Toward the Bolzano Weierstrass theorem from scratch

Understand that mathematical systems are built on 'taken for granted' axioms (or accepted truths. These axioms are 'accepted' to be true without any proof. Any proof of any mathematical law is built on a set of axioms.

We will state one such axiom for real number system called the Completeness Axiom. From wikipedia we quote "The axiom states that every non-empty subset S of R that has an upper bound in R has a least upper bound, or supremum, in R"

Let us understand the different terms in the axiom or given truth:

1. R is bold = This symbol denotes the set (that is collection) of real numbers.
2. Real Number = collect all the fractions, integers, square-roots of non-square numbers and other irrational numbers like e, \(\pi\), and put them in one set. This makes the set of real numbers.
3. subset of R = collection of some (distinct) numbers from the set of real numbers.
4. upper bound = take the literal meaning
5. least upper bound = again take the literal meaning
6. supremum = another word for 'least upper bound'

Now we are set to proof a simple form of Bolzano - Weierstrass theorem. Understand that theorems are based on logical conclusions derived from axioms by reasoning. So we will use our Completeness Axiom to build the proof of the theorem.

Statement: If \(a__k\) is a monotone sequence of real numbers (e.g., if \(a__k ≤ a_{_{k+1}\}), then this sequence has a finite limit if and only if the sequence is bounded.

Please note that this is NOT the original statement of the Bolzano Weierstrass theorem. But for the purpose of high school mathematics we can use this clone. Stay tuned and we will continue

understanding limit and continuity - brief sketch of a Cheenta Session

The limit and continuity is considered as the most important 'idea' in Analysis (for our purpose 'calculus'). Unfortunately deep understanding of these to concepts are beyond the reach of school level mathematical culture of India. Cheenta presents a series of intensive classes that puts the knowledge thirsty student in the right track.

1. Understanding monotonic and bounded sequences by definition and examples.
2. Considering the optional existence of analytical expression (eg. series of primes) and graphical expression (Dirichlet Sequence).
3. Considering the presence of monotonicity and boundedness in the same sequence (Xeno's paradox, eye of Horas). Considering the case of \((1+ \frac{1}{n})^n\) and  \((1+ \frac{1}{n})^{n+1}\)
4. The epsilon definition of limit of a sequence
1. The number a is said to be the limit of a sequence ( \(y_n\) ) if for any positive number \(\epsilon\) there is a real number N such that for all n > N the following inequality holds: | \(y_n\) - a | < \(\epsilon\).
5. Some problems concerning trivial limit calculations.
6. Some theorems related to limit of a sequence:
1. If a sequence has a limit it is bounded.
2. If a sequence is both bounded and monotonic, it has a limit. (Weierstrass)
3. A convergent sequence has only one limit.
7. A schematic approach to handle boundedness, monotonicity, convergency.
8. Some more theorems on limit of a sequence:
1. If a sequence (\(y_n\) ) and (\(z_n\) are convergent (we denote their limits by a and b respectively), a sequence (\(y_n + z_n\) is convergent too, it's limit being a+b.
2. Converse of the above theorem is not necessarily true. Very important point and is illustrated by an example. (if sum is convergent, then both are convergent or both are divergent but not otherwise).
3. \(\displaystyle\lim_{n\to\infty}c y_n = c \displaystyle\lim_{n\to\infty}y_n\)
9. Understanding infinitesimal sequences (definition)
10. Some theorems concerning infinitesimals
1. If (\(y_n\)) is a bounded sequence and (\(\alpha_n\) is infinitesimal, then (\(y_n \alpha_n\) is infinitesimal as well.
2. A sequence (\(y_n z_n\)) is convergent to ab if the sequences (\(y_n\)) and (\(z_n\)) is convergent to a and b respectively.
3. If a sequence (\(y_n\) ) and (\(z_n\) are convergent (we denote their limits by a and b respectively when b is not 0), a sequence (\(\frac{y_n}{z_n}\) is convergent too, it's limit being \(frac{a}{b}\).
11. Understand elimination, or addition, and any other change of a finite number of terms of a sequence do not affect either it's convergence or it's limit (if the sequence is convergent). We consider the case for the change of infinite terms as well.
12. Finally we take up problem solving from I.A. Maron (testing convergence of sequences).

This discussion concludes the first session of Limit and continuity concept building.

I.S.I. 10+2 Subjectives Solution (2 problems)

P164. Show that the area of the bounded region enclosed between the curves \(y^3=x^2\) and \(y=2-x^2\), is \(2\frac{2}{15}\).

Solution:

Note that \(y=x^{\frac{2}{3}}\) is an even function (green line).


P165. Find the area of the region in the xy plane, bounded by the graphs of \(y=x^2\), x+y = 2 and \(y=-\sqrt {x}\)

Solution:



The parabola and straight line intersects at (1,1) (we find that by solving the \(y=x^2\) and x+y=2)
Thus the area is found by adding area under parabola (from 0 to 1) and area under straight line (from 1 to 2).
\(\int^1_0 x^2\,dx=\left[\frac{x^3}{3}\right]^1_0=\frac{1}{3}\) (area under parabola)
area under straight line above 'x' axis is the triangle with height 1 unit and base 1 unit (from x=1 to x=2, area under x+y=2)
that area = \(\frac{1}{2}\times 1\times 1=\frac{1}{2}\)
Thus total area above x axis (of the required region) is \(\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
Now we come to the region below 'x' axis.
x+y = 2 and \(y=-\sqrt{x}\) intersect at (4, -2) (found by solving the two equations). We calculate the area under the curve \(y=-\sqrt{x}\) from x=0 to x=4 and subtract from it the area of the triangle with base from x=2 to x=4 and height =2 (hence the area of the triangle to be subtracted is 2 sq unit).
Area under the square root curve is
\(|\int^4_0 -\sqrt{x}\,dx|=\int^4_0 \sqrt{x}\,dx= \left[\frac {x^{\frac{1}{2}+1}} {\frac{1}{2}+1}\right]^4_0\).
=\(\frac {2}{3}\times 8=\frac {16}{3}\)
Delete 2 square unit from this and add the area computed before (above 'x' axis).
Area = \(\frac {16}{3} - 2 + \frac{5}{6} = \frac{25}{6}\) (ANS)