# Sergei Yakovenko's blog: on Math and Teaching

## Wednesday, December 23, 2009

### אוסף בעיות מספר 4

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

## Continuation of continuity

Let $f\colon [0,1]\times[\alpha,\beta]\to\mathbb R^2$ be a continuous function defined on the closed rectangle $[0,1]\times[\alpha,\beta]$. Then the functions $f_a(x)=f(x,a)$ are all defined on the segment $[0,1]$ and “converge” to $f_{a_*}(x)$ as $a\to a_*\in[\alpha,\beta]$. If $a_1,a_2,\dots, a_n,\dots$ is a sequence of values of $a$ converging to $a_*$, $\lim a_n=a_*$, then we have a sequence of functions $f_n=f_{a_n}\colon[-1,1]\to\mathbb R$, all defined on the same segment, which also “converges” to the limit $f_*=f_{a_*}$

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

Problem. Prove that in the above notations, for any point $x\in[0,1]$ the sequence of numbers $f_n(x)$ converges to the number $f_*(x)$. This type of convergence is called pointwise convergence (התכנסות נקודתית).

Example. Prove that the functions $f_n(x)=x^n,~n=1,2,\dots$, converge pointwise to a certain limit function $f_*\colon[0,1]\to\mathbb R$. Find this function.

Example. Let $\phi(x)=\frac1{1+x^2}$ be the function defined on the entire axis $\mathbb R$, and denote $f_n(x)=\phi(x-n)\colon\mathbb R\to\mathbb R$ (translation to the right by $n$). The same question: does a limit exist?

Formally, the pointwise convergence is defined by the following formula: $\lim f_n=f_*$, if $\forall x\in [0,1],~\forall \varepsilon >0,~\exists N=N(x,\varepsilon),~\forall n\ge N,~|f_n(x)-f_*(x)|<\varepsilon$. Clearly, this is the same as the formula

$\forall \varepsilon >0,~\forall x\in [0,1],~\exists N=N(x,\varepsilon),~\forall n\ge N,~|f_n(x)-f_*(x)|<\varepsilon.$

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

Theorem. If $f(x,y)$ is a continuous function on the rectangle, as above, then for any $\varepsilon>0$ there exists a positive $\delta>0$ such that for any $a$ with $|a-a_*|<\delta$, the functions $f_a$ and $f_{a_*}$ differ at most by $\varepsilon$ on the entire segment $[0,1]$: $\forall x\in[0,1],~|f_a(x)-f_{a_*}(x)|<\varepsilon$.

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

$\forall\varepsilon>0,~\exists N=N(\varepsilon):~\forall x\in[0,1],~\forall n\ge N,~|f_n(x)-f_*(x)|<\varepsilon.$

### 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 $\mathbb R^3$. We can also define the “distance” between the functions as follows, $d(f_1,f_2)=\max_{x\in[0,1]}|f_1(x)-f_2(x)|$. This number is always non-negative and finite:

1. $d(f_1,f_2)=0\iff f_1=f_2$;
2. $d(f_1,f_2)=d(f_2,f_1)$ (symmetry);
3. $d(f_1,f_3)\le d(f_1,f_2)+d(f_2,f_3)$ (the triangle inequality).

Proposition/Definition. The sequence of functions $\{f_n\}$ converges uniformly to $f_*$, if $d(f_n,f_*)\to0$.

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 $\gamma\colon [0,1]\to \mathbb R^2$ 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 $\gamma_n\colon [0,1]\to\mathbb R^2$. By construction, each next curve $\gamma_{n+1}$ differs from its predecessor $\gamma_n$ 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 $\gamma_2$ differs from  the curve $\gamma_2$ by no more than $\sqrt 2$. In the same way $\|\gamma_3-\gamma_2\|\le \sqrt 2/2$$\|\gamma_4-\gamma_3\|\le \sqrt 2/2^2$ etc.

Thus it is clear that for any $t\in[0,1]$ the limit $\lim_{n\to\infty}\gamma_n(t)$ exists, and $|\gamma_n(t)-\gamma_*(t)|\le \sqrt{2}/2^{n-3}$. 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 $\gamma_n$ passes through centers of all squares of the $n$th level. Any point $a_*$ of the initial square can be approximated by centers of the squares of sufficiently high level. There exists an infinite sequence of points $a_n\in\gamma_n([0,1])$  which converges to $a_*$, i.e., the infinite sequence of moments $t_n\in[0,1]$ such that $\lim_{n\to\infty}\gamma_n(t_n)=a$. Passing to a subsequence if necessary, we can assume (since the segment [0,1] is compact) that the sequence $\{t_n\}$ converges to a limit $t_*$. One can easily check that because of the uniform continuity, $\gamma_*(t_*)=a_*$.

Problem. Write the detailed accurate exposition of this proof.

## Topology of the plane

1. Functions of two variables as maps $f\colon \mathbb R^2\to\mathbb R$. Domains in the plane. Open and closed domains.
2. Convergence of planar points: $\lim_{n\to\infty}(x_n,y_n)=(X,Y)\in\mathbb R^2\iff\lim x_n=X\ \&\ \lim y_n=Y$. Alternative description: any square $Q_\varepsilon(X,Y)=\{|x-X|<\varepsilon,\ |y-Y|<\varepsilon\}$ contains almost all elements of the sequence.
3. Limits of functions: $f\colon U\to\mathbb R$, $Z=(X,Y)\in\mathbb R$. We say that $A=\lim_{(x,y)\to Z}f(x,y)$, if for any sequence of points $\{(x_n,y_n)\}$ converging to $Z$, the sequence $f(x_n,y_n)$ converges to $A$ as $n\to\infty$. We say that $f$ is continuous at $Z$, if $A=f(Z)$.
4. Exercise: $f$ is continuous at $Z$, if the preimage $f^{-1}(J)$, of any interval $J=(A-\delta,A+\delta),\ \delta>0$, contains the intersection $Q_\varepsilon\cap U$ for some sufficiently small $\varepsilon>0$.
5. Exercise: if $f$ is continuous at all points of the rectangle $\{x\in I,\ y\in J\}\subseteq\mathbb R^2$, then for any $y\in J$ the function $f_y\colon I\to\mathbb R$, defined by the formula $f_y(x)=f(x,y)$ is continuous on $I$. Can one exchange the role of $x$ and $y$? Formulate and think about the inverse statement.
6. Exercise: Check that the functions $f(x,y)=x\pm y$ and $g(x,y)=xy$ are continuous. What can be said about the continuity of the function $f(x,y)=y/x$?
7. Exercise: formulate and prove a theorem on continuity of the composite functions.
8. Exercise: Give the definition of a continuous function $f\colon I\to\mathbb R^2$ for $I\subseteq\mathbb R$. Planar curves.  Simple curves. Closed curves.
9. Intermediate value theorem for curves. Connected sets, connected components. Jordan lemma.
10. Rotation of a closed curve around a point. Continuity of the rotation number. Yet another “Intermediate value theorem” for functions of two variables.

## Topology on the real line

Definition. A homeomorphism between two subsets $X,Y\subseteq\mathbb R$ is a one-to-one map $\Phi\colon X\to Y$ such that both $\Phi$ and its inverse $\Phi^{-1}\colon Y\to X$ are continuous. If such a map exists, the two subsets are called homeomorphic (or topologically equivalent) and we write $X\simeq Y$.

• Homeomorphisms between $(a,b),~(0,+\infty)$ and $\mathbb R$.
• The property “Every continuous function is bounded” is invariant by homeomorphism: if it is true for a set $X$ and $Y\simeq X$, then this property also holds for $Y$. Non-homeomorphy of $[0,1]$ and $\mathbb R$.
• The properties “Closedness” and “Openness” are invariant by homeomorphism: if $X$ is closed/open and $Y\simeq X$, then $Y$ is closed/open respectively.
• Is the property “Boundedness” invariant by homeomorphism?
• The disjoint union $(-1,0)\cup (0,1)$ is homeomorphic to $(-2,-1)\cup(1,2)$ yet non-homeomorphic to $(-1,1)$ or $(-1,1]$.

Definition. A set $K\subseteq\mathbb R$ is called compact, if from any covering of $K$ by open sets (intervals), $K\subseteq\bigcup_i U_i$, one can choose a finite sub-covering $U_{i_1}\cup U_{i_2}\cup\dots\cup U_{i_n}\supseteq K$.

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

Theorem. The segment $[0,1]$ is compact.

Proof. Assume that it is not compact, and $\bigcup U_i$ is an open covering that does not admit a finite subcovering. Then out of the two halves $[0,\frac 12]$ and $[\frac12, 1]$ 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 $[0,1]=K_0\supset K_1\supset K_2\supset\cdots\supset K_n\supset\dots$ of lengths decreasing to 0, which has a (unique) common point $a\in K_0$: none of these segments by construction admits a finite subcovering by the sets $U_i$. But the open interval $U_{i_k}$ covering $a$, covers also all sufficiently small segments $K_n$. The resulting contradiction proves that the initial segment $K_0$ was compact. ♦

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

Proof. For each point $a\in X$ there exists a neighborhood $U_a\subset\mathbb R$ such that $f$ is bounded on $U_a$ (e.g., takes values on the interval $(f(a)-1,f(a)+1)$. Choose a finite sub-covering and pick the maximal out of the (finitely many!!) local bounds. ♦

Definition. A subset $X\subseteq\mathbb R$ is called connected, if together with any pair of points $a,b\in X$ it contains the entire segment $[a,b]$.

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

There are only four non-empty connected subsets on $\mathbb R$, not homeomorphic to each other: a point, an open interval, a closed segment and a semi-interval $(0,1]$.

## The Cantor set

From the closed segment $[0,1]$ delete the (open) middle third $(\frac 13,\frac 23)$. To each of the two remaining closed segments $[0,\frac13],\ [\frac23,1]$ apply the same procedure (deleting the middle third), and repeat it.

As a result, we obtain an infinite collection (family) of nested closed subsets $[0,1]=K_0\supset K_1\supset\cdots\supset K_n\supset\cdots$, each of which is a finite union of $2^n$ closed segments of total length $(\frac 23)^n$ going to zero.

The intersection $\mathbf K=\bigcap_{n=0}^\infty K_n$ is the Cantor set.

• $\mathbf K$ is closed. For any $\varepsilon>0$ it can be covered by finitely many intervals of total length less than $\varepsilon$.
• $\mathbf K$ contains no open subset (everywhere holes!), yet no point of $latex\mathbf K$ is isolated.
• $\mathbf K$ is in one-to-one correspondence with the entire segment $[0,1]$.

These properties can be seen from the alternative description of $\mathbf K$ as points on the segment $[0,1]$, 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 $a=0.a_1a_2a_3\dots\in\mathbf K$ is not isolated, we show that each interval of length $3^{-n}$ centered at $a$, contains another point of $\mathbf K$. For this, it is sufficient to change a “ternary digit” $a_k$ with the number $k$ greater than $n$ 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 $0.0222222222\dots$ and $0.10000\dots$ correspond to the same point! how does this affect the above arguments?

Difficult question. Construct a continuous  function $f\colon \mathbf K\to[0,1]$ such that the preimage $f^{-1}(b)$ of any point $b\in[0,1]$ always consists of only one or two points from $\mathbf K$.

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