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Such an ideal does not exist. Indeed, suppose the contrary, and let $I$ be such an ideal. The sequence $(x_n)=(1,0,1,0,\dots)$ is almost convergent, and therefore $I$-convergent, to $1/2$. So, $$\m …

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 …

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, …

$\newcommand\abs[1]{\lvert#1\rvert}$It already suffices that $f$ and $g$ be even and nondecreasing on $[0,1]$ (which of course is the case if $f$ and $g$ are even and convex). Indeed, then the identit …

$\newcommand\LL{\mathcal L}$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 $\LL_D$ …

$\newcommand\R{\mathbb R}\newcommand\F{\mathcal F}$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 …

$\newcommand\R{\mathbb R}$ Let $T$ be any countable nonempty set. Let $B$ be the Borel $\sigma$-algebra over $\R^T$ generated by the Tikhonov product topology on $\R^T$. Take any $t_0\in T$. For each …

$\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 …

$\newcommand\R{\mathbb R}$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 …

$\newcommand{\R}{\mathbb R}$Here is an elementary (albeit somewhat longish) solution, without using the Jordan curve theorem. Let us borrow the inversion idea from Olivier Bégassat, so that the two ar …

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 …

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 …

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 …

$\newcommand{\ep}{\varepsilon} \newcommand{\de}{\delta} \newcommand{\B}{\mathcal B} \newcommand{\K}{\mathcal K} \newcommand{\NN}{\mathcal N} \newcommand{\PP}{\mathcal P} \newcommand{\supp}{\operatorna …