Theorems on the sets
1) If it exists a maximum for with this is only.
It is reasoned for absurdity, that is one supposes that m1 and m2 is 2 maximums, remembering that a maximum is not other that a making maggiorante part of with from increased it, achieves some that 1 ism that m2 must belong to together and therefore being m1 maximum m2 precedes or is equal to m1, analogous being m2 maximum m1 precedes or he is equal to m2, achieves some that the 2 can only coexist if m1 = m2.
2) an entirety totally ordered Is X and is To X, with To ended and not empty To it admits maximum and minimum.
It is demonstrated for semiinduction:
* if I have a single element then it is max and min.
* if I have 2 elements then I make the comparison and I find max and the min.
* if I have n elements then I make n-1 it confronts and it determines max and the min.
This is true single if To it is ended (that is constituted from a n° of elements n, with n ? ?), and the ordering is total.
3) Property of density on Q:
Description: " x, y with x < y $ infinites such elements z that x < z < y
One takes the average between x and y and a point z 1 is found after which it takes the average between x and z1 and they find z2 and therefore via.
4) Property of Archimedes on Q:
Description: " x, y > 0 $ n ? ? such that nx ³ y
X = is placed p/q ed y = r/s to this point enough to choose n = qr in order to make that nx = pr ³ r/s.
5) Property of thoroughness of " :
Description: is To " , To ¹ 0. If To it is limited advancedly(it admits at least a maggiorante) $ Sup To then? " (that is the smallest $ of the maggioranti). Analogous if To it is limited inferiorly(a minorante) $ Inf To (that is largest admits then at least of the minoranti).
6) Property of density on " :
Description: " x, y with x < y $ infinites numbers rations them z such that x < z < y and infinites irrazionali elements with the same property.
The property of Archimedes concurs us to say that a n existsor ? ? such that the distance between i points x and y is > 2*10 - n0
therefore an alignment will exist decimates comprised them between x and y - 10 - n0 . It will be enough to add figures to the last one of this alignment in order to find other numbers comprised between x and y, the added figures can create is alignments limits to you or periodic giving life therefore to numbers it rations them, than alignments he does not limit and not periodic giving life to irrazionali numbers, all to you rigorously comprised between x and y.
7) Inequality of Bernoulli:
Description: " h ? " , with h > -1 ³ 1 is hadche( 1 h ) n nh
It is demonstrated for induction.
For n = 0 1 ³ 1 is obtained which verification the inequality.
It is supposed true for n and it is demonstrated for n 1 in particular decomposes the power in (1 h)n 1the ³ 1 (n 1)h and it is obtained to the first member (1 h)n (1 h), to this point can be written 2° the member through the inequality demonstrated to the step n and compensating the term (1 h) multiplying it also to 2° the member, is obtained quindi
(1 h)n 1 ³ 1 (n 1)h nh2 ³ 1 (n 1)h
where the term centers them is obtained from the product (1 nh)(1 h). Therefore the inequality is verified also for n 1.
8) How many permutations are possible with n objects?
Description: Pn = n!
It is demonstrated considering that in order to order 1 object in 1ª the case there are n various ways, therefore in order to order 1 object in 2ª the case are n-1 various ways. To k-esima the case we have placed n-1 objects, if k = n then Pn = n!
9) How many permutations are possible with n objects of which k1 equal between they and k2 equal ones between various they but from kthe 1 ?
Evidently if some equal elements are that it renders some permutations identical and therefore of it it reduces the total number that is estrinsecato from n! . In the fattispecie the restriction happens dividend n! for the number of combinations invalidated equal to (n_elementi_uguali)!
10) They are To and B 2 sets countable. Then AxB is numerabile(can be placed in correspondence biunivoca with ?).
It is demonstrated placing the elements of the 2 sets in a matrix and using the diagonal procedure of Cantor for its scansion, in such a way every element comes caught up a single time and therefore is had one correspondence biunivoca between together and ?.
11) the entirety " is not countable.
It is demonstrated for absurdity asserting that the interval (0.1) equipollente to is countable " and demonstrating that a n° can be created that does not belong to with therefore constructed obtained changing the i-esima number of i-esimo the number, therefore not correspondence biunivoca between together is one and ? in how much the function is not suriettiva.
12) Demonstrate that if f are crescent in To and g it is crescent in f(A) g°f is crescent in To.
It is demonstrated observing that if f is crescent in To if x1³ x2 follow that f(x1) ³ f(x2), moreover if g it is crescent in f(A) g(f(x1)) ³ g(f(x2)) and therefore g°f it is crescent.
13) Demonstrate that the composition of a function for its inverse one is the identical application.
It is immediate.
14) Inequality of Cauchy - Schwarz:
Description: " x, y ? " has che 2|xy| £ x2 y2
The remarkable products are used, are had:
(x y)2 ³ 0 x2 y2 2xy ³ 0 x2 y2 ³ -2xy
(x - y)2 ³ 0 x2 y2 - 2xy ³ 0 x2 y2 ³ 2xy
and remembering that |xy| = xy if xy > 0 e |xy| = - xy if xy < 0 the inequality is demonstrated.
15) Inequality of Young:
Description: " x, y, and ? " has che 2|xy| £ andx2 y2/and
ed is placed x = u/a y = av replacing is obtained to 1° member 2|uv| to which the inequality of Cauchy-Schwarz can be applied and therefore the 2|uv| £ u2 v2 replacing so as to to obtain one disequazione in x and y it is obtained:
2|uv| £ to2x2 y2/a2 in which it can be replaced to =
.
16) Varying of the triangular inequality:
Description: | ||x|| - ||y|| | £ || x - y ||
Part is given ||x|| to its inside one joins and it embezzles y dopodichè applies the triangular inequality ||x y|| £ ||x|| ||y|| and ottiene ||x|| - ||y|| £ || x - y || while leaving they give ||y|| it is reached ||y|| - ||x|| £ || x - y || = || y - x || and finally taking advantage of the characteristics of the module, it is reached to demonstrate the task.
17) x are of accumulation for and > every around of x it contains infinites heads of and.
All the affirmations with > are demonstrated demonstrating the 2 backs e separately ? .
? is banal in fact defines accumulation point x having in every its around a ¹ and one point point x
if x are of accumulation for and then in every its around is at least a point of and, a point is found of and inside to the 1° around and is assumed that ||x-x1|| it is the beam of the new one around, to its inside a new element will find itself of and which different from the 1° it is found in how much around is a structure that excludes the points that they are on the edge.
After all therefore the definition of accumulation point says to us that in every its around there is at least a point of and various from x, this precise theorem saying that in truth ce n' it is not one single but infinites points.
18) If F is one family of sets opened of "n then ?F with is opened of "n .
If x belong to the entirety union of the family of sets x it must belong to one of these sets that they constitute the family, as an example to with To which is an open that it wants to say that every its point is stung inner and therefore also x, must therefore exist one around of x completely contained in To and therefore completely contained in ?F. Therefore x are stung inner to ?F and for the arbitrariness with which we have chosen x follows that all i points of ?F is stung inner and therefore the entirety ?F is opened.
19) If F is one family ended of sets opened ? "n then ?F with is opened of "n .
Can be considered the case of 2 sets opened To and B, takes one point x that belongs is to To that to B and it is demonstrated that it entire is contained in To ?B, is that is inner point and therefore the entirety ?F is an open. So that x are contained in To?B is necessary to choose around of the minimal beam between the beam that concur with x of being stung inner of To and the beam that concurs to x of being stung inner of B.
20) If and it is a closed entirety and it contains its ¶ frontierand and.
If x do not belong to and then unavoidablly it belongs to the complementary Cand which with it is opened (being and closed) and therefore constituted alone inner points it achieves some that x are inner to Cand and therefore does not belong to ¶and, in ¶ conclusionand and.
21) If and it contains its frontier ¶and E and contains all its heads of accumulation.
Remembering that a point of accumulation for and is or a point of frontier or an inner point and observing that and contains also its frontier of it achieves that and contains all its heads of accumulation.
22) If and it contains all its it heads of accumulazione and is closed.
It is necessary to demonstrate that the complementary Cand with is opened that is that it contains solo heads inner that is obvious because if x do not belong to and then do not belong not even to the frontier of and (inasmuch as to say that and it contains its accumulation points is equivalent to say that ¶and and) then x must be inner point to the complementary Cand that therefore it is with open.
23) Theorem of Bolzano - Weierstrass:
Description: Is and "n limited (pu² is enclosed in around of the origin) and infinitely (constituted from n° a infinitely of elements) Esiste in "n at least a point of accumulation for and.
to) a candidate to being stung of accumulation finds itself:
Being and limited then a rectangle T 0 in a position toenclosing it exists, since and it is infinite, this rectangle contains infinites heads of and, the 2 following steps are executed then ricorsivamente:
1) subdivides the rectangle in 4 parti.
2) chooses between the 4 created rectangles a rectangle that still contains infinites heads of and.
3) joins to the S succession of the left edges of the rectangles an element that is ³ of the previous one.
4) joins to the D succession of the edges rights of the rectangles an element that is £ of the previous one.
5) joins to the B succession of the low edges of the rectangles an element that is ³ of the previous one.
6) joins to the succession To of the high edges of the rectangles an element that is £ of the previous one.
From the observation that the sets therefore constructed they are limits to you and therefore they moreover admit at least a maggiorante and a minorante finding to us on ", for the thoroughness property the minimum of the maggioranti exists that is the Sup and the maximum of the minoranti that is the Inf gains the condition of escape from the ricorsione that is must be obtained che Sup(S) = Inf(D) and that Sup(B) = Inf(A) values that come caught up in the distance between the sup and the inf it is given from the distance between the edges it begins them uniform for the n° arbitrary n of the executed subdivisions. The point that possesses these coordinates turns out therefore candidate to being an accumulation point.
b) it is demonstrated that the found point is of accumulation
It must be demonstrated that such point is of
accumulation that is that in every its around of beam and exists infinites heads of the
entirety, in order to make that enough to enclose the rettangolino in
around of centered beam and
in point x, this is obtained centering in the point candidate
the square [ sup(S) - 
, sup(S) ]x[sup(B) - , sup(B) ].
Registering such squared in a circle, it will be had that
the circle has beam and , is
centered in x and contains infinites heads of with for pact pushing
the process of the subdivision until the point in which the distance
between 2 edges of the rettangolino is smaller of .
24) If and "n it is a closed and limited entirety exists Max(E) and Min(E)
Limited, for the thoroughness property, it implies the existence of the Sup and of the Inf which are of the frontier points and since a closed entirety contains also its frontier of it it achieves that and it contains the Inf and the Sup which therefore are respective Min and Max.