The Peano curve: continuity can be counter-intuitive

The Peano curve is obtained as the limit of piecewise-linear continuous (even closed) curves $\gamma_n$. Denote by $K=\{|x|+|y|\le 1\}$ the square (rotated by $\frac \pi/4$) and by $\mathbb Z^2=\{(x,y):x,y\in\mathbb Z$ the grid of horizontal and vertical lines at distance 1 from each other, then one can construct a family of piecewise-linear continuous curves $\gamma_n:[0,1]\to\mathbb R^2$ which visits all points of the intersection $K\cap\frac1{2^n}\mathbb Z^2$ in such a way that $|\gamma_n(t)-\gamma_n(t)|<\frac1{2^n}$ uniformly on $t\in[0,1]$.

This sequence of curves converges uniformly to a function (curve) $\gamma_*:[0,1]\to\mathbb R^2$ and this curve is closed and continuous for the same reasons that justify continuity of the Koch snowflake curve.

What are the properties of the images $C_n=\gamma_n([0,1])$ and of the limit curve $C_*=\gamma_*([0,1])$?

• Each curve $C_n$ for any finite $n$ is piecewise-linear. It has zero area in the sense that for any $\varepsilon > 0$ the curve $C_n$ can be covered by a finite union of (open) rectangles with the total area less than $\varepsilon$;
• Each curve $C_n$ has finite length (although it grows to infinity as $n\to\infty$, – check it!).
• The limit curve $C_*$ has no length (that’s the same as saying that it has infinite length). Moreover, unlike many other curves of infinite length (say, the straight line $\{y=0\}\subseteq\mathbb R^2$), no part $\gamma([a,b]),\ a, of $C_*$ has finite length!
• The limit curve $C_*$ coincides with the square $K$, hence fills the area equal to 2.

All these assertions are easy except for the last one. Let’s prove it.

Consider the images $C_n=K\cap \frac1{2^n}\mathbb Z^2$. The union of these images is dense in $K$: by definition, this means that any point $P\in K$ can be approximated by a sequence of points $P_n\in C_n$ which converge to $P$ as $n\to\infty$. Being in the image of $\gamma_n([0,1])$, each point $P_n$ is the image of some point in [0,1]: $\exists a_n\in[0,1]:\ \gamma(a_n)=P_n$. Such point may well be non-unique, and in any case we have absolutely no knowledge of how the points $a_1,a_2,\dots$ are distributed over [0,1].

However, we know that the sequence $a_n\in [0,1]$ must have an accumulation point $a_*\in [0,1]$, which is by definition a limit of some infinite subsequence. (This won’t be the case if instead of [0,1] we were dealing with the curves defined on the entire real line!). Replacing the sequence by this subsequence, we see that it still converges to the same limit, $P_n=\gamma(a_n)\to a_*=\gamma_*(a_*)=P$. Thus we proved that an arbitrary point in $K$ lies in the image: $P\in C_*$.

Topology: the study of properties preserved by continuous maps (functions, applications, …)

Definition. A neighborhood of a point $a\in\mathbb R^n$ in the Euclidean space is any set of the form $\{x:|x-a| 0$, where $| ??? |$ is a distance function satisfying the triangle inequality. Examples:

• $|x|=\sqrt{x_1^2+\cdots+x_n^2}$ (the usual Euclidean distance on the line, on the plane, …) for $x=(x_1,\dots,x_n)\in\mathbb R^n$;
• $|x|=\max\{|x_1|, \dots, |x_n|\}$ (in the above notation);
• $|x|=|x_1|+\cdots+|x_n|$.

Definition. A subset $A\subset\mathbb R^n$ of the Euclidean space (OK, plane) is called open, if together with any its point $a\in A$ it contains some neighborhood of $a$.
A subset is called closed, if the limit of converging infinite sequence $\{a_n\}\subset A$ again belongs in $A$.

Theorem. A subset $A$ is open if and only if its complement $\mathbb R^n\smallsetminus A$ is closed.

Theorem. The union of any family (infinite or even uncountable) of open sets is open. Finite intersection of open sets is also open (for infinite intersections this is wrong).
Corollary. Intersection of any family (infinite or even uncountable) of closed sets is closed. Finite union of closed sets is also closed (for infinite intersections this is wrong).

One can immediately produce a lot of examples of open/closed subsets in $\mathbb R^n$. It turns out that any property that can be formulated using only these notions, is preserved by maps which are continuous together with their inverses. The corresponding area of math is called topology.

Continuity of functions

Let $f\colon D\to\mathbb R,\ D\subseteq\mathbb R$ be a function of one variable, and $a\in D$ a point in its domain. The function is said to be continuous at $a$, if for any precision $\varepsilon>0$ the function is $\varepsilon$-constant (equal to its value $f(a)$) after restriction on a sufficiently short interval around $a$.

Formally, if we denote by $\bold I=\bold I^1$ the (open) interval, then the continuity means that $\forall \varepsilon>0\ \exists\delta>0\ f(a+\delta\bold I)\subseteq f(a)+\varepsilon\bold I$ (check that you understand the meaning of the notation $u+v\bold I$ for a subset of $\mathbb R$).

A function $f\colon D\to\mathbb R$ is said to be continuous on a subset $D'\subset D$, if it is continuous at all points $a\in D'$ of this subset. Usually we consider the cases where $D'=D$, that is, functions continuous on their domain of definition.

Remarks.
1. If $a$ is an isolated point of the domain $D$, then any function is automatically continuous at $a$ for a simple reason: for all sufficiently small $\delta>0$ the intersection $(a+\delta\bold I)\cap D$ consists of a single point $a$, so the image is a single point $f(a)$.

2. If $a\notin D$ but $\inf_{x\in D} |x-a|=0$ and there exists a limit $A=\lim_{x\to a}f(x)$, then one can extend $f$ on $a$ by letting $f(a)=A$ and obtain a function defined on $D\cup\{a\}$ which is continuous at $a$.

Obvious properties of the continuity

The sum, difference and product of two functions continuous at a point $a$, is again continuous. The reciprocal of a continuous function $\frac1{f}$ is continuous at $a$, if $f(a)\ne 0$.

This is an obvious consequence of the rules of operations on “approximate numbers” (קירובים). When dealing with the sum/difference, one has to work with absolute approximations, when dealing with the product/ratio – with the relatice approximations, but ultimately it all boils down to the same argument: if two functions are almost constant on a sufficiently small interval around $a$, then application of the arithmetic operations is almost constant.

Not-so-obvious property of continuity

When the continuity is compatible with transition to limit? More specifically, we consider the situation where there is an infinite number of functions $f_n\colon:D\to\mathbb R$ defined on the common domain $D$. Assume that for any $a\in D$ the values (numbers!) $f_n(a)\in\mathbb R$ form a converging sequence whose limit is denoted by $f_*(a)$ (it depends on $a$!). What can one say about the function $f_*\colon a\mapsto f_*(a)$?

Example.
Assume that $f_n(x)=x^n$ and $D=[0,1]$. All of them are continuous (why?). If $a<1$, then $\lim_{n\to\infty} a^n=0$. If $a=1$, then for any $n\ a^n=1$. Thus the limit $\lim_{n\to\infty}f_n(a)$ exists for all $a$, but as a function of $a\in[0,1]$ it is a discontinuous function. Thus without additional precautions a sequence of continuous functions can converge to a discontinuous one.

Distance between the functions.
The distance between real numbers $a,b\in\mathbb R$ is the nonnegative number $|a-b|$ which is zero if and only if $a=b$. Motivated by that, we introduce the distance $\|f-g\|$ between two functions $f,g\colon D\to \mathbb R$ as the expression $\sup_{a\in D}|f(a)-g(a)|$.

Exercise. Prove that any three functions $f,g,h$ defined on the common domain $D$, the “triangle inequality” $\|f-g\|\leqslant \|f-h\|+\|h-g\|$.

Definition. A sequence of functions $f_n\colon D\mathbb R$ is said to be uniformly converging to a function $f_*\colon D\to\mathbb R$, if $\lim_{n\to\infty}\|f_n-f_*\|=0$.

Theorem. If a sequence of continuous functions converges uniformly, then the limit is also a continuous function.

Indeed, denote by $g$ the limit function, $a\in D$ any point, and let $\varepsilon>0$ be any “resolution”. We need to construct a small interval around $a$ such that $g$ on this interval is $\varepsilon$-indistinguishable from the value $g(a)$. We split the resolution allowance into two halves. The first half we use to find $N$ such that $\|f_n-g\| < \frac\varepsilon2$ for all $n\ge N$. The second half we spend on the continuity: since $f_N$ is continuous, there exists a segment $a+\delta\bold I$ on which $f_N$ is $\frac\varepsilon2$-indistinguishable from $f_N(a)$. Collecting all inequalities we see that for any point $x\in a+\delta\bold I$ we have three inequalities: $|f_N(a)-g(a)|<\frac\varepsilon2, \ |f_N(x)-g(x)|<\frac\varepsilon2,\ |f_N(x)-g(x)|<\frac\varepsilon2$. By the triangle inequality, $|g(x)-g(a)|< \frac{3\varepsilon}2$. Ooops! we were heading for $\varepsilon$! One should rather divide our allowance into unequal parts, $\frac{2\varepsilon}3$ for the distance and $\frac\varepsilon3$ for the continuity of $f_N$ if we thought ahead of the computation! 😉 in any case, the outcome is the same.

Curves

The notion of continuity, the distance between functions etc. can be generalized from functions of one variable to other classes of functions.

For instance, functions of the form $\gamma\colon [0,1]\to\mathbb R^2$ can be called (parametrized) curves. Here the argument $x\in[0,1]$ can be naturally associated with time, so the $\gamma(t)$ is the position of the moving point at the moment $t$. We can draw the image $\gamma([0,1])$: this drawing does not reflect the timing: to indicate it, we can additionally mark the images, say, $\gamma (\frac k{10}),\ k=0,1,\dots,10$.

To define continuity for curves, denote by $\bold I^2$ the unit square $\{|x|<1,\ |y|<1\}$. A curve $\gamma$ is continuous at a point $a\in [0,1]$ if $\forall\varepsilon >0 \ \exists \delta>0$ such that $\gamma (a+\delta\bold I^1)\subseteq \gamma(a)+\varepsilon \bold I^2$. (Do you understand this formula? 😉 )

The distance between two points $a=(a_1,a_2),\ b=(b_1,b_2)\in\mathbb R^2$ is usually defined as $\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}$, but this difference is not very much different from the expression $|a-b|=\max_{i=1,2}|a_i-b_i|$ (this definition can be immediately generalized for spaces $\mathbb R^n$ of any finite dimension $n=3,4,\dots$. The distance between two curves has a very similar form: $\|f-g\|=\sup_{x\in [0,1]}|f(x)-g(x)|$.

Remark. If the functions $f,g$ are continuous, we can replace the supremum by maximum (which is always achieved).

Koch snowflake revisited

Now we can return to one of the examples we discussed on Lecture 1, the Koch snowflake. In contrast with that time, we now have an appropriate language to deal with it.

The process of constructing the curve actually produces a sequence of closed curves. The image of the first curve is an equilateral triangular, the second one gives the Star of David, the third one has no canonical name.

In all cases the new curve $\gamma_{n+1}$ is obtained by taking the previous curve $\gamma_n$ and modifying it on a subset of its domain: instead of traversing a line segment with constant speed, one takes a middle third of this segment and forces $\gamma_{n+1}$ to detour. This requires increasing the speed, but we don’t care as long as the trajectory remains continuous. The distance between $\gamma_n$ and $\gamma_{n+1}$ is $\frac{\sqrt3}2$ times the size of the size of the segment $\frac1{3^n}$.

This observation guarantees that $\|\gamma_n-\gamma_{n+1}\|< C(1/3)^n$. This implies that the sequence of maps $\gamma_n\colon [0,1]\to\mathbb R^2$ converges uniformly. The result is continuous curve $\gamma_*\colon[0,1]\to\mathbb R^2$ which has "infinite length" (in fact, it has no length at all).

Wednesday, November 11, 2015

Lecture 3, Nov 10, 2015

Filed under: Rothschild course "Analysis for high school teachers" — Sergei Yakovenko @ 5:18
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Limits

First, what’s the problem?
Assume we want to calculate the derivative of the function $f(x)=x^2$, say, at the point $x=2$. This derivative is the number, defined using the divided difference $\displaystyle\frac{(2+h)^2-4}{h}=4+h$ when $h$ is “very small”. What does it mean “very small”? We cannot let $h$ be exactly zero, since division by zero is forbidden. On the other hand, if $h\ne 0$, then the above expression is never equal to 4 (as expected) precisely, so in any case the “derivative” cannot be 4, as we want. To resolve this controversy, Leibniz introduced mysterious “differentials” which disappear when added to usual numbers, but whose ratio has precise numerical meaning.

The approach by Leibniz can be worked out into a rigorous mathematical theory, called nonstandard analysis, but historically a different approach, based on the notion of limit (of sequence, function, …), prevailed.

Limit of a sequence

Consider an (infinite) sequence $\{a_n:n\in\mathbb N\}=\{a_1,a_2,a_3,\dots,a_n,\dots\}$ of real numbers. We say that it stabilizes (מתיצבת) at the value $A\in\mathbb R$, if only finitely many terms in this sequence can be different from $A$, and the remaining infinite “tail” consists of the repeated value $A$. Since among the finitely many numbers $n\in\mathbb N$ one can always choose the maximal one (denoted by $N$), we say that the sequence $latex\{a_n\}$ stabilizes, if

$\exists N\in\mathbb N\ \forall n>N\ a_n=A.$

Obviously, stabilizing sequences are not interesting, but their obvious properties can be immediately listed:

1. Changing any finite number of terms in a stabilizing sequence keeps it stabilizing and vice versa;
2. If a sequence $\{a_n\}$ stabilizes at $A$, and another sequence $\{b_n\}$ stabilizes at $B$, then the sum-sequence $\{a_n+b_n\}$ stabilizes at $A+B$, the product-sequence $\{a_nb_n\}$ at $AB$.
3. The fraction-sequence $\{a_n/b_n\}$ may be not defined, but if $B\ne 0$, then only finitely many terms $b_n$ can be zeros, just change them to nonzero numbers and then the fraction-sequence will be defined for all $n$ and stabilize at $A/B$.
4. Exercise: formulate properties of stabilizing sequences, involving inequalities.

Blurred vision

As we have introduced the real numbers, to test their equality requires to check infinitely many “digits”, which is unfeasible. All the way around, we can specify a given precision $\varepsilon >0$ (it can be chosen rational). Then one can replace the genuine equality by $\varepsilon$-equality, saying that two numbers $a,b\in\mathbb R$ are $\varepsilon$-equal, if $|a-b|<\varepsilon$. This is a bad notion that will be used only temporarily, since it is not transitive (for the same fixed value of $\varepsilon$). Yet as a temporary notion, it is very useful.

We say that an infinite sequence $\{a_n\}\ \varepsilon$-stabilizes at a value $A$ for a given precision $\varepsilon$, if only finitely many terms in the sequence are not $\varepsilon$-equal to $A$. Formally, this means that

$\exists N\in\mathbb N\ \forall n>N\ |a_n-A|<\varepsilon.$

The choice of the precision $\varepsilon>0$ is left open so far. In practice it may be set at the level which is determined by the imperfection of our measuring instruments, but since we strive for a mathematical definition, we should not set any artificial threshold.

Definition. A sequence of real numbers $\{a_n\}$ is said to converge to the limit $A\in\mathbb R$, if for any given precision $\varepsilon>0$ this sequence $\varepsilon$-stabilizes at $A$. The logical formula for the corresponding statement is obtained by adding one more quantifier to the left:

$\forall\varepsilon>0\ \exists N\in\mathbb N\ \forall n>N\ |a_n-A|<\varepsilon.$

If we want to claim that the sequence is converging without specifying what the limit is, one more quantifier is required:

$\exists A\in\mathbb R\ \forall\varepsilon>0\ \exists N\in\mathbb N\ \forall n>N\ |a_n-A|<\varepsilon.$

Of course, this formula is inaccessible to anybody not specially prepared for the job, this is why so many students shuttered their heads over it.

Obvious examples

1. $a_n=\frac 1n$, or, more generally, $a_n=\frac1{n^p},\ p>0$. The limit is zero.
2. $a_n=c,\ c\in\mathbb R$. Joke.
3. $a_n =n^p,\ p>0$. Diverges.
4. These rules (plus the obvious rules concerning the arithmetic operations) allow to decide the convergence of any sequence $a_n$ whose general term is a rational function of $n$.
5. Exceptional cases are very rare: e.g., when $a_n=\displaystyle\left(1+\frac 1n\right)^n$

Limits of functions

Let $f\colon D\to\mathbb R$ be a function defined on a subset $D$, and $a\in\mathbb R\smallsetminus D$ a point outside the doomain of $f$. We want to “extend” the function to this point if this makes sense.

For a given precision $\varepsilon>0$ we say that $f$ is $\varepsilon$-constant on a set $S\subseteq D$, if there exists a constant $A\in\mathbb R$ such that $\forall x\in S\ |f(x)-A|<\varepsilon$.

Definition. The function $f\colon D\to\mathbb R$ is said to have a limit equal to $A$ at a point $a\notin D$, if

1. all intersections between $D$ and small intervals $I_\delta=\{|x-a|<\delta\}$ are non-empty and
2. $\forall\varepsilon>0\ \exists\delta>0$ such that the function restricted on $D\cap \{|x-a|<\delta\}$ is $\varepsilon$-constant.

In other words, the function is $\varepsilon$-indistinguishable from a constant on a sufficiently small open interval centered around $a$.

Remark. One can encounter situations when the function is defined at some point $a\in D$, but if we delete this point from $D$, then the function will have a limit $A$ at this point. If this limit coincides with the original value $f(a)$, then the function is well-behaved (we say that it is continuous). If the limit exists but is different from $f(a)$, then we understand that the function was intentionally twisted, and if we change its value at just one point, then it will become continuous. If $f$ has no limit at $a$ if restricted on $D\smallsetminus \{a\}$, we say that $katex f$ is discontinuous$at $a$. Clearly, such extension by taking limit is possible (if at all) only for points at “zero distance” from the domain of the function. For more detail read the lecture notes from the past years. Series As was mentioned, the problem of calculating limits of explicitly given (i.e., elementary) functions is usually not very difficult. The real fun begins when there is no explicit formula for the terms of the sequence (or the function). This may happen if the sequence is produced by some recurrent (inductive) rule. The most simple case occurs where the rule is simply summation (should we say “correction”?) of a known nature: $a_{n+1}=a_n+\text{something explicitly depending on }n$. If we denote the added value by $b_n$, then the sequence will take the form $a_1, a_1+b_1, a_1+b_1+b_2,a_1+b_1+b_2+b_3,a_1+b_1+b_2+b_3+b_4,\dots$. If we can perform such summations explicitly and write down the answer as a function of $n$, it would be great. Example. Consider the case where $b_n=\frac1{n(n+1)}=\frac1n-\frac1{n+1}$. Then we get a “telescopic” sum which can be immediately computed. But this is rather an exception… Another example is the geometric progression where $b_n=cq^n$ for some constant $c,q$. In general we cannot write down the sum as a function of $n$, which makes the task challenging. Definition. Let $b_n$ be a real sequence. We say that the infinite series $\sum_{k=1}^\infty b_k$ converges, if the sequence of its finite sums $a_n=\sum_{k=1}^n b_k$ has a (finite) limit. Examples. 1. The geometric series $\sum_{k=1}^\infty q^k$ converges if $|q|<1$ and diverges if $|q|\ge 1$. 2. The harmonic series $\sum_{k=1}^\infty \frac1k$ diverges. 3. The inverse square series $\sum_{k=1}^\infty \frac1{n^2}$ converges. The last two statements follow from the comparison of the series with patently diverging or patently converging series (fill in the details!). Later on we will concentrate specifically on the series of the form $\sum_{k=0}^\infty c_k q^k$ with a given sequence of “Taylor coefficients” $c_0,c_1,c_2,\dots$ which contain a parameter $q$. Considered as the function of $q$, these series exhibit fascinating properties which make them ubiquitous in math and applications. Tuesday, January 10, 2012 Lectures 11-12, January 3, 10 (2012) Properties of continuous functions. Basic notions of topology The standard list of properties of functions, continuous on intervals, includes theorems on intermediate value, on boundedness, on attainability of extremal values etc. We explain that these results are manifestations of the following phenomenon. There are several properties of subsets of $\mathbb R^1$ (and, in general, arbitrary subsets of the Euclidean space), which can be defined using only the notions of limit and proximity. Such properties are called topological. Examples of such properties are openness/closeness, connectedness (arc-connectedness) and compactness. The general principle then can be formulated (vaguely) as follows: the topological properties are preserved by continuous maps (or their inverses). Wednesday, December 28, 2011 Lecture 10, December 27, 2011 Continuity and limits of functions of real variable In the first lecture we introduce the notion of continuity of a function at a given point in its domain and a very close notion of a limit at a point outside of the “natural” domain. This notion is closely related to the notion of sequential limit as introduced earlier. This paves a way to generalize immediately all arithmetic and order results from numeric sequences to functions. The novel features involve one-sided limits, limits “at infinity” and continuity of composition of functions. The (unfinished) notes, to be eventually replaced by a more polished text, are available here: follow the updates, this temporary link will eventually be erased. Monday, December 21, 2009 Lecture 8 (Dec 22, 2009) 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. Monday, December 14, 2009 Lecture 7 (Dec 15, 2009) 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. Tuesday, December 8, 2009 Lecture 6 (Dec 8, 2009) 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 $\mathbf K$.

Continuity

A function $f\colon X\to\mathbb R$ defined on a subset $X\subseteq\mathbb R$ is continuous (in full, continuous on $X$), if it is continuous at each point $a\in X$. Continuity is equivalent to the requirement that the preimage $f^{-1}(U)$ of any open set $U$ is open in $X$ (i.e., is an intersection between $X$ and an open subspace $V\subseteq \mathbb R$).

• Continuity at a point is a local property. Automatic continuity at isolated points of $X$.
• Examples of discontinuity points.
• Continuity and arithmetic operations. Continuity of the composition. Continuity of elementary functions.
• Dirichlet function: the ugly beast. Further pathologies.
• Continuity and global properties:
1. Continuity and boundedness
2. Continuity and existence of roots
3. Continuity and monotonicity
• Continuity and functional equations $f(x+y)=f(x)+f(y)$, $f(x+y)=f(x)\cdot f(y)$, $f(xy)=f(x)+f(y)$, $f(xy)=f(x)\cdot f(y)$.

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