לפוסט זה מצורף אוסף בעיות מס’ 4 אם עולות שאלות בקשר לבעיות לפני מפגש התרגול הבא, אפשר לשאול כאן.

## Wednesday, December 23, 2009

## Monday, December 21, 2009

### Lecture 8 (Dec 22, 2009)

## Continuation of continuity

Let be a continuous function defined on the closed rectangle . Then the functions are all defined on the segment and “converge” to as . If is a sequence of values of converging to , , then we have a sequence of functions , all defined on the same segment, which also “converges” to the limit

The meaning of the “convergence of functions” is not yet defined.

**Problem**. Prove that in the above notations, for any point the sequence of numbers converges to the number . This type of convergence is called *pointwise convergence* (התכנסות נקודתית).

**Example**. Prove that the functions , converge pointwise to a certain limit function . Find this function.

**Example**. Let be the function defined on the entire axis , and denote (translation to the right by ). The same question: does a limit exist?

Formally, the pointwise convergence is defined by the following formula: , if . Clearly, this is the same as the formula

However, the “convergence” which comes from the continuity in two variables, is stronger.

**Theorem**. *If is a continuous function on the rectangle, as above, then for any there exists a positive such that for any with , the functions and differ at most by on the entire segment : .*

**Corollary**. In the above notation, converges to “simultaneously” (uniformly, במידה שווה):

### Functions as “points” (vectors)

Continuous functions defined on the same interval, can be added, subtracted, multiplied by constants etc, which makes them similar to vectors in . We can also define the “distance” between the functions as follows, . This number is always non-negative and finite:

- ;
- (symmetry);
- (the triangle inequality).

**Proposition/Definition**. The sequence of functions converges uniformly to , if .

**Question**. In which of the examples above the convergence is uniform?

### Can a limit of continuous functions be disconjugate?

A sequence of *pointwise* converging continuous functions may converge to a discontinuous function (see the first example above).

**Theorem**. *The limit of uniformly converging on a closed sequence continuous functions is again continuous. *

**Example**. The snowflake is a continuous closed curve of “infinite length”.

### Peano (plane-filling) curve

This is a continuous map which fills completely (i.e., passes through each point of) the unit square. There are many examples of such curves, below is the example constructed by D. Hilbert. Giuseppe Peano was an Italian mathematician

### Sketch of the justification

In both cases the “malicious” curve is obtained as the limit of an infinite sequence of planar polygonal (hence continuous) curves . By construction, each next curve differs from its predecessor by an explicitly limited “modification”. In the example with the Peano curve, we subdivide the initial unit square into for small squares of the “first level” and draw the curve crossing each of the four squares in the specified order. Then each 1st level square is again subdivided into 4 tiny sub-squares of the “second level” and each segment of the curve is modified so as it (1) remains inside the same 1st level square, yet (2) crosses each of the four 2nd level squares as plotted. Hence the curve differs from the curve by no more than . In the same way , etc.

Thus it is clear that for any the limit exists, and . Thus the convergence of the functions is uniform, and therefore the limit is also a continuous function.

How to show that it indeed passes through each point of the unit square? Note that the curve passes through centers of *all* squares of the th level. Any point of the initial square can be approximated by centers of the squares of sufficiently high level. There exists an infinite sequence of points which converges to , i.e., the infinite sequence of moments such that . Passing to a subsequence if necessary, we can assume (since the segment [0,1] is compact) that the sequence converges to a limit . One can easily check that because of the uniform continuity, .

**Problem. **Write the detailed accurate exposition of this proof.

## Monday, December 14, 2009

### Lecture 7 (Dec 15, 2009)

## Topology of the plane

- Functions of two variables as maps . Domains in the plane. Open and closed domains.
- Convergence of planar points: . Alternative description: any square contains almost all elements of the sequence.
- Limits of functions: , . We say that , if for any sequence of points converging to , the sequence converges to as . We say that is continuous at , if .
**Exercise**: is continuous at , if the preimage , of any interval , contains the intersection for some sufficiently small .**Exercise**: if is continuous at all points of the rectangle , then for any the function , defined by the formula is continuous on . Can one exchange the role of and ? Formulate and think about the inverse statement.**Exercise**: Check that the functions and are continuous. What can be said about the continuity of the function ?**Exercise**: formulate and prove a theorem on continuity of the composite functions.**Exercise**: Give the definition of a continuous function for . Planar curves. Simple curves. Closed curves.- Intermediate value theorem for curves. Connected sets, connected components. Jordan lemma.
- Rotation of a closed curve around a point. Continuity of the rotation number. Yet another “Intermediate value theorem” for functions of two variables.

## Tuesday, December 8, 2009

### Lecture 6 (Dec 8, 2009)

## Topology on the real line

**Definition.** A *homeomorphism* between two subsets is a one-to-one map such that both and its inverse are continuous. If such a map exists, the two subsets are called *homeomorphic* (or *topologically equivalent*) and we write .

- Homeomorphisms between and .
- The property “
*Every continuous function is bounded*” is invariant by homeomorphism: if it is true for a set and , then this property also holds for . Non-homeomorphy of and . - The properties “
*Closedness*” and “*Openness*” are invariant by homeomorphism: if is closed/open and , then is closed/open respectively. - Is the property “
*Boundedness*” invariant by homeomorphism? - The disjoint union is homeomorphic to yet non-homeomorphic to or .

Finite point sets are compact. Union of two compacts is a compact.

**Theorem**. The segment is compact.

**Proof**. Assume that it is not compact, and is an open covering that does not admit a finite subcovering. Then out of the two halves and one is non-compact and the same covering does not admit finite subcovering of this bad half. The bad half-segment can be again divided by two etc. This process produces a nested family of segments of lengths decreasing to 0, which has a (unique) common point : none of these segments by construction admits a finite subcovering by the sets . But the open interval covering , covers also all sufficiently small segments . The resulting contradiction proves that the initial segment was compact. ♦

**Proposition**. A function continuous on a compact set, is bounded (and achieves its extremal values).

**Proof**. For each point there exists a neighborhood such that is bounded on (e.g., takes values on the interval . Choose a finite sub-covering and pick the maximal out of the (finitely many!!) local bounds. ♦

Definition. A subset is called connected, if together with any pair of points it contains the entire segment .

The property “Being connected” is invariant by homeomorphisms (what does this mean?)

There are only four non-empty connected subsets on , not homeomorphic to each other: a point, an open interval, a closed segment and a semi-interval .

## The Cantor set

From the closed segment delete the (open) middle third . To each of the two remaining closed segments apply the same procedure (deleting the middle third), and repeat it.

As a result, we obtain an infinite collection (family) of nested closed subsets , each of which is a finite union of closed segments of total length going to zero.

The intersection is the *Cantor set.*

- is closed. For any it can be covered by finitely many intervals of total length less than .
- contains no open subset (everywhere holes!), yet no point of $latex\mathbf K$ is isolated.
- is in one-to-one correspondence with the entire segment .

These properties can be seen from the alternative description of as points on the segment , whose infinite “ternary” (i.e., on the basis 3) representation does not involve “ones” (see the warning below), only zeros and “twos”.

**Example**. To show that any point is not isolated, we show that each interval of length centered at , contains another point of . For this, it is sufficient to change a “ternary digit” with the number greater than by “the other” digit, i.e., 0 by 2 or 2 by 0.

**Warning**. The infinite ternary representation is non-unique (in the same way as the decimal representation): the two expressions and correspond to the same point! how does this affect the above arguments?

**Difficult question**. Construct a continuous function such that the preimage of any point always consists of only one or two points from .