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Arzela's lemma(in Russian!)

Jan 31, 2012
54
Hello!

I've a problem understanding the following lines(Arzela lemma, and first two sentences of a proof) from Fichtengoltz's book.

I know, that some(2 members?) of you know Russian, help me please translate these line into English, with a short explanation on bold lines.

Пусть конечном промежутке $[a,b]$ содержатся системы $D_1,D_2,...,D_k,...$ промежутков, каждая из которых состоит из конечного числа не налегаюших друг на друга замкнутых промежутков. Если сумма длин промежутков каждой системы $D_k$ $(k=1,2,3,...)$ больше некторого постояного положительного числа $\delta$, то найдется, по крайней мере, одна точка $x=c$, принадлежащая бесконечному множеству систем $D_k$

Доказательство:
Если промежуток какой-нибудь системы $D_k$ $(k>1)$ налегает на промежутки предшествующих систем $D_1,D_2,...,D_{k-1}$и их концами делится на части, то эти части мы впредь будем расматривать как отдельные промежутки системы $D_k$

Thank you!
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,488
Let the finite interval [a, b] contain the systems (i.e., sets) $D_1,D_2,\dots,D_k,\dots$ of intervals, each of which consists of a finite number of non-overlapping closed intervals. If the sum of interval lengths of each system $D_k$ ($k=1,2,3,\dots$) is greater than some fixed positive number $\delta$, then there exists at least one point x = c that belongs to an infinite set (i.e., number) of systems $D_k$.

Proof:
If an interval of some system $D_k$ (k > 1) overlaps with intervals from the preceding systems $D_1,D_2,\dots,D_{k-1}$ and is divided into parts by their ends, then we will consider these parts as separate intervals of the system $D_k$. (End of translation.)

At first I thought that "не налегаюшие друг на друга промежутки" means "disjoint intervals," but probably the right translation is "non-overlapping intervals." In the proof, when a point (the end of an interval from a preceding system) divides an interval from $D_k$ into two parts, both parts are considered elements of $D_k$. But since $D_k$ by assumption contains closed intervals, the two parts will not be disjoint, but will not "налегать друг на друга," i.e., will not overlap.