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0

$G=\{0,1,2,3\}$ , $\tau=\{\emptyset, G, \{0,2\},\{1,3\}\}$. Then $(G,+_4,\tau)$ is disconnect toplogical group.


0

Keeping the definiton of the events $A$ and $B$ and applying Bayes' Theorem twice $$ P(A | B) = \frac{P(A\cap B)}{P(B)} = \frac{P(B|A)P(A)}{P(B)}. $$ I am answering the question: is it possible to have $P(A|B) > P(A)$ ? From the equation above, that is indeed possible if $$ P(B|A) > P(B) $$ or, in terms of the original random variables, if $$ P(X_2 ...


2

Let $X = \mathbb{R}^n$ and $d$ the usual euclidean metric. Define $d'(x,y) = \frac{d(x,y)}{d(x,y) + 1}$ for $x,y \in \mathbb{R}^n$. As the map $\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}, x \mapsto \frac{x}{x + 1}$ is monotone and concave, it can be shown that $d'$ is a metric on $X$. Notice that $B_{d'}(x_0,r) = X$ for all $r \geq 1$. We thus ...


3

It seems the following. We shall use this question by Amathstudent and its answer by Brian M. Scott. We prove that each weak Hausdorff compactly generated $T_1$ space is $KC$. Since there is a weak Hausdorff compact $T_1$ space, which is not $KC$ (see the space $\Bbb Q^*\times\Bbb Q^*$ in the answer by Brian M. Scott), this space is not compactly ...


2

Every open subset of $B$ is evenly covered if and only if the covering $C\to B$ is trivial. Duh!


0

Proof that $(P2) \implies (P1)$: Given any $x,y\in G$, take $K = \{x, y \}$. Now, there exists $s\in G$ such that $sK$ commutes with $K$. Therefore in particular, $sxx = xsx$, so by cancellation $sx = xs$ and $x$ commutes with $s$. From here we get: $(sy)x = x(sy) = (xs)y = (sx)y$, so $yx=xy$ by canceling $s$. $\Box$ Note that this proof holds for any ...


1

The first part of the question (about uniform convergence on the $A_{N}$) turns out to be false. To see this, let us consider the set $$ \Omega:=\left\{ \left(x_{n}\right)_{n\in\mathbb{N}}\in c_{0}\left(\mathbb{N}\right)\,\mid\, x_{n}\geq0\text{ for all }n\in\mathbb{N}\right\} $$ of all non-negative null-sequences. It is easy to see that this is a closed ...


1

Let $f(x)$ be 0 everywhere but at $x_1$, where $f(x)=1$ Let $g(x)= 1$ everywhere but at $x_1$ where $g(x)= 0$ Then $f+g$= 1 everywhere and $f*g$= 0 everywhere


2

One of the simplest functions that is discontinuous at a point is a step function. Can you see how to get the discontinuities to cancel in the sum? For the product if one or the other is $0$ everywhere, what will the product be?


1

When given such questions, it is easy to think of piecewise defined functions that by themselves have discontinuities (most commonly the discontinuity is at the point where the piecewise definition "splits"), but their sum / product ends up being constant, and thus continuous. For example, let $$ f(x) = \begin{cases} 1 & \text{if }x \ge 0 \\ -1 & ...


2

There are no such examples, for any Lie group $G$. For, suppose $H$ is an open subset of a Lie group $G$. Because $H$ is open in $G$, $T_e H = T_e G$, so $\mathfrak{h} = \mathfrak{g}$. Using the fact that a connected Lie subgroup of $G$ is completeley determined by its Lie algebra, this implies that the identity component of $H$ and $G$, denoted, $H^0$ ...


1

$$(17\sqrt2-24)(17\sqrt2+24)=17^2\times2-24^2=289\times2-576=2$$ so $$0<\sqrt2-{24\over17}={2\over17(17\sqrt2+24)}\dot={1\over\sqrt217^2}$$ where $a\dot=b$ means $a$ is very close to $b$; in particular, $0<\sqrt2-{24\over17}<{1\over17^2}$. But the convergents to the continued fraction for $\sqrt2$ are $1/1,3/2,7/5,17/12,41/29,\dots$ so $24/17$ is ...


1

First of all, if $\int_0^\infty f(x)dx$ exists, you can not conclude that $\lim_{x\to\infty} f(x)$ exists, let alone equal to zero. There are several proofs of this fact across math.SE. Second, a hint: take $a_n=(-1)^n n^{-1/2}$, then the sum $\sum_{n\ge 1}a_n$ converges, yet the sum $\sum_{n\ge 1}a_n^2$ does not.


1

The conclussion that $f(x)\to0$ is wrong. $f$ may even be unbounded. Counterexample: $f\colon[1,\infty)\to[0,\infty)$ defined as equal to $n^2$ on $[n,n+1/n^4]$, and zero everywhere else. $f$ is nonnegative and integrable, but $f^2$ is not.


4

Hint: Consider $$f(x) = \left\{ \begin{array}{cc} \frac{1}{x-b}, & x < b \\ 1, & x \ge b\end{array}\right.$$


2

Hint: since $[b,c]$ is compact and $f$ is continuous, $f$ must be bounded on $[b,c]$; but this doesn't need to be true on $[a,b)$. Can you think of a function that wouldn't be integrable on $[a,b)$?


1

Your approach is certainly right: superharmonic functions form a convex cone, that is if $f$ and $g$ are superharmonic, then $\alpha f + \beta g$ is superharmonic for non-negative $\alpha$ and $\beta$. Moreover, if $f$ and $-f$ are superharmonic then $f$ is harmonic, so the whole question is to construct an example when there are superharmonic functions that ...


3

Yes. The set of discontinuity points can even be the whole space. For instance, $1_{\mathbb{Q}}$ is Lebesgue integrable, with its integral being $m(\mathbb{Q})=0$.


1

Let us assume that $f$ is continuous (otherwise it is wrong, obviously). As $f$ is concave, $f$ has a (maybe infinite) right and left derivative at every point (let us use the notation $f'_+ / f'_-$ for these). Plus, this derivative is decreasing. In particular, $x>2\implies f'_\pm(x) \le f'_+(2) $. As $f$ is positive, $f'_\pm$ remains positive because ...


1

Simply draw your lines in such a way they all intersect in distinct points. This will give rise to $\binom{5}{2}=10$ intersections and since each line intersects the other 4 in distinct locations, each line will have 4 intersections. Place your points on these intersections and you will have your solution. Since $\binom{5}{2}=10$ is the maximum value for ...


1

Here is an unconventional but instructive proof of the extreme value theorem for a continuous function $f$ on the interval $[0,1]$. Let $H$ be (gasp) an infinite hypernatural. Partition the interval into $H$ equal subintervals, each of infinitesimal length. Among the partition points $p_i$, choose the one, say $p_{i_0}$, with the maximal value of $f$. Now ...


-2

First some (counter-)examples. a) $f$ need not be continuous at $1$. Consider the concave function $f(1)=0$, $f(x)=1$, $x>1$. b) Assuming that $f$ is continuous at $1$, it may still happen that the derivative at $1$ does not exist. Consider the concave function $$ f(x)=\begin{cases}\sqrt{1-(x-2)^2},&\mbox{if }1\leq x\leq 2\\ 1&\mbox{if } 2\leq ...


1

A hint provided that $f$ is of class $C^1$: assume without loss of generality that $f'(1)$ exists (otherwise replace $1$ with any greater number and recall that continuity on compact subsets implies uniform continuity), and pick $1 \leq x<y$. Then, since $f'$ is decreasing, $$ f(x)-f(y) = \int_x^y f'(t) \, dt \leq f'(1) (x-y). $$ Exchanging the rôle of ...


0

It's interesting to try to prove the basic results on derivatives without using the mean value theorem. See for instance this previous MO discussion of the role of MVT in first-year calculus, a related post I made on the Expii "Mean Value Theorem" page, and (also on Expii) a MVT-free proof (inspired by the proof (given e.g. in Stein & Shakarchi) of ...


4

Instead of just giving an example, I'll show a typical path to it. $L^p$ norms are invariant under rearrangement of the function: informally, we can move its values around, for example sorting them in decreasing order. So, it does not seem too restrictive to focus on decreasing functions. If $f$is decreasing, then $\int_{E} f\, dm\leq \int_{0}^{ m(E)} ...


1

Disclaimer I will give only a sketch!! Reference This example is taken from: MVT: Logarithm Counterexample Given the cutted unit circle: $\Omega:=\mathbb{S}\setminus\{-1\}$ Consider the main branch of the logarithm: $\ln(re^{i\varphi}):=\ln r+i\varphi$ Assuming the estimate holds: $$\Delta\downarrow0:\quad2\pi\leftarrow|2\pi-2\Delta|=\left|\ln ...


1

HINTS: As Arthur Fischer said in the comments, the deleted Tikhonov plank is not the space that you described, but rather the space obtained by deleting the point $\{\langle\omega,\omega_1\rangle$ from the product space $[0,\omega]\times[0,\omega_1]$. Show that the deleted Tikhonov plank is a non-closed subset of a compact Hausdorff space. The deleted ...


2

The estimate is always true, as long as $\Omega$ is convex. Because in this case, \begin{eqnarray*} f(b) - f(a) &=& \int_0^1 (\frac{d}{dt}f(a + t(b-a))) \, dt\\ & = &\int_0^1 f'(a + t(b-a)) \cdot (b-a) \, dt. \end{eqnarray*} Now simply take the absolute value and use that: $|\int f \, dt| \leq \int |f|\,dt$ Hence, this has not so much to ...


2

The exponential function $f(x)=e^x$. $$ f(2\pi i)-f(0) = 0 $$ but $f'(c)\cdot 2\pi i=e^c\cdot 2\pi i$ can never be zero. For the estimate I don't know yet.


0

HINT: For (a) you can show that if $U$ is a non-empty open set, then $U$ contains a line segment $L$. $L$ is a closed, discrete set in the whole space, and it has cardinality $\mathfrak{c}$. And $\operatorname{cl}U$ is separable, so by Jones's lemma it is not normal. It Hausdorff, however, so it cannot be compact; why? For (b) you can use the same basic ...


11

I always love to prove that: If $\{a_n\}_{n\in\mathbb{N}}$ is a bounded real sequence, it has a converging subsequence. with the Erdos-Skeres', or Dilworth's, theorem: (Erdos-Szekeres, finite version) Every sequence with $n^2+1$ terms admits a weakly monotonic subsequence with $n+1$ terms. (Dilworth, infinite version) Every infinite POset ...


1

Equip $\Bbb R$ with the topology $\tau=\{U\cup(V\setminus K)\mid U,V\text{ open in the Euclidean topology}\}$, where $$K=\left\{1,\tfrac12,\tfrac13,\dots\right\}$$ This is the coarsest topology containing the Euclidean one and having $K$ as a closed set, and it is generated by the sets $(a,b)$ and $(a,b)\setminus K$ as a basis, where $a$ and $b$ range over ...


3

Let $p(x) \in F[x]$ be such that $p(ab) = 0$. Let $$p(x) = \sum_{i=0}^{n}\alpha_{i}x^{i}$$ where $\alpha_{i} \in F$. Now consider $$g(x) = \sum_{i=0}^{n}\alpha_{i}a^{i}x^{i} \in F(a)[x]$$ What is $g(b)$?


2

I think the classic example of this is the whole field of non-standard analysis. It took 300 years to make infinitesimals rigorous (finally realized in the 1960's), but once equipped with such a toolkit you can derive all the basic calculus results (and much more) in just a couple of lines of infinitesimal algebra.


2

No. Let $S \neq \emptyset$ be as in your question. Let $a \in S$ be arbitrary, $T_a : S \to S$ be defined by $T_a(x) = a + x$ and let $0 = T_a^{-1}(a)$. Then for all $x \in S$, $$T_a(x+0) = a + x + 0 = x + (a+0) = x + T_a(T_a^{-1}(a)) = x + a = T_a(x)$$ Since $T_a$ is a bijection, $x+0=x$, and this is for any $x \in S$.


2

Assume $S$ is nonempty, containing the element $x$, and is such that addition of any element gives a bijection. Then since addition of $x$ is a bijection, there exists a $y$ such that $x+y=x$. We claim that $y$ is an identity element. Let $z\in S$. Then, since addition of $x$ is a bijection, there exists a $w$ such that $z=w+x$. But then ...


2

For $a \in \Bbb C \setminus \Bbb R$, we indeed have $\Bbb R(a) = \Bbb C$; indeed, we have $i \in \Bbb R(a)$; this may be seen as follows: With such $a$, we have $0 < \bar a a \in \Bbb R \subset \Bbb R(a); \tag{1}$ since $\Bbb R(a)$ is a field and $a \ne 0$, we further have $\bar a = a^{-1} (\bar a a) \in \Bbb R(a); \tag{2}$ thus $a - \bar a \in ...


2

The most trivial one would be $$S_k= (0,\frac{1}{k}).$$ Note that $\bigcap_{k=1}^\infty S_k$ is empty. If only finitely many of them are not closed, then the result still hold.


0

From Rockafellar's Convex Optimization book, Theorem 23.3: Let $f$ be a convex function, and let $x$ be a point where $f$ is finite. If $f$ is subdifferentiable at $x$, then it is proper. If $f$ is not subdifferentiable at $x$, there must be some infinite two-sided directional derivative at $x$, i.e. there must exist some $y$ such that $f'(x;y) = ...


4

Let $$g(x)=\cases {\exp\left(-\frac{1}{x(1-x)}\right) & if $0\le x \le 1$ , $x$ irrational \\ 0 & elsewhere}$$ Then, in the interval $[0,1]$, $g(x)$ is only continuous and differentiable at $x=0$ and $x=1$ (and all the derivatives are zero). The copy and paste the $y$-scaled/$x$-scaled-translated copies of it: $$ f(x)=\sum_{n=1}^\infty \frac{g(x \, ...


2

Here is a hint to get started: define $$f(x) = \left\{ \begin{array}{ll} 1 & x \text{ irrational}, \\ 0 & x \text{ rational}. \end{array} \right.$$ Then $x^2 f(x)$ is differentiable at $0$ and nowhere else, $x^2(x-1)^2f(x)$ is differentiable at $0$ and $1$ and nowhere else, $x^2(x-1)^2(x-1/2)^2 f(x)$ is differentiable at $0$ and $1$ and $1/2$ and ...


1

Suppose $f$ is continuous onto from a separable metrisable $X$ to $Y$ which is compact Hausdorff. As $\operatorname{nw}(Y) = \operatorname{nw}(f[X]) \le \operatorname{nw}(X) = \aleph_0$, $Y$ has a countable network and so a countable base (as $Y$ is compact Hausdorff). [Lemma: if $f: X \rightarrow Y$ is continuous and onto, and $\mathcal{N}$ is a network ...


5

The category of Kan complexes is a full subcategory of the category of simplicial sets that is closed under filtered colimits and small products, but it fails to have equalisers. Consider the ordinal $\omega + 1 = \{ 0, 1, 2, \ldots, \omega \}$. Let $\mathcal{C}$ be the poset of $\sup$-closed subsets of $\omega + 1$. It is not hard to see that ...


3

Take a field $k$, put $R := k[x]/(x^2)$ and consider the full subcategory ${\mathscr C}$ of $R\text{-Mod}$ consisting of projective (=flat=injective, see below) $R$-modules. Directed limits of flat modules are always (i.e., for any ring) flat, so ${\mathscr C}$ inherits them from $R\text{-Mod}$. Products of injectives are always (i.e, in any category) ...


0

Why does differentiability imply continuity, but continuity does not imply differentiability? Why are all differentiable functions continuous, but not all continuous functions differentiable? For the same reason that all cats are animals, but not all animals are cats. Differentiability also implies a certain “smoothness”, apart from mere continuity. It ...


1

Here is an intuitive explanation. Continuity requires that $f(x)-f(y)\to 0$ as $x - y \to 0$. Differentiability requires that $f(x)-f(y)\to 0$ as $x - y \to 0$, and that $f(x)-f(y)\to 0$ at least as fast as $x - y \to 0$ (in the sense that the ratio still has a limit). In particular, the conditions for differentiability include the condition for ...


1

$$f(x,y)=y\sin\left(\frac1x\right)$$ at $(0,0)$. (You can define $f(0,0)=0$, for example).


3

At $(x_0,y_0)=(0,0)$,$$(x,y)\mapsto \frac{xy}{x^2 + y^2}\ ?$$ Edit: Following Matthias's comment -- this does not answer the question. I was thinking of a non-existing limit when $(x,y)\to (x_0,y_0)$.


2

No. Jeremy already commented that we may assume the center to be trivial. Now look at the involutions (= elements of order 2). Their centralizers are finite, so there are countably many conjugacy classes of involutions. Two involutions generate a dihedral group. If both involutions are not conjugate (in $G$ and hence within the dihedral group), the dihedral ...



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