dg.differential geometry – Closeness of a family of functions

I am studying chapter 4 of Geometric Measure Theory book by H. Federer and I have some questions from the following part:

Assuming that $X, Y$ are Banach spaces with $operatorname{dim} X<infty$ and that $U$ is an open subset of $X$ we let
$$
mathscr{E}(U, Y)
$$

be the vectorspace of all functions of class $infty$ mapping $U$ into $Y$. For each nonnegative integer $i$ and each compact subset $K$ of $U$ we define the seminorm
$$
v_{K}^{i}(phi)=sup left{left|D^{prime} phi(x)right|: 0 leq j leq i text { and } x in Kright}
$$

whenever $phi in mathscr{E}(U, Y)$.

  1. Why is $v^i _K$ a seminorm? For this, I have to prove that if $v_{K}^{i}(phi)=0$, then the mentioned $sup$ is also $0$, but how can I show this?

The family of all seminorms $v_{K}^{i}$ induces a locally convex, translation invariant Hausdorff topology on $mathscr{E}(U, Y) ;$ basic neighborhood of any $psi in mathscr{E}(U, Y)$ are the sets
$$
mathscr{E}(U, Y) capleft{phi: v_{K}^{i}(phi-psi)<rright}
$$

corresponding to all $i, K$ and all $r>0 .$ We let
$$
mathscr{E}^{prime}(U, Y)
$$

be the vectorspace of all continuous real valued linear functions on $mathscr{E}(U, Y),$ and we endow $mathscr{E}^{prime}(U, Y)$ with the weak topology generated by the sets
$$
mathscr{E}^{prime}(U, Y) cap{T: a<T(phi)<b}
$$

corresponding to all $phi in mathscr{E}(U, Y)$ and all $a, b in mathbf{R} .$ Defining
$$
operatorname{spt} phi=U cap operatorname{Clos}{x: phi(x) neq 0} text { for } phi in mathscr{E}(U, Y)
$$

spt $T=U sim bigcup{W: W$ is open, $T(phi)=0$ whenever
$$
phi in mathscr{E}(U, Y) text { and } operatorname{spt} phi subset W}
$$

for $T in mathscr{E}^{prime}(U, Y),$ we observe that spt $T$ is compact because
$T leq M cdot v_{K}^{i}$ for some $i, K$ and some $M<infty$
Thus we find that $mathscr{E}^{prime}(U, Y)$ is the union of its closed subsets
$$
mathscr{E}_{K}^{prime}(U, Y)=mathscr{E}^{prime}(U, Y) cap{T: operatorname{spt} T subset K}
$$

corresponding to all compact subsets $K$ of $U$. It may also be shown that all members of any convergent sequence in $mathscr{E}^{prime}(U, Y)$ belong to some single set $mathscr{E}_{K}^{prime}(U, Y)$
For each compact subset $K$ of $U$ we define
$$
mathscr{D}_{K}(U, Y)=mathscr{E}(U, Y) cap{phi: operatorname{spt} phi subset K}
$$

and observe that $mathscr{D}_{K}(U, Y)$ is closed in $mathscr{E}(U, Y) .$

  1. How can I observe this? Actually, I don’t understand the topology of these spaces. I read about the topology induced by seminorms, but still don’t understand how to prove that this is close!

Then we consider the vectorspace $mathscr{D}(U, Y)=bigcupleft{mathscr{D}_{K}(U, Y): Kright.$ is a compact subset of $left.Uright}$
with the largest topology such that the inclusion maps from all the sets $mathscr{D}_{K}(U, Y)$ are continuous; accordingly a subset of $mathscr{D}(U, Y)$ is open if and only if its intersection with each $mathscr{D}_{K}(U, Y)$ belongs to the relative topology of $mathscr{D}_{K}(U, Y)$ in $mathscr{E}(U, Y)$.

  1. And finally, why is $mathscr{D}(U, Y)$ a vectorspace? For this, I’m gonna take $f in mathscr{D}_{K_1}(U, Y)$ and $g in mathscr{D}_{K_2}(U, Y)$. Now, to prove that $af+g in mathscr{D}(U, Y)$, can I say that it lies in $K=K_1 cup K_2$?

pr.probability – “Relative compactness of a family of probability measures” and relative compactness & sequential compactness of sets

I’m studying Billingsley’s convergence of probability measures, and wondering why the definition of “Relative compactness of a family of probability measures” reasonable.

In the discussion the set $X$ is always assumed to be a metric space, and let $mathcal{P}(X)$ be the space of all Borel probability measures on $X$ equipped with weak convergence topology.

  • In Billingsley’s textbook, a family of Borel probability measures $Msubsetmathcal{P}(X)$ is said to be relatively compact if every sequence in $M$ has a convergent subsequence with limit in $mathcal{P}(X)$.

Here we also have relative compactness and sequential compactness for general topological spaces:

  • A set $Asubset X$ is relatively compact if $bar{A}$ is compact.
  • A set $Asubset X$ is sequentially compact if every sequence in $A$ has a convergent subsequence with limit in $A$.

Billingsley’s relative compactness is different from the relative compactness in general topology (and also from sequential compactness), so I cannot see why we say such families of measures are relatively compact. So far, I have found a related question, which assumes $X$ is a polish space.

https://math.stackexchange.com/questions/3640221/prokhorovs-theorem-the-statement-precompact-sequentially-compact-relativel

In that question user87690 argued that the term “relatively sequentially compact” is more appropriate. Also since in that question $X$ was assumed to be a polish space, sequential compactness is equivalent to compactness so that we can simply say “relatively compact”. One problem for me is that $mathcal{P}(X)$ is not always metrizable, although it is metrizable if $X$ is separable as polish spaces are.

So I may guess that, as user87690 suggested, “relatively compact” stands for “relatively sequentially compact”, and one shortened the term because either we usually deal with metric spaces $X$ which are at least separable, or simply “relatively sequentially compact” is too long. But I’m still looking for more persuasive explanations.

Any answers will be appreciated. Thank you!

france – How does UK quarantine work for visits specifically to see family?

I would like to see my parents who I haven’t seen since February. They live within an hour’s drive of Dover, I live within an hour’s drive of Calais.

With the current Covid restrictions, people arriving from France (plus other countries) must self-quarantine for two weeks while they are in the UK, but they are allowed to leave the UK again before the two weeks is up – going to the UK for a day trip is therefore technically possible.

Seeing as my trip would be for the day and the only time I would leave the car would be to sit in my parents’ back garden, am I breaking the quarantine restrictions? Is it even allowed? I have no other address to go to in the UK so can’t isolate somewhere else as there is nowhere else to isolate.

Is there any official guidance on this sort of situation? What is the British interpretation of quarantine in the current context? Can I visit my parents or does quarantine make this impossible?

Name of the family of curves

I tried to generalize the Lemniscate of Gerono.

Instead of $x = dfrac{t^{2} – 1}{t^{2}+1}$ and $y = dfrac{2t(t^{2}-1)}{(t^{2} – 1)^{2}}$, I used $x = dfrac{a(t^{2} – 1)}{t^{2}+1}$ and $y = dfrac{bt(t^{2}-1)}{(t^{2} – 1)^{2}}$.

When eliminating $t$, I used the positive root instead of both. What I got is this:

$y = dfrac{bx}{a}left(dfrac{a – x}{2a}right)sqrt{dfrac{a + x}{a – x}}$

Out of curiosity, I replaced $y$ by $y^{2}$. Can I ask what is the name of the family of the curve
$y^{2} = dfrac{bx}{a}left(dfrac{a – x}{2a}right)sqrt{dfrac{a + x}{a – x}}$?

ag.algebraic geometry – Family over the coarse moduli space of curves

Let $k$ be an algebraically closed field. As the coarse moduli space of curves $M_g$ of genus $g$ over $k$ is not a fine moduli space, it does not have a universal family. But I am wondering if it has a family (proper and flat) such that the fiber over every point $(C)$ of $M_g$ is isomorphic to the curve $C$.

As a disclaimer: I am not that familiar with the language of stacks. As far as I understand the situation in this context, the stack $mathcal{M}_g$ has an universal family $mathcal{C}_g$. The corresponding coarse moduli space of $mathcal{C}_g$ is $M_{g,1}$, so the coarse moduli space of curves with one marked point. The morphism $pi colon M_{g,1} to M_g$ on the level of quasiprojective varieties is just forgetting about the marked point. This family has the property that the fiber over a point $(C)$ is isomorphic to $C$, at least if $C$ has no nontrivial automorphisms. In all other cases the fiber is isomorphic to $C/operatorname{Aut}(C)$. Is it possible to get something better than that?

Family dentist La Grange KY

Visiting a family dentistry is a very beneficial thing. A single place where the whole family can get dental services and get rid of their dental issues easily. A family dentist will help to deal with all the dental issues and also figure out what kind of family dental issues the whole family is suffering from and what could be the reasons behind it. One should consult a family dentist from time to time.

.

probability theory – Is the family $mathcal{U} = leftlbrace Uleft(left[0, nright]right): n in mathbb{N}rightrbrace$ conditionally compact?

Is the family $mathcal{U} = leftlbrace Uleft(left(0, nright)right): n in mathbb{N}rightrbrace$ conditionally compact?

I got hint, that I should use Prokhorov’s theorem (probability calculus), but I don’t know how to proceed from this at all…

All help will be appreciated.