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2

Let $S$ be the set of isolated points of $M$. If $S$ is infinite, then, since isolated points are open, writing $S$ as the disjoint union of two infinite sets gives what's needed. Suppose $S$ is finite. Then since $M$ is infinite, there are distinct elements $m_1$ and $m_2$ of $M\setminus S$. Choose disjoint open nhoods $N_1$ of $m_1$ and $N_2$ of $m_2$. ...


0

I think that the question you raise is linked to general function sequences convergence. You should study first pointwise convergence, i.e. for a given point $x$ if $$\lim\limits_{n \to \infty} f_n(x)$$ exists or not. Then study uniform converge. Based on that you'll discover the properties of the limit when it exists.


0

as fast answer I would suggest to construct by induction such set. Being that an open set is a set where every point contains an open neightbourhood completely contained insided the original set. I will try first to demonstrate that such set with an added point is a new set still open. Then by induction. The key point is the induction step


0

The answer is no. See this construction. Take $A = U$. Then $A$ is open, so $A = A_2$ and $cl(A_2) = cl(A)$. But $cl(A) \setminus A$ contains a fat Cantor set of positive measure. Hence, $m(A) = m(A_2)$ (that's the Lebesgue density theorem) but in general $m(A) \leq m(cl(A_2))$.


1

What about the set of points where $f$ has local extrema is the finite sum of points and intervals?


1

Since $\mu$ is Radon, it is inner regular on open sets, in particular on $X$. Thus, there is a sequence of compact sets $K_n$ with $\mu(K_n) > 4^n$. By the Uryshon-Lemma, there is a function $f_n \in C_c(G)$ with $0 \leq f_n \leq 1$ and with $f_n \equiv 1 $ on $K_n$. Now the $M$-test shows that $$ f := \sum_n 2^{-n} f_n $$ converges uniformly and ...


2

Let us consider the measurable space $\left(\left(0,\infty\right),\mathcal{B}\right)$, where $\mathcal{B}$ is the Borel $\sigma$-algebra. Let $\mu$ be the usual Lebesgue measure on this space. Let $M_{n}:=\left(n,n+2^{-n}\right)$ for $n\in\mathbb{N}$. Note that the sets $\left(M_{n}\right)_{n}$ are pairwise disjoint with ...


0

I think your basic problem comes from the standard equations $$e^{1/x} = a \Leftrightarrow a^{x} = e$$ which is perfectly valid. Note that from the above it follows that $a$ depends on $x$. When you take limit of $e^{1/x}$ as $x \to \infty$ the result is not something which depends on $x$, rather it is independent of $x$. You have understand this very ...


0

You have an obvious (very obvious in fact) flaw in your argument. If $M_{i} - m_{i} \leq 2|f(a)|$ and $\Delta x_{i} \leq \epsilon/(2|f(a)|)$ then we clearly have $$S(P) - s(P) \leq \sum_{i = 1}^{n}2|f(a)|\cdot\frac{\epsilon}{2|f(a)|} = n\epsilon$$ and normally when $\epsilon \to 0$ the variable $n \to \infty$ and therefore we can't say that $n\epsilon$ is ...


0

Say $x^*$ is a limit point of $\Omega$. That is, for each $\epsilon > 0$ there exists some $x\in\Omega\setminus\{ x^* \}$ with $|x-x^*| < \epsilon.$ Thus, the set $$ X_{\epsilon} = \{ x\in\Omega\setminus\{ x^* \} : |x-x^*| < \epsilon \} $$ infinite. Now, you can define the indices for the subsequence recursively: $$ k_1 = \min\{ k \ge 1 : x_k \in ...


0

Since $k!\binom{x}{k}$ is a degree $k$ polynomial in $x$ with lead coefficient $1$ it is pretty easy see that it is possible to write $\newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}}$ $$ x^k=\sum_{j=0}^kj!\stirtwo{k}{j}\binom{x}{j}\tag{1} $$ where $\stirtwo{k}{k}=1$. The numbers $\stirtwo{k}{j}$ are the Stirling Numbers of the Second Kind. Using ...


4

I would write it as: Let $P(n)$ stand for the expression: $$\forall x\leq n(x\not\in A)$$ Then use the assumption that $A$ has no least element to prove that $P(1)$ and $P(n)\implies P(n+1)$. Thus, we've shown that $\forall x:x\not\in A$, which means $A$ is empty. That's essentially the same as your proof, but uses less set notation.


4

If $P$ is a perfect subset of the real numbers, show that there exist $a < b < c < d$ such that $P \cap [a,b]$ and $P \cap [c,d]$ are both perfect. Let $P_1 = (P \cap [a,b]) \cup (P \cap [c,d])$. Repeat this process with each of $P \cap [a,b]$ and $P \cap [c,d]$, obtaining $P_2 \subset P_1$ consisting of intersections of $P$ with 4 intervals, $P_3 ...


-1

I doubt you can prove this in the most general case. Take any finite set $S$ with the indiscrete topology. Then $S$ will be a perfect subset, indeed a perfect space. Now define a Cantor Set to be a set that is topologically isomorphic to the usual construction of removing thirds of the real line. Every Cantor set must then be uncountable, so there can be no ...


11

Denote $I=[a,b]$ and let $f:I\to\mathbb R$ be a differentiable function whose differential vanishes identically. Let $\epsilon>0$. For any $x\in I$ we have $f'(x)=0$, so by definition of derivatives, there is $\delta_x>0$ such that $|f(x)-f(y)|<\epsilon|x-y|$ whenever $|y-x|<\delta_x$ and $y\in I$. The open intervals $(x-\delta_x,x+\delta_x)\cap ...


5

This is not possible. Consider a function defined on two disjoint closed intervals. If the Heine-Borel theorem implied the result on a closed interval, then the proof should work equally well on the disjoint union of two closed intervals. But in that case the result is false-the function taking the value $1$ on the first interval and $3$ on the second ...


0

Your observation is correct. The fact that \begin{equation} \limsup_{n\to\infty} \sqrt[n]{a_n} = 1 \end{equation} does imply that the root test gives no information, according to Theorem 3.35 (c). Since \begin{equation} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert \geq 1, \quad \forall n\in \mathbb{N}, \end{equation} i.e., \begin{equation} ...


0

Denote by $X_p\in\mathbb R^n$ the vector $(\Vert f_1\Vert_{L_p(0,1)},\dots ,\Vert f_n\Vert_{L_p(0,1)} )$. Then the quantity you are looking at, say $I_p$, is equal to $\Vert X_p\Vert_{\ell_p}$. As $p\to\infty$, you know that $X_p\to X=(\Vert f_1\Vert_\infty,\dots ,\Vert f_n\Vert_\infty)$ component-wise, hence for any norm on $\mathbb R^n$; and you also know ...


3

Ooh, a tauberian theorem with a simple elementary proof, cool. Assume $x$ is always so large below that $xf(x)$ is increasing. Fix $\delta>0$. It follows that $$\int_x^{(1+\delta)x}f(y)\,dy\sim A((1+\delta)^\alpha-1)x^\alpha.$$In particular, if $x$ is large enough we have $$\int_x^{(1+\delta)x}f(y)\,dy\le(1+\delta)A((1+\delta)^\alpha-1)x^\alpha.$$But ...


0

First, recall the definitions: Closure: The union of a set and its limit points, or the union of a set and its boundary points, or even more broadly, $x$ is in the closure of $S$ if every open interval containing $x$ intersects $S$ Boundary point: Every neighborhood of $x$ intersects both $S$ and the complement of $S$ Interior point: at least one ...


1

Let $S=\{x_1,\ldots,x_n\}\subset \mathbb R$. If $x_j\in S$, then any open ball centered at $x_j$ contains infinitely many points, so clearly it contains a point not in $S$, and so $x_j$ isn't an interior point of $S$. Hence the interior is empty. A singleton set $\{x\}$ is closed as $x=\bigcap_{n=2}^\infty \left[x-\frac1n,x+\frac1n\right]$ is the ...


0

None of the points in your finite set $A \subset \mathbb{R}$ are interior points: no open interval around any point $x \in A$ is contained in $\{ x \}.$ This shows that the interior of $A$ is the empty set. On the other hand the set of boundary points of $A$ consists exactly of $A$: every point $x \in A$ is a boundary point since every open interval around ...


0

An open set contains an open interval and thus has infinitely many points. Thus the interior of a finite set is empty. Any point outside the set has a positive minimum distance from points in the set, so there is an open interval around it not intersecting the set. Thus the set is closed because its complement is open. A closed set with empty interior is ...


2

Certainly $A\subset B.$ To see $A$ can be a proper subset of $B,$ take $U=(0,1),F=[0,1].$ Then $f(x) = x\sin (1/x)$ is $C^\infty$ on $(0,1)$ and continuous on $[0,1].$ So $f\in B.$ But $f'$ is unbounded on $(0,1),$ hence $f\not \in A.$ I would guess that $A$ is always a proper subset of $B.$ To have any chance for $A=B$ we should at the very minimum define ...


2

Note that $$ \lim_{n\to\infty}\sum_{k=n}^\infty\frac1{k^2}=0 $$ and for $m\gt n$, $$ a_m\le a_n+\sum_{k=n}^\infty\frac1{k^2} $$ First, take the $\limsup\limits_{m\to\infty}$: $$ \limsup_{m\to\infty}a_m\le a_n+\sum_{k=n}^\infty\frac1{k^2} $$ which must be non-negative. Then take the $\liminf\limits_{n\to\infty}$: $$ ...


1

Heh. If anyone's wondering how to construct those independent sets, here's a cute way to organize the construction: Say $0<\alpha<1$. For $S\subset[0,1]$ define $T(S)\subset[0,1]$ by $$T(S)=\alpha S\cup(\alpha+(1-\alpha)S).$$(Here $cS=\{ct:t\in S\}$ and $c+S=\{c+t:t\in S\}$.) Define $$E_1=[0,\alpha]$$and $$E_{n+1}=T(E_n).$$


2

It means (ii), which is clearly equivalent to (iii). It's also clear that (ii) implies (i): Given $\delta>0$, if $P$ has mesh less than $\delta$ and $P'$ is a refinement of $P$ then $P'$ has mesh less than $\delta$. But in this generality (i) does not imply (ii). Silly example: Say $[a,b]=[-1,1]$. Define $F(P)$ by saying that $F(P)=1$ if $0\in P$, ...


0

If $f$ is of bounded variation, then it is the difference of two increasing functions. Hence, it is Riemann integrable.


0

Suppose not i.e., $F$ is differentiable at some rational point $x_0 \in [0,1]$ and therefore $f$ is continuous at this $x_0$. Let $x_0 \in [0,x]$ with $0\le x\le 1$. Since $0\le x_0 \le x$ then $x-x_0\ge 0$ so by Archimedean property there exists a positive integer $n \ge 1$ such that $n(x-x_0)>1$ i.e., $x-x_0 > \frac{1}{n}$. Also, since $f$ is ...


3

Consider a sequence of sets $E_n$ of measure $t \in (0,1)$ whose indicator functions are all independent random variables. That is, given $E_1, \ldots, E_n$, consider each of the $2^n$ sets $G_j = F_1 \cap \ldots F_n$ where $F_i$ is either $E_i$ or its complement $E_i^c$, and let $E_{n+1}$ have fraction $t$ of each $G_j$. Then the intersection of any $k$ ...


0

Given a claimed function $f:A\to B$ that is given by a formula or algorithm for $f(x)$, we show it is well-defined by: Showing that if $x\in A$ then the claimed method for finding $f(x)$ always gives a value. This prohibits things like $\frac 1x$ where $x=0$. This shows possibility and the correct domain. Show that $f$ is consistent, i.e. that $x\in A\land ...


0

Smoothness certainly does not imply boundedness. The function $f(x) = e^x$ is smooth (analytic, even) but is unbounded along with all of its derivatives. In the link you provided the assumption is that the functions are smooth and compactly supported. The key is the last assumption. We know from real analysis that any continuous function with compact ...


0

Any totally ordered set $(M,<)$ has a natural topology related to the order and which is a good way to express its second order properties. This is the order topology, of which open intervals form a basis $\mathcal{I}$. Let's assume for simplicity that $M$ has no extrema. If there is a countable sub-basis $\mathcal{N}$ of $\mathcal{I}$, then the Nested ...


1

On the very next page, Proposition 5.11(e) shows that it is always a unit vector field.


0

$$ \phi'(x) = \begin{cases} 1 & \mbox{for} & x \in [1,2) \\ 2 & \mbox{for} & x \in (2, 3] \end{cases} = \left. f(x) \right\vert_{[1,3]\setminus \{2 \}} $$ Then use the fundamental theorem of calculus and note that difference in one point, here $x=2$, does not matter for the integral value. Note: The derivative does not exist at $x=2$ because ...


2

Just to elaborate a bit. First of all, the above example of Conrado works perfectly, but perhaps the following one is a bit easier to memorize, and it may clearly deliver the main idea that Conrado formulated in the last message of his answer. Let $\Omega = \{a,b\}$ and $\mathcal C = (\{a\}, \Omega)$, then $\mathcal C$ is a $\pi$-system that generates ...


0

I think here you probability need $\Omega$ to be bounded as well. (however since $u\in W_0^{1,2}(\Omega)$, you do not need smooth boundary condition) And I will assume by $d\lambda^n$ you just mean $x\in \mathbb R^n$, the standard integration notation. For your first question, since $u\in W_0^{1,2}(\Omega)$, by Sobolev embedding you have $u\in L^{p^*}$ ...


3

Let $x\in\Bbb R$ and $\delta>0$. For every $\epsilon\in(0,\delta)$ there is some integer $n$ such that $x-\delta<n\epsilon<x+\delta$. Proof: Using the Archimedean property, we see that there exists some $u\in\Bbb Z$ such that $u\epsilon\ge x+\delta$ and some $v_0\in\Bbb Z$ such that $v\epsilon\le x-\delta$ for any integer $v\le v_0$. Then the ...


2

Here is how I would prove questions 1. and 2. 1. $h$ is an homeomorphism By the mean value theorem you have for $x,y \in \mathbb R^b$ $$\Vert g(x) - g(y) \Vert = \Vert (h(x) - h(y)) - (x-y)\Vert \le M \Vert x-y \Vert$$ hence by the reverse triangle inequality $$(1-M)\Vert x- y \Vert \le \Vert h(x)-h(y)\Vert$$ which implies that $h$ is injective. According ...


1

Oh well, I will answer the question just because it seems to me that it is better to have as many as possible questions on the site that are answered. I would like to thank Erick Wong who pointed me in the right direction with his comment. In this paper Louis Joel Mordell proves much more general result in a theorem which he states in this way: A ...


0

Those are not equivalent. For $f(x)$ a real function, the limit of $f$ as $x\to\infty$ being $a$ means that, for all $\varepsilon>0$, there exists $\bar x\in\mathbb{R}$ such that $|f(x)-a|<\varepsilon$ whenever $x>\bar x$. That is: $$\lim_{x \to \infty} f(x) = a \Leftrightarrow \forall \varepsilon > 0, \ \exists \bar x\in\mathbb{R} \ \mbox{ ...


1

You're misunderstanding the notation. The $\lim$ notation works on expressions, not entire equations. So $$ \lim_{x\to\infty} e^{1/x} = a \qquad\text{means}\qquad \left(\lim_{x\to\infty} e^{1/x} \right) = a $$ And $$ \lim_{x\to\infty} e = a^x \qquad\text{would mean}\qquad \left(\lim_{x\to\infty} e \right) = a^x $$ but that is nonsense because $x$ does not ...


0

They are not equivalent simply because the in first limit you wrote is $a=1$, in the second the limit is $\infty$ if $|a|>1$ $0$ if $|a|<1$ $1$ if $a=1$ doesn't exist if $a=-1$


0

If you already know that $\lim\limits_{n\to\infty} y_n$ exists, then you can apply the order limit theorem twice to obtain the result $$\lim_{n\to\infty} x_n = \lim_{n\to\infty} y_n = \lim_{n\to\infty} z_n.$$ But part of the squeeze theorem is the conclusion that (under the hypotheses of the squeeze theorem) $\lim\limits_{n\to\infty} y_n$ exists, it is not ...


1

(I'm probably going into topics you haven't covered, so bear with me.) Have you heard of a "topology"? A topology is probably the most general "useful" object for doing analysis. Basically, a topology on a set $S$ (it can be any general set) is a set of subsets of $S$ satisfying three properties, which you can research yourself if you're interested. These ...


1

After what you wrote and just as user84413 commented, for a given tolerance equal to $\epsilon$ you need to find the first $n$ such that $\frac{1}{2^{n}n^2}<\epsilon$ or $2^n n^2 >\frac{1}\epsilon$. In the case where $\epsilon$ is not too small, inspection could suffice to find the appropriate value of $n$ where to stop. But, just for your curiosity, ...


2

The subsets \begin{align} A &= \{ (x, \frac{1}{x}): x > 0\}\,, \\ B &= \{ (x, \frac{1}{-x}): x < 0\} \end{align} of $\Bbb{R}^2$.


2

Let one set be $\mathbb{Z}$. Let the other one be $\{i+1/(i+1);\,i\in\mathbb{Z},i>0\}$. The main point here is that the two closed sets must both be unbounded to get what you want. Otherwise they will both be compact. (I'm talking about within $\mathbb{R}$ here.)


0

Try to use a good norm, for instance$||F(x,t)-F(x_0,t_0)||=\max\{||f(x,t)-f(x_0,t_0)||,||g(x,t)-g(x_0,t_0)||\},$ and then use the continuity of the functions $f, g$. And remember that all norms in finite dimensional vectors spaces are equivalents. Then continuity in one is continuity in another.



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