Can someone give me an example in real numbers where the outer dimension and the inner dimension are the same?

If E is contained in [a, b], then the outer dimension of E is defined as g.l.b the length of G, taking g.l.b over all open sets G that contain E.

The inner measure of E is defined as l.u.b the length of F, where l.u.b is taken over all closed sets F contained in E.

Now my question in real numbers is: Give an example in which outer dimension and inner dimension are the same.
Explain it in detail

Find the outer measure and the sigma algebra generated by the caratheodal measure

There is the question

Enter image description here

Where $ U _ {*} $ is defined by outer dimension
$$
u _ {*} = inf_ {A_k in mathcal {A} \ E subset cup A_k} sum_ {k} u_ {0} (A_k)
$$

With $ Omega $ denotes the room $ U_0 $ Premass is and $ mathcal {A} $ Algebra is in $ Omega $ and
$$
sum_ {c} = {E in Omega | E , , {text is caratheodary measure} , , (u _ {*} – measure) }
$$

Thank you very much.

Outer Dimension – Prove that there is a limited set $ A $ such that $ m ^ * (F) le m ^ * (A) -1 $ for each closed set $ F subset A $

Prove that there is a limited amount $ A subset mathbb {R} $ so that $ m ^ * (F) le m ^ * (A) -1 $ for every closed sentence $ F subset A $,

Here $ m ^ * $ is the outer dimension.

I used the limited and immeasurable Vitali set $ V $and defined $ A = tV $, from where $ t> 0 $, but I could only prove that
$$ m ^ * (A setminus F) ge1 $$
Since then, however $ A $ is also not measurable, I have $ m ^ * (A) le m ^ * (F) + m ^ * (A setminus F) $So I don't know how to end the evidence.

Oracle – How do I use values ​​from the outer query in the sub-subquery?

I can not query data in subqueries based on the value of the outer query. How does that work in this query?

SELECT
    d.txt_advcatdtlcode ,
    c.txt_description,
    COUNT(DISTINCT(j.int_vacancynum) ) AS vacnum,
    (
        SELECT
            COUNT(*)
        FROM
            (
                SELECT
                    job.txt_appkey,
                    COUNT(*) repeats
                FROM
                    data_tbl data,
                    jobapplication_tbl job
                WHERE
                    job.int_vacancynum = data.id
                    AND data.txt_advcatdtlcode = d.txt_advcatdtlcode //error in this line
                    AND job.dat_appdate > TO_DATE('01-may-2018','DD-MON-YYYY')
                GROUP BY
                    job.txt_appkey
                HAVING
                    COUNT(*) > 1
            )
    )
FROM
    jobapplication_tbl j,
    data_tbl d,
    jobcategorydtl_tbl c
WHERE
    j.int_vacancynum = d.id
    AND d.txt_advcatdtlcode = c.txt_code
    AND j.dat_appdate >= TO_DATE('01-MAY-2018','DD-MON-YYYY')
GROUP BY
    c.txt_description,
    d.txt_advcatdtlcode
HAVING
    COUNT(DISTINCT(j.int_vacancynum) ) > 1

the mistake

"D". "TXT_ADVCATDTLCODE": Invalid identifier

I added a comment to the corresponding line in the error stack.
Many thanks .

Algorithm – Calculation of the outer polygon

I need to compute an outer polygon from a self-intersecting polygon. I've found that Mathematica has recently implemented the following feature that could solve this problem. Does anyone know how it could be implemented?

https://reference.wolfram.com/language/ref/OuterPolygon.html

Any book or article dealing with this topic would also be helpful.

To illustrate my problem, I have a self-cutting polygon (blue) and want to get the outer polygon (red):

Auto-cutting polygon and outer polygon

If it helps, this polygon is created by "displacing" * a previous polygon that did not overlap itself:

Polygon-term development

Could there be a way to increase the polygon that considers self-overlapping?

Thanks a lot!

* not technically balancing, as only some parts of the polygon move in the normal direction.

Move the outer key inside in an association

I have a list that looks like this:

{15000001-><|"Loss" -> 2.85396*10^8, "Exposure" -> 6.61052*10^10|>,
 15000002 -> <|"Loss" -> 1.25297*10^8, "Exposure" -> 1.57863*10^11|>,
 15000003 -> <|"Loss" -> 2.05979*10^8, "Exposure" -> 6.88024*10^10|>}

I would like to come to the following conclusion:

{<|"Event"->15000001,"Loss" -> 2.85396*10^8, "Exposure" -> 6.61052*10^10|>,
 <|"Event"->15000002,"Loss" -> 1.25297*10^8, "Exposure" -> 1.57863*10^11|>,
 <|"Event"->15000003,"Loss" -> 2.05979*10^8, "Exposure" -> 6.88024*10^10|>}

I tried to make two lists, one with keys and one with values, and put them together, but I could not quite get that to work. I would be happy about suggestions.

linear algebra – The various products on $ k $ -altering forms and their relationship to the outer product

To let $ V $ be a vector space and $ k in mathbb {N} $, Describe $ Lambda ^ k V $ the outer $ k
$
-Power of $ V $,

To let $ f: Lambda ^ k V ^ * to ( Lambda ^ k V) ^ * $ if the card is such that a $ k $-Covector $ eta_1 wedge cdots wedge eta_k $ will be sent to the $ k $– alternating form (identified as a dual element of $ Lambda ^ k V $) $ (v_1, cdots, v_k) mapsto sum_ { sigma} epsilon ( sigma) eta_1 (v_ { sigma (1)}) cdots eta_k (v_ { sigma (k) }) $, (It seems that there is another convention with one factor $ frac {1} {k!} $ in this formula.)

There are two ways (see here) of defining natural products in between $ k $alternating forms. Let's call $ wedge_1 $ the product is based on the formula with the $ Old $ Operator (in the previous link) and $ wedge_2 $ the product is based on the formula with a sum via shuffle permutations (in the previous link).

Question: What is true and what is wrong in the following statements? To let $ omega in lambda ^ k V ^ * $ and $ omega in lambda ^ {k & # 39; v ^ * $,

1) $ f ( omega wedge omega & # 39;) = f ( omega) wedge_1 f ( omega & # 39;) $

1 & # 39;) As 1), but with the $ 1 / k! $ Factor in the definition of $ f $,

2) $ f ( omega wedge omega #) = f ( omega) wedge_2 f ( omega #) $

2 & # 39;) As 2), but with the $ 1 / k! $ Factor in the definition of $ f $,

postgresql – Lateral / Outer Apply Join speed

I have the following Postgres database & query. It takes some time items and multiply his hours by a role tariff if a matching value for a exists Employee& # 39; s roleotherwise it will multiply by a personal rate if there is a match.

Processing 300,000 lines takes about 7 minutes, and I wonder if I can do something to speed it up.

Delete table if data cascade exists;
Create the time_entries table (id int, staff_id integer, entry_date date, hours numeric).
insert in time_entries
values
(1, 1, to_date (& # 39; 15 -01-2019 & # 39 ;, & # 39; tt-mm-yyyy & # 39 ;, 1),
(2, 1, to_date (& # 39; 15 -02-2019 & # 39 ;, & # 39; d-mm-yyyy & # 39 ;, 2),
(3, 1, to_date (& # 39; 15 -03-2019 & # 39 ;, & # 39; d-mm-yyyy & # 39 ;, 3)),
(4, 2, to_date (& # 39; 15 -01-2019 & # 39 ;, & # 39; tt-mm-yyyy & # 39;) 4),
(5, 2, to_date (& # 39; 15 -02-2019 & # 39 ;, & # 39; tt-mm-yyyy & # 39;), 5),
(6, 2, to_date (& # 39; 15 -03-2019 & # 39 ;, & # 39; DD-MM-YYYY & # 39 ;, 6);

Create a table staff_rates (id int, staff_id integer, start_date date, rate numeric);
insert in staff_rates
values
(1, 1, to_date (# 01 -01-2019,, # t -t-mm-yyyy &), 1),
(2, 1, to_date (# 01 -02-2019,, # t -t-mm-yyyy &), 2),
(3, 1, to_date (& # 39; 01 -03-2019 & # 39 ;, & # 39; t-mm-yyyy & # 39 ;, 3),
(4, 2, to_date (# 01 -01-2019,, # t -t-mm-yyyy &), 4),
(5, 2, to_date (# 01 -02-2019,, # tt-mm-yyyy #), 5),
(6, 2, to_date (# 01 -03-2019,, TT DD-MM-YYYY)), 6);

Create the role_rates table (id int, role_id integer, start_date date, rate numeric).
insert in role_rates
values
(1, 2, to_date (# 01 -01-2019,, # t -t-mm-yyyy)), 7),
(2, 2, to_date (# 01 -02-2019,, # t -t-mm-yyyy &), 8),
(3, 2, to_date (& # 39; 01 -03-2019 & # 39 ;, & # 39; dd-MM-YYYY & # 39 ;, 9);

Create table staff (id int, role_id integer);
to insert in staff
values
(1, 1),
(2, 2);

CHOOSE
t.staff_id,
staff.role_id,
t.entry_date,
t.hours * (at rr.rate> 0 then rr.rate else sr.rate end) AS total_rate
FROM time_entries t, staff
LEFT JOIN LATERAL (
SELECT sr.rate
FROM staff_rates sr
WHERE sr.staff_id = t.staff_id
AND sr.start_date <= t.entry_date - "at or before this time entry"
ORDER BY sr.start_date DESC NULLS LAST
LIMIT 1
) sr ON true
LEFT JOIN LATERAL (
SELECT rr.rate
FROM role_rates rr, staff
WO rr.role_id = staff.role_id AND t.staff_id = staff.id
AND rr.start_date <= t.entry_date - "at or before this time entry"
ORDER BY rr.start_date DESC NULLS LAST
LIMIT 1
) rr ON true
WO t.staff_id = staff.id