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.

a>=0, b>0

Prove: sqrt(a/b) <= 1/2 * (a + 1/b).

Prove: sqrt(a/b) <= 1/2 * (a + 1/b).

hint: 0 <= (a*sqrt(b) - 1/sqrt(b))^2

a[n] >= 0

series a[n]*a[n] converges

show that series a[n] / n converges

series a[n]*a[n] converges

show that series a[n] / n converges

hint: sqrt(a/b) <= 1/2 * (a + 1/b)

Prove that in a metric space, if every finitely bound
collection of closed sets has nonempty intersection, then the
space is sequentially compact.

Take any sequence. Suppose that it contains no convergent
subsequence. Let A[n] = {x[n], x[n+1], ...}. {A[n]} is a
sequence of closed sets having the finite intersection
property. Hence there is a point belonging to all A[n]. It must
be a subsequential limit. Contradiction.

Prove that in a topological space, if every finitely bound
collection of closed sets has nonempty intersectiom, then every
open cover of this space contains a finite subcover.

Theorem 57, page 144 in the palace notebook

(The proof is very short.)

(The proof is very short.)

Prove that in a topological space, if every open cover of that
space contains a finite subcover, then every finitely bound
collection of closed sets has nonempty intersection.

Argue by contradiction. This is easy.

Theorem 57, page 144 in the palace notebook

Theorem 57, page 144 in the palace notebook

Let (X,d) be a compact metric space.

Prove that

\/(a,b:-X) /\(x,y:-X) [ d(x,y) <= d(a,b) ]

Prove that

\/(a,b:-X) /\(x,y:-X) [ d(x,y) <= d(a,b) ]

Define g:(X*X) * (X*X) -> R by

g( (x,y), (a,b) ) = d(x,a) + d(y,b).

Show that (X*X,g) is a compact metric space.

Show that d is a continuous function on that space.

Deduce that it attains its maximum.

page 146 in palace

g( (x,y), (a,b) ) = d(x,a) + d(y,b).

Show that (X*X,g) is a compact metric space.

Show that d is a continuous function on that space.

Deduce that it attains its maximum.

page 146 in palace

What is a precompact metric space?

In a precompact metric space every sequence has a Cauchy
subsequence. It is equivalent to total boundedness.

Let a[n] be a Cauchy sequence. Let x[n] be a sequence.

Suppose that d(a[n],x[n]) -> 0.

Show that x[n] is also Cauchy.

Suppose that d(a[n],x[n]) -> 0.

Show that x[n] is also Cauchy.

=

Let (X,d) be a metric space. Let A c X.

Show that: Clo(A) is compact => A is precompact.

Also, show that the converse need not be true.

Show that: Clo(A) is compact => A is precompact.

Also, show that the converse need not be true.

Let A be the whole space X, and let X be non-complete.

Let (X,d) be a metric space. Let A c X.

Show that: A is precompact <=> clo(A) is precompact.

Show that: A is precompact <=> clo(A) is precompact.

To prove (=>) use the fact that if a[n] is Cauchy and

d(a[n],x[n]) -> 0, then x[n] is also Cauchy.

(<=) is trivial.

d(a[n],x[n]) -> 0, then x[n] is also Cauchy.

(<=) is trivial.