# 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 W c P(X) be an algebra of sets. Let M c X.
Find the algebra generated by W u {M}.
K(W u {M}) = ???
K( W u {M} ) = { (A n M) u (B \ M) : A,B :- W }
page 51 in 2nd measure
Define the generalized Cartesian product.
Let J be a non-empty set. Let X be a set.
Let I : J -> P(X). Now {I(j)}j:-J is a collection of subsets of X.
The Cartesian product of sets I(j) indexed with j:-J is defined as follows:
{ f: f : J -> Uj:-J I(j) and /\j:-J f(j):-I(j) }
Let J be a set. Let /\j:-J I(j) be a set. Now, let /\j:-J /\i:- I(j) A[i,j] be a set.
Complete:
\\//j:-J //\\i:-I(j) A[i,j] = ???
Let K be the Cartesian product of sets I(j) indexed with j:-J.
\\//j:-J //\\i:-I(j) A[i,j] = //\\f:-K \\//j:-J A[f(j),j]
page 55 in 1st measure
Let J be a set. Let /\j:-J I(j) be a set. Now, let /\j:-J /\i:- I(j) A[i,j] be a set.
Let K be the Cartesian product of sets I(j) indexed with j:-J.
Complete:
//\\f:-K \\//j:-J A[f(j),j] = ???
//\\f:-K \\//j:-J A[f(j),j] = \\//j:-J //\\i:-I(j) A[i,j]
page 55 in 1st measure
Let J be a set. Let /\j:-J I(j) be a set. Now, let /\j:-J /\i:- I(j) A[i,j] be a set.
Complete:
//\\j:-J \\//i:-I(j) A[i,j] = ???
Let K be the Cartesian product of sets I(j) indexed with j:-J.
//\\j:-J \\//i:-I(j) A[i,j] = \\//f:-K //\\j:-J A[f(j),j]
page 55 in 1st measure
Let J be a set. Let /\j:-J I(j) be a set. Now, let /\j:-J /\i:- I(j) A[i,j] be a set.
Let K be the Cartesian product of sets I(j) indexed with j:-J.
Complete:
\\//f:-K //\\j:-J A[f(j),j] = ???
\\//f:-K //\\j:-J A[f(j),j] = //\\j:-J \\//i:-I(j) A[i,j]
page 55 in 1st measure
Let (Y,g) be a complete metric space. Let f : X -> Y be 1-1 and onto.
Let d : XxX -> |R be defined by d(x,u) = g(f(x),f(u)).
What can we conclude?
(X,d) is a complete metric space
page 67 in OLDTIMER
If W is a collection of sets,
let Ws denote the collection of all finite unions of sets from W,
let Wd denote the collection of all finite intersections of sets from W.
Prove that Wsd c Wds.
Use
//\\j:-J \\//i:-I(j) A[i,j] = \\//f:-K //\\j:-J A[f(j),j]
where K is the Cartesian product of sets I(j) indexed with j:-J.
page 57 in 1st measure
If W is a collection of sets,
let Ws denote the collection of all finite unions of sets from W,
let Wd denote the collection of all finite intersections of sets from W.
Prove that Wds c Wsd.
Use
\\//j:-J //\\i:-I(j) A[i,j] = //\\f:-K \\//j:-J A[f(j),j]
where K is the Cartesian product of sets I(j) indexed with j:-J.
page 57 in 1st measure
Let A be a collection of sets.
Can we conclude that M(R(A)) = S(A) ?
Yes.
1) R(A) c S(A)
2) S(R(A)) c S(A)
3) S(R(A)) = S(A)
4) M(R(A)) = S(R(A)) page 54 in 1st measure
5) M(R(A)) = S(A)