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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. ...

3

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

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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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