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4

The statement is true, but your argument is false: what if $S$ is uncountable? The key is that a union of arbitrarily many open sets is still open - can you see a way of writing $S$ as a union of open sets? HINT: each $\{x\}\subseteq S$ is open . . .


3

$\displaystyle\int_0^\infty \left( \int_y^\infty \cdots \,dx\right) \,dy$ means $y$ is a positive number, and for any particular value of $y$, $x$ must be bigger than $y$. $\displaystyle\int_0^\infty \left( \int_0^x \cdots\,dy\right) \,dx$ means $x$ is a positive number, and for any particular value of $x$, $y$ must be smaller than $x$ but still positive. ...


2

The interchanging of two $\int$ signs is justified using the Fubini theorem. Moreover, notice that $$(x,y)\in [y,\infty)\times[0,\infty)\iff 0\le y\le x\iff (x,y)\in[0,\infty)\times[0,x]$$ and this explains how the domain has changed.


2

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


2

There's a good reason you're having trouble! As written, that's not a topology: for example, let $\mathcal{I}$ be the ideal generated by all finite unions of open intervals of the form $(x, x+{1\over 2})$ for $x\in\mathbb{Z}$. Then the set $\bigcup_{x\in\mathbb{Z}} [x+{1\over 2}, x+1)$ is a union of sets of the form $O\setminus I$ ($O$ open, $I$ in the ...


2

Yes, the fact that this holds for rationals and irrationals separately (but consistently) does allow you to conclude differentiability. In particular, you are trying to show that the limit $$\lim_{h\rightarrow 0}\frac{f(h)}h=1$$ and you have that $f(x)$ at every point is equal to either $f_1(x)=x$ or $f_2(x)=\sin(x)$ at every point. You have ...


2

I assume you want this $\forall K>0$, not all $t$. Since $f(t)^Tf(t)\geq0$, you can take $\beta=(\int_0^{\infty}f(t)^Tf(t)\,dt)^{1/2}.$


1

The upper and lower sums defined with respect to a partition $(a =x_0,x_1, \ldots, x_{n-1},x_n = b)$ of the interval $[a,b]$ are given by $$S(f,P) = \sum_{j=1}^n \sup_{x \in [x_{j-1},x_j]}f(x)(x_j - x_{j-1}), \\s(f,P) = \sum_{j=1}^n \inf_{x \in [x_{j-1},x_j]}f(x)(x_j - x_{j-1}). $$ In this case, $a = 0$, $b = 10$, $n = 10$, and $x_j = 0 + (10-0)(j/n) = j$ ...


1

Just note that $\|f\|_K^2=\int_0^K f(t)^Tf(t) \, dt\leq \int_0^\infty f(t)^Tf(t) \, dt$ for all $K$, since $f(t)^Tf(t)$ is always nonnegative.


1

$\text{“}E_i$ and $F_i$ are disjoint$\text{''}$ could be construed to mean $E_i\cap F_i=\varnothing$, and that is not true. It is true that $E_1,E_2,E_3,\ldots$ are pairwise disjoint. Suppose $x\in\bigcup_i F_i$. Then there is some smallest index $i_0$ such that $x\in F_i$. For that smallest index $i_0$ we have $x\in E_{i_0}$; therefore $x\in\bigcup_i ...


1

Using the Spectral Theorem, $$ Ax=\int_{0}^{\infty}\lambda dE(\lambda)x \\ \mathcal{D}(A) = \left\{ x : \int_{0}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty \right\}. $$ Then the positive square root $\sqrt{A}$ is $$ \sqrt{A}x = \int_{0}^{\infty}\sqrt{\lambda}dE(\lambda)x \\ \mathcal{D}(\sqrt{A}) = \left\{ x : ...


1

Fix $x \in [a,b]$. If $x+h \in [a,b]$, and $h \neq 0$, then $$\frac{1}{h}(\Phi(x+h) - \Phi(x)) = \int_c^d \frac{f(x+h,y)-f(x,y)}{h}dy.$$ For each $c \leqq y\leqq d$, we can apply the mean value theorem to the map $t \mapsto f(t,y)$ to obtain $0<\theta_y<1$ such that $$f(x+h,y) - f(x,y) = h\frac{\partial f }{\partial x}(x+\theta_yh,y).$$ Let $\epsilon ...


1

Let $\gamma_n = |g_1|+\cdots+ |g_n|$. Then $\int \gamma_n = \sum_{k=0}^n \int |g_k|$ and you are given that $\lim_n \int \gamma_n < \infty$, hence the monotone convergence theorem shows that if $\gamma = \lim_n \gamma_n$, then $\int \gamma = \lim_n \int \gamma_n $, and hence $\gamma$ is in $L^1$. In particular, $\gamma(x) < \infty$ for ae. [$\mu$] $x$, ...


1

Yes. A corollary to the Rellich Kondrachov theorem says that for all $p\ge 1$, we have $W^{1,p}$ compactly contained in $L^p$. Evans PDE book has a nice treatment of this. EDIT: Actually, if I recall correctly, for $\Omega \subseteq \mathbb R^n$, to prove that $W^{1,p}(\Omega)$ is compactly contained in $L^p(\Omega)$, we only need the Rellich Kondrachov ...


1

It seems the quoted result immediately implies what you want to show: Let $F$ be a finite set and $A \subseteq F.$ Then the inclusion map $I:A\to F$ defined by $f(a)=a$ for $a \in A$ is an injective map from $A$ into $F.$ So by the quoted result $A$ is finite.


1

I got it. Any feedback on my solution would be appreciated. Fix $\varepsilon>0$. Since $\sum_{n=1}^{\infty}a_n b_n<\infty$, there exists some $N_0\in\mathbb N$ such that $$\sum_{n=N_0+1}^{\infty}a_n b_n<\frac{\varepsilon}{2}.$$ Since $\lim_{n\to\infty} b_n=0$, there is some $N_1\in\mathbb N$ such that if $N\in\mathbb N$ and $N\geq N_1$, then ...


1

The only way this is possible is when $S$ is empty, since then the set $\{d(p_0,p) : p \in X\}$ is empty. Because if $S$ is nonempty, then the minimum is attained for a point in $S$ PS: closed and bounded is not the same as compact in an arbitrary metric space.


1

EDIT: I wrote this assuming the problem asks for distinct $a, b\in A$ with $a-b\in\mathbb{Z}$. See John Dawkins' comment. HINT: For a real $r$, let $[r]$ be the fractional part of $r$: that is, $[r]$ is the least nonnegative real $s$ such that for some integer $z$, $z+s=r$. For example, $[2.38]=0.38$, and $[\pi]=\pi-3$. Let $B=\{[a]: a\in A\}$. Then what ...


1

Suppose for all $x,y\in A$, $x-y\in \mathbb{Q}$ and suppose $A$ is non-empty(otherwise trivially $m^{*}(A)=0$), pick any $x\in A$, then $A\subset x+\mathbb{Q}$ therefore at most countable. For higher dimension, see All distances are rational prove the set is countable


1

How to solve $\tan(y)$ to $y$? I can't solve this. Hint. You may use $$ \arctan (\tan (y))=y,\quad y \in \left(-\frac{\pi}2,\frac{\pi}2\right). $$ Edit. There is a mistake in your steps above, you rather have $$ \int \frac1{\tan (y)}\:dy=\int \frac{\cos (y)}{\sin (y)}\:dy=\ln \left|\:\sin (y)\:\right| $$ giving $$ \ln \left|\:\sin (y)\:\right|=\ln ...


1

In the accepted answer to the linked question $f(x)$ was shown to be a monotone function which was strictly increasing on the open interval of those $x$ such that $0<f(x)<|A|$. Remove connectedness and $f$ may not be strictly increasing on a given open subset of such $x$. In particular, there can be infinitely many values with $f(x)=|A|/2$. A simple ...


1

For any $s \in (\alpha,\beta)$ and for $h$ small enough, the interval $[s,s+h]$ is strictly contained in $(\alpha,\beta)$. Therefore by the hypotheses, $f$ is continuous on $[s,s+h]$ and differentiable on $(s,s+h)$.


1

Note that $$|x-x_0| <x_0/2 \implies x_0/2 < x < 3x_0/2$$ Hence $$(x+x_0)/x^2 < (x_0 + 3x_0/2)(2/x_0)^2 =10/x_0^3$$ The choice is somewhat arbitrary. You simply want to get an upper bound for $1/x^2$ in terms of $x_0$ when $x$ is sufficiently close to $x_0$.


1

Your bound is wrong, but here is an other approach similar : $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\dots>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\dots=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\dots$$ More formaly if ...



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