# Math ASCII Notation Demo

Mathematical content on Apronus.com is presented in Math ASCII Notation which can be properly displayed by all Web browsers because it uses only the basic set of characters found on all keyboards and in all fonts.

The purpose of these pages is to demonstrate the power of the Math ASCII Notation. In principle, it can be used to write mathematical content of any complexity. In practice, its limits can be seen when trying to write complicated formulas (containing, for example, variables with many indexes or multiple integrals).

Despite its limitations the Math ASCII Notation has much expressive power, as can be seen from browsing through these pages.

Let A,B c X.
(X\A) \ (X\B) = ???
(X\A) \ (X\B) = B\A
(B\C) \ (A\D) c ???
(B\C) \ (A\D) c (B\A) u (D\C)
Let F be a s-algebra in X.
Let f : X -> |R*.
Suppose that |f| is measurable.
Does f have to be measurable?
NO.
Let E c X such that E does not belong to F.
Define f : X -> |R* as follows: f = 1 on E and f = -1 on X\E
{x:-X : f(x)>0} = E does not belong to F
Hence f is not measurable.
But |f| is constant hence measurable.
Let R be a ring of sets in X.
Describe K(R) - the algebra generated by R.
K(R) = {EcX : E:-R or X\E:-R}
page 67 in 2nd measure
Let A[1],A[2],A[3],... be a decreasing sequence of sets, having empty intersection.
What can we conclude from this?
A[1] = \\*//(n=1 to oo) A[n] \ A[n+1]
/\m:-|N A[m] = \\*//(n=m to oo) A[n] \ A[n+1]
page 97 in 2nd measure
Let A[1],A[2],A[3],... be a decreasing sequence of sets, having empty intersection.
Decide if we can conclude that
A[1] = \\*//(n=1 to n=oo) A[n] \ A[n+1]
YES
page 97 in 2nd measure
Let A[1],A[2],A[3],... be a sequence of sets, having empty intersection.
Decide if we can conclude that
A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]
NO
1) Let A[1] = {-7}.
For n>=2, let A[n] = {n,n+1,n+2,n+3,...}.
2) Let A[1]=O, A[2]={7}, A[n] = O, for n>=3
Let A[1],A[2],A[3],... be a decreasing sequence of sets.
Decide if we can conclude that
A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]
NO
Let A[n] = {1} for all n:-|N.
Let A[1],A[2],A[3],... be a sequence of sets, having empty intersection.
Suppose that /\n:-|N A[n] c A[1].
Decide if we can conclude that
A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1]
YES
The right side is easily seen to be contained in A[1].
To prove the inverse inclusion one argues by contradiction.
Suppose that x:-A[1] and /\n:-|N ( x:-A[n] => x:-A[n+1] ).
Then it follows that x :- /\n:-|N A[n].
This is a contradiction because the intersection is assumed to be empty.
Let A[1],A[2],A[3],... be a sequence of sets, having empty intersection.
Suppose that /\n:-|N A[n] c A[1].
Decide if we can conclude that
A[1] = \\*//(n=1 to n=oo) A[n] \ A[n+1]
NO.
Although we can conclude that A[1] = \\//(n=1 to n=oo) A[n] \ A[n+1],
we cannot conclude that the union is disjoint.
Let A[1] = A[3] = {1}. Let A[2] = A[4] = O. Let A[n] = O for n>4.
Then (A[1]\A[2]) n (A[3]\A[4]) is not empty.