De nos jours, je trouve la façon dont ils l’ont écrit assez difficile à
comprendre. Je suis plus à l’aise avec $\sup$ que “b.s.” que désigne “borne
supérieure”. Ils se sont servi de $M[f]$ pour $\lVert f \rVert_{\rm Lip}$, où
La mesurabilité de l’application dans le sous-titre est basée sur l’égalité
suivante.
$$
\Bor{\R}{\OXT} \cap \CO = \Bor{\CO}
$$
J’ai passé quatres heures pour comprendre
pourquoi ça entraîne la mesurabilité ?
pourquoi l’égalité elle-même est vraie ?
Réponses
Mesurabilité de la trace sur $\CO$ de $\Bor{\R}{\OXT}$
A la première lecture, je ne connaisais même pas la définition de la trace
d’une tribu sur un emsemble. En effet, c’est une définition universaire sur
des ensembles, selon une question sur la trace sur Math.SE.
Posted on October 7, 2018
(Last modified on February 16, 2021)
| 1 min
| Vincent Tam
|
1 comment
Motivation
The books that I read in the past didn’t explain what a dataframe meant.
Definition
Dataframe
A table of data in which the values of each observed variable is contained in
the same column.
Counterexample
I’ve difficulty in reading long lines of text like the above definition, so
let’s illustrate this definition with a counterexample.
We have carried out repeated experiments with four types of things and obtaine
some data. (Say, poured some liquid into an empty cup and take the temperature.)
Posted on October 5, 2018
(Last modified on February 16, 2021)
| 2 min
| Vincent Tam
|
0 comment
Ultra filter
A filer $\mathcal{F}$ containing either $Y$ or $Y^\complement$ for any
$Y \subseteq X$.
Two days ago, I spent an afternoon to understand Dudley’s proof of this little
result.
A filter is contained in some ultrafilter. A filter is an ultrafilter iff
it’s maximal.
At the first glance, I didn’t even understand the organisation of the proof!
I’m going to rephrase it for future reference.
only if: let $\mathcal{F}$ be an ultrafilter contained in another filter
$\mathcal{G}$. If $\mathcal{F}$ isn’t maximal, let $Y \in \mathcal{G}
\setminus \mathcal{F}$. Since $\mathcal{F}$ is an ultrafilter, either $Y \in
\mathcal{F}$ or $Y^\complement \in \mathcal{F}$. By construction of $Y$, only
the later option is possible, so $Y^\complement \in \mathcal{G}$ by hypothesis,
but this contradicts our assumption $Y \in \mathcal{G}$: $\varnothing = Y \cap
Y^\complement \in \mathcal{G}$, which is false since $\mathcal{G}$ is a
filter.
Posted on October 3, 2018
(Last modified on February 16, 2021)
| 3 min
| Vincent Tam
|
0 comment
Problem
To show that a measure $\mu$ defined on a metric space $(S,d)$ is regular.
outer regularity: approximation by inner closed sets
inner regularity: approximation by outer open sets
Discussion
Since this problem involves all borel sets $A \in \mathcal{B}(S)$, the direct
way $\forall A \in \mathcal{B}(S), \dots$ won’t work. We have to use the
indirect way: denote $$\mathcal{C} = \lbrace A \in \mathcal{B}(S) \mid
\mathinner{\text{desired properties}} \dots \rbrace.$$ Show that
Posted on October 3, 2018
(Last modified on February 16, 2021)
| 2 min
| Vincent Tam
|
0 comment
Motivation
$$ \gdef\vois#1#2{\mathcal{V}_{#1}(#2)} $$
Nets and filters are used for describing convergence in a non-metric space $X$.
Denote the collection of (open) neighbourhoods of $x \in X$ by $$\vois{X}{x}$$.
Definitions and examples
Directed set
A partially ordered set $I$ such that
$$\forall i, j \in I: i \le j, \exists k \in I: k \ge j.$$
Net
A function in $X^I$, where $I$ is a directed set.
example: any sequence in $X^\N$
Convergence of nets to a point
$x_i \to x$ if
$$\forall A \in \vois{X}{x}, \exists j \in I: \forall k \ge j, x_k \in A.$$
example: absolute convergence of series ($I$ is the collection of finite
subsets of $\N$, finite sum $\Sigma \in \R^I$.)
example: Riemann integral ($I$ is the collection of tagged partitions,
the partial order doesn’t depend on tags, $\int \in \R^I$.)
Filter base
A nonempty collection $\mathcal{F} \subseteq \mathcal{P}(X) \setminus
{\varnothing}$ such that
$$\forall F,G \in \mathcal{F},\exists H \in \mathcal{F}: H \subseteq F \cap G.$$
(contains nonempty part of intersection)
Difference with topological basis: sets have to be nonempty here
Filter
A filter base $\mathcal{F}$ so that
contains supersets: $\forall F \in \mathcal{F}, \forall G \supseteq F, G \in \mathcal{F}$
contains intersection: $\forall F, G \in \mathcal {F}, F \cap G \in \mathcal{F}$
The image of a filter $\mathcal{F}$ under a function $f$ is also a filter,
denoted by $f[[\mathcal{F}]]$.
I didn’t plan to test whether Staticman v3 work on GitHub since it’s
proprietary. However, from Staticman issues #222 and #227,
we know that the official server doesn’t respond to
GET /v2/connnect/<USERNAME>/<REPONAME>
To help others, I self-advertised my own Staticman API instance and the
migration to GitLab pages. Unfortunately, nobody had managed to
create a GitHub repo running on my API instance. To convince others that it’s
also working on GitHub, I decided to create a minimal GitHub repo.