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You can parametrize such sets and then consider the differentiability with respect to the parameters. The differentiability property will be invariant with respect to diffeomorphisms: if two parametri …

Yet a bit more on the positive side: If $Y$ is a convex subset of an open convex subset $X$ of a normed space, then the set $Z:=\bigcup_{y\in Y}B_y^X$ will do, where $B_y^X$ is the largest open ball c …

$\newcommand{\R}{\mathbb{R}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\de}{\delta}\newcommand{\ol}{\overline}$The property of $\R$ that you want to prove is that the topological space $\R$ is normal. All …

$\newcommand\R{\mathbb R}\newcommand\F{\mathcal F}$A counterexample is as follows. Let $X=(0,\infty)$ and $\tau:=\{\emptyset\}\cup\{(t,\infty)\colon t\in[0,\infty)\}$. Then $\F$ is the usual Borel $\s …

$\newcommand\LL{\mathcal L}$Brief answer: The Scott topology is not Hausdorff, and therefore we have to deal here with the set of limits, rather than with the (unique) limit. Here, for the set $\LL_D$ …

$\newcommand\M{\mathcal M}\newcommand\F{\mathcal F}\newcommand\si{\sigma}\newcommand\th{\theta}$I think the answer is no. Here is the idea of how to show this: Suppose that such a map $\phi$ exists. I …

Apparently, there are no such nice properties. Even $K=[0,1]$ will not be such a compact Hausdorff space. Indeed, consider the countable double-indexed family $(U_{n,k}\colon n\in\Bbb N,k\in\{0,\dots, …

Yes, any regular Borel outer measure $m$ is topologically additive. Indeed, take any $S,T \subseteq X$ such that $S\subseteq U$ and $T\subseteq V$ for some disjoint open subsets $U$ and $V$ of $X$. Ta …

$\newcommand\R{\mathbb R}$$\newcommand\LBV{\mathrm{LBV}}$As was noted in comments, the set of all monotone functions (or your version of it, $\mathrm{CM}^+(\R)$) is not a linear space. An appropriate …

$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$No. Consider first the case $d=1$. Let $\J$ denote the set of all subsets of $\Z$. For $J\in\J$, let $$f_J:=\sum_{j\in J}1_{[ …

$\newcommand\R{\mathbb R}$No. E.g., suppose that $n=k=1$ and $$F(f)=\min\Big(1,\int_0^2 |f(x)|\,dx\Big)$$ for $f\in C(\R;\R)$. Then $F$ is continuous and bounded on $C(\R;\R)$. For natural $m$, let $$ …

$\newcommand\de\delta\newcommand\ep\varepsilon$You wrote Could you help me to show that under Ass1 and Ass2 $$d_H(A, A_n)\rightarrow_{a.s.} 0$$ Of course, this is not true in such generality. For in …

The answer is no. Indeed, let $h$ be the periodic function with period $2$ such that $h(x)=2|x|$ if $|x|\le1$. Let $$f(x):=x+h(x)=x+2 \left| x-2 \left\lfloor \frac{x+1}{2}\right\rfloor \right| $$ for …

By the Riesz–Markov theorem, your $C^*$ is the space of all finite signed Borel measures on $Q:=[0,1]^2$ endowed with the total variation norm $\|\cdot\|$, which is the same as your dual norm $\|\cdot …

This is impossible, because for any complete normed space $X$ and any nonincreasing sequence $(B_n)$ of nonempty closed balls in $X$ we have $\bigcap_n B_n\ne\emptyset$. Indeed, take any nonincreasi …