of $S_3$ and certainly does not inject into $C_2$. The correct universal property of the abelianization is as follows: Let $G$ be a group, let $H$ be an abelian group and let $\phi\colon G\to H$ be a homomorphism. Then there is a unique homomorphism $\hat{\phi}\colon ... 1 Let$X=\mathbb{N}$and define$f_n:X^n\to X$by$f_n(\tilde{x})=n$for all$\tilde{x}$. Then this trivially satisfies your extension properties, but cannot come from an associative operation. On the other hand, if you require$f_1$to be the identity map, then every such family does come from an associative operation. Indeed, for any$x,y,z\in X$, we have ... 5 This would be true if$g$were continuous with no other assumptions. If you make$g$discontinuous then "generically" this convergence fails. To be more specific, here is a simple approach: assume$X_n$converges a.s. to some fixed constant$c$, but that the$X_n$are never equal to$c$. Then define$g(c)$to have one value and$g(x)$to have some other ... 1 Take$g(x)=-\frac{7 x^3}{2}+\frac{21 x^2}{2}-\frac{85 x}{8}+\frac{33}{8}$, and set$f(x)=\begin{cases}g^{-1}(x)&\text{if $0\le x\le 4$}\\ h(x)&\text{if $x>4$}\end{cases}$We choose$h$without inflexion points, positive and decreasing, such that$f$is smooth. This is possible, since$g^{-1}(4)\approx 0.0119041928166$and$(g^{-1})'(4)\approx ...

0

(1) associative and commutative. Beyond ordinary addition and multiplication you could take a binary operation $x*y$ defined by: $x*y=x+y+1$ (or any other constant eg $x*y=x+y+2$). Similarly, you could take $x*y=2xy$ or $x*y=3xy$ or $x*y=-xy$ etc. Another idea would be $x*y=\max(x,y)$ or $x*y=\min(x,y)$. (2) associative but not commutative. You could ...

-1

Matrix multiplication is associative, but not commutative. A different operation that is associative and commutative... addition and multiplication are really the canonical examples because if a binary operation "looks" like addition (or multiplication), we usually name it that. For example, the composition of two elements in a group, we call it ...

2

The answer is negative. All such functions are analytic. For totally monotone functions it was proved by Bernstein, and for the general case see J. A. M. McHugh "A proof of Bernstein's theorem on regularly monotonic functions".

6

There's also a cylinder. That's it. You can prove this by fully classifying 2-dimensional Lie groups. It's much easier to classify 2-dimensional Lie algebras, of which there are two up to isomorphism, and hence 2 simply connected 2-dimensional Lie groups up to isomorphism: $\Bbb R^2$ and $\text{Aff}(1)$, the affine transformations of the line. Now one ...

1

Let $X$ be a Hilbert space with orthonormal basis $\{e_n\}_{n\in\mathbb{N}}$. Let $x_n=e_{2n}$, and let $y_n=e_{2n}+\frac{e_{2n+1}}{n+1}$. Take $M$ to be the closed span of the $x_n$ and $N$ to be the closed span of the $y_n$. Note that $M+N$ contains $e_n$ for all $n$, so the closure of $M+N$ is all of $X$. However, I claim that $M+N$ does not contain ...

3

No. The Cantor set is probably the easiest example of an uncountable null set. Of course, there are many others. For instance, every Lebesgue measurable set is a union of a Borel set and a null set.

1

Define $f:[0,1]\to\mathbb{R}$ by $f(x)=1$ if $x\in\mathbb{Q}$ and $f(x)=-1$ if $x\not\in\mathbb{Q}$. Then $f$ is not Riemann integrable on $[0,1]$, but $f(x)^2$ is the constant function $1$, hence is integrable on $[0,1]$.

0

To elaborate on Anthony Carapetis's comment, let $f,g:\mathbb{R}\to\mathbb{R}$ be given by $f(x)=x$ and $g(x)=\begin{cases}x &\text{if$x\leq 0$}\\ x+x^{r+1}&\text{if$x>0$}\end{cases}$. These are both homeomorphisms, and $f^{-1}\circ g=g$ is $C^r$ but not $C^{r+1}$, with nonvanishing derivative everywhere (as long as $r>0$). So they are ...

6

I think Expected outcome for repeated dice rolls with dice fixing qualifies. You roll $n$ standard six-sided dice and may fix any non-empty subset of them, then re-roll the others and again fix at least one of them, and so on until all dice are fixed. The intuition that you should always fix any $6$s you roll if you want to maximise the expected sum of the ...

2

How about $$f(x,y)=\begin{cases}\frac{xy}{x^2+y^2},&(x,y)\neq(0,0)\\ 0,&(x,y)=(0,0)\\ \end{cases}$$

1

Standard example: $$f(x,y) = \dfrac{xy}{x^2 + y^2}$$ with $f(0,0) = 0$.

0

I have built a counter-example for this statement. Because the building process is quite complicated (it has images too). So I post my counter example in the following link: https://hongnguyenquanba.wordpress.com/2016/05/12/problem-6/

0

A convex function must be subdifferentiable on the relative interior of its domain (where it is finite). So the only points where one can have an issue is the boundary of its domain. The function constructed by Kevin Holt is one where all of the domain is boundary--hence Theorem 23.4 in Rockafellar's Convex Analysis cannot be used. As another illustration ...

2

Let $G=(V,E)$ be a (simple) triangle-free graph with $n$ vertices. I will show that $|E|\le\lfloor\frac{n^2}4\rfloor,$ with equality only when $G=K_{\lceil\frac n2\rceil,\lfloor\frac n2\rfloor}.$ Let $k$ be the independence number of $G,$ the maximum cardinality of an independent set of vertices. Since $G$ is triangle-free, every vertex $v$ has degree ...

1

$A = \mathbb{Q}, B = \mathbb{R} \setminus \mathbb{Q}$. Then $\operatorname{Int}(A) = \operatorname{Int(B)} = \emptyset$, while $\operatorname{Int}(A \cup B) = \operatorname{Int}(\mathbb{R}) = \mathbb{R}$. This is the most extreme example of what can go wrong...

1

You can take $A=(-\infty,0]$ and $B=[0,\infty)$, as subspaces of $\mathbb R$ with the usual topology.

5

Let $A=(0,1)$ and $B=(1,2)$. Then $A\cap B=\emptyset$ so the LHS is empty. But the RHS equals $\{1\}$ since $\overline A=[0,1]$ and $\overline B=[1,2]$.

2

There are $2$ ways of handling your example: In axiomatic set theory (normally using the ZFC axioms), everything in that universe is a set, in particular $x$, so if you want to calculate $\bigcup A$, you have to specify the nature of the set $x$, for example, if $x=\emptyset$, or $x = \{\emptyset, \{\emptyset\}\}$, and so on. There is a concept in set ...

1

These are wildly false even for very simple examples. Consider $x_n = \frac{1}{\ln n}$ for $n>1$, or, for any nonconstant polynomial $P(x)$, $x_n = \frac{1}{P(n)}$. We have $x_n \rightarrow 0$, $\left| \frac{x_{n+1}}{x_n} \right| \rightarrow 1$. For $n \geq 0$, let $x_n = \prod_{j=1}^n \left( \frac{7}{6} + (-1)^j\frac{5}{6} \right)$ so our sequence ...

6

Put $x_n = 1/n$ if $n$ is even and $x_n = 2/n$ if $n$ is odd. Then $$\left \lvert \frac{x_{n+1}}{x_n} \right \rvert = 2^{(-1)^n}\frac{n}{n+1}$$ which is less than $1$ for $n$ odd and greater than $1$ for $n$ even and has no limit as $n \to \infty$.

4

Consider $u_n$ such that $u_{2n}=\frac 1 {2n}$ and $u_{2n+1}=\frac 1 {\sqrt{2n}}$. $u_n\rightarrow 0$ but $\frac {u_{2n+1}} {u_{2n}}=\sqrt{2n}\rightarrow\infty$

9

Consider the sequence $$\frac{1}{1},-\frac{1}{1},\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\ldots$$ The terms go to zero, but consecutive terms have ratio that is $1$ in absolute value, half of the time.

5

I'll assume $S$ not empty. Since the function $f\colon S\to\mathbb{R}$, $f(p)=d(p_0,p)$ is continuous, when $S$ is compact its image is compact, hence closed and bounded; therefore the image of $f$ contains its minimum. If $S$ is only assumed to be closed and bounded, but not compact, the statement is not generally true. Consider $X=\{0\}\cup (1,2]$, with ...

3

The function $x\mapsto d(p_0,x)$ is continuous on $X$, hence attains its minimum on $S$ since $S$ is compact.

0

One may consider $$\sum_{n=1}^\infty e^{-(1-|x|)n^2-n}$$ If $x \in [-1,1]$ then $$\left| e^{-(1-|x|)n^2-n}\right|\leq e^{-n}$$ and the initial series converges by comparison. If $x \notin [-1,1]$ then, as $n \to \infty$, using $-(1-|x|)n^2-n=n((|x|-1)n-1)\to +\infty$, one gets $$\lim_{n \to +\infty}e^{-(1-|x|)n^2-n}=+\infty \neq0$$ and the ...

3

$$\sum \frac{x^n}{n^2}$$ $$\ \$$

1

How about a bounded quasinilpotent operator $T \in \mathscr{L}(L^2[0,1])$ defined by $$(Tf)(x) = \int_{0}^{x}f(t)dt.$$ This operator has $\sigma(T)=\{0\}$. The resolvent cannot have a pole at $0$ because $T$ is not nilpotent of any order. So the resolvent has an essential singularity. You can solve for the resolvent by solving ...

8

Some points. 1) An oriented manifold $M$ which occurs as the boundary of an oriented manifold $W$ necessarily has $\chi(M)$ even. This means that not only are all of your surfaces not going to work, but neither is anything that's the boundary of a manifold. 2) Every odd-dimensional manifold has $\chi(M) = 0$ (even non-oriented ones). So you can't try that. ...

8

By the classification of surfaces, every closed oriented surface is a connected sum of tori, so you won't be able to find a $2$-dimensional example. By Poincare duality, any closed odd-dimensional manifold has Euler characteristic $0$, so the smallest dimension where you can find an example is $4$. To find a $4$-dimensional example, you can start with ...

1

Just pick a strictly negative function, let $f(x) = -1$ for all $x$. Then $$-1=\int_0^1 f\ge \int_0^{2/3}f+\int_{1/3}^1f = -{4\over 3}$$ Your positive function will not do the trick since $$\int_E1=m(E)$$ So if $\displaystyle\bigcup_{n=1}^\infty E_n=E$ $$\int_Ef\le \sum_{n=1}^\infty \int_{E_n}f \iff m(E)\le \sum_{n=1}^\infty m(E_n)$$ and the latter ...

0

As suggested by daw, take $u(x) = 1$. If you have some regularity of $\Omega$, then $$\int_{\mathbb R^n} u \nabla v \, \mathrm{d}x = \int_{\partial\Omega} \frac\partial{\partial n} v \, \mathrm{d}s$$ for $v \in C_c^\infty(\mathbb R^n)$. The right hand side cannot be written as $\int_{\mathbb{R}^n} w \, v \, \mathrm{d}x$ for some $w \in ... 1 If$\Omega$is bounded, it suffices to take$u(x)\equiv 1$. Then$\hat u$cannot be in$W^{1,p}(\mathbb R^n)$for$p>n$since it is discontinuous. (Sobolev embedding) 1 In any infinite-dimensional Hilbert space$\mathcal H$, an orthonormal basis is not a basis in the algebraic sense. Suppose not: let$B$be such an orthonormal basis. Let$b_1, b_2, \ldots$be a sequence of distinct members of$B$and$x = \sum_{j=1}^\infty b_j/j$. If$x = \sum_{b \in B} c_b b$(where only finitely many$c_b \ne 0$), then$c_b = \langle b, ...

1

Values of Fabius function or it's first prime in dyadic rationals are rational, for example. Concernig non dyadic rational it seems to be an open question indeed.

26

Let $g \colon \mathbb{C} \rightarrow \mathbb{C}$ be a function that is discontinuous everywhere and bounded. Consider $f(z) := z^2 g(z)$. Then $f$ is continuous only at $z = 0$, and $f$ is differentiable at $z = 0$ as $$\lim_{z \to 0} \frac{f(z) - f(0)}{z} = \lim_{z \to 0} z g(z) = 0.$$

19

Consider $f(z)=|z|^2\cdot \mathbf 1_{\Bbb Q[i]}(z)$

Top 50 recent answers are included