# Two properties preserved by continuous maps: compactness and connectivity

These two properties are key to existence of solutions to infinitely many problems in mathematics and physics.

## Compactness

Compactness (of a subset $A\subseteq \mathbb R^n$) is the “nearest approximation” to finiteness of $A$. Obviously, if $A$ is a finite set of points, then

1. Any infinite sequence $\{a_n\}_{n=1}^\infty\subseteq A$ has an infinite stationary (constant) subsequence;
2. $A$ is bounded and closed;
3. If $\bigcup_\alpha U_\alpha\supseteq A$ is an arbitrary covering of $A$ by open subsets $U_\alpha$, then one can always choose a finite subcovering $U_{\alpha_1}\cup\cdots\cup U_{\alpha_N}\supseteq A$.

The first two properties are obvious, the third one also. For each point $a_1,\dots,a_N\in A$ it is enough to find just one open set $U_{\alpha_i}$ which covers this point. Their union (automatically finite) covers all of $A$.

Definition.The following three properties of a set $A\subseteq \mathbb R^n$ are equivalent:

1. Any infinite sequence $\{a_n\}_{n=1}^\infty\subseteq A$ has a partial limit (i.e., the limit of an infinite subsequence), which is again in $A$;
2. $A$ is bounded and closed;
3. If $\bigcup_\alpha U_\alpha\supseteq A$ is an arbitrary covering of $A$ by open subsets $U_\alpha$, then one can always choose a finite subcovering $U_{\alpha_1}\cup\cdots\cup U_{\alpha_N}\supseteq A$.

Example. The closed segment, say, $[0,1]\subset\mathbb R^1$ possesses all three properties.

1. The standard trick of division into halves and choosing each time the half that contains infinitely many members of the sequence allows to construct a partial limit for any sequence confined to $[0,1]$.
2. Obvious.
3. Assume (by contradiction) that there exists a very perverse covering of $[0,1]$, which does not allow for a choice of finite subcovering. Then at least one of the two segments, $[0,\frac12],\ [\frac12,1]$, also suffers from the same problem (if both admit finite subcovering, one would easily construct a finite subcovering for the initial segment $[0,1]$). Continuing this way, we construct an infinite nested sequence of closed intervals which do not admit a finite subcovering. Their intersection is a point $a\in[0,1]$ which must be covered by at least one open set. But then this set covers also all sufficiently small segments from our nested sequence. Contradiction.

Problem. Prove (using the Example) that the three conditions are indeed equivalent. Hint: any bounded set can be confined to a cube $x_i\in [-C_i,C_i],\ i=1,\dots, n$. Use the closedness of $A$ to prove that the partial limit of any sequence is again in $A$.

Theorem. If $f\colon A\to \mathbb R^m$ is a continuous map and $A$ is compact, than $f(A)$ is also compact.

Corollary. Any continuous function restricted on a compact is bounded and attains its extremal values.

## Connectivity

A subset $A\subseteq \mathbb R^n$ is called connected, if it cannot be split into two disjoint parts “apart from each other”. How this can be formalized?

Example (proto-Definition). A subset $A\subseteq [0,1]$ is called connected, if together with any two points $a,b\in A$ it contains all points $x$ such that $a\le x\le b$.

All connected subsets of the real line can be easily described (Do it!).

How can we treat subsets $A\subseteq \mathbb R^n$ for $n>1$? Two ways can be suggested.

Definition. A set $A\subseteq \mathbb R^n$ is called path connected, if for any two points $a,b\in A$ there exists a continuous map $f\colon [0,1]\to A$ such that $f(0)=a,\ f(1)=b$.

This definition mimics the one-dimensional construction. However, this is not the only possibility to say that a set cannot be split into smaller parts.

Definition. A subset $A\subseteq \mathbb R^n$ is called disconnected, if there exist two open disjoint sets $U_1,U_2\subseteq\mathbb R^n, \ U_1\cap U_2=\varnothing$, such that the two parts $A\cap U_i, \ i=1,2$ are both nonempty. If such partition is impossible, then $A$ is called connected.

Problem. Prove that for subsets on the real line the two definitions coincide.

Problem. Consider the subset of the plane $A$ which consists of the graph $y=\sin \frac1x,\ x>0$ and the point $(0,0)$. Prove that it is connected but not path connected.

Chapter 3 from Abbot, Understanding Analysis. Especially sections 3.2 (open/closed sets), 3.3 (compact sets) and 3.4 (connected sets). Pay attention to the exercises!

## 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).

## 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 $\mathbf K$.

## 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 $f(x+y)=f(x)+f(y)$, $f(x+y)=f(x)\cdot f(y)$, $f(xy)=f(x)+f(y)$, $f(xy)=f(x)\cdot f(y)$.

## Sunday, March 23, 2008

### 2-Sphere eversion in 3D-space

Filed under: links — Sergei Yakovenko @ 12:23
Tags: , , ,

If a smooth curve  embedded in the plane $\mathbb R^2$  is deformed allowing self-intersections but remaining smooth, then there is a natural integral invariant, the rotation number, which prevents eversion of a circle (deformation of the oriented circle into another circle with an opposite orientation). For two-dimensional surfaces smoothly embedded in $\mathbb R^3$ a similar invariant of deformations exists, yet this invariant does not preclude eversion of the sphere inside out.

The possibility of such deformation was discovered bt S. Smale in 1958. Relatively recently W. Thurston invented a general algorithm of smoothening, which yields an explicit sphere eversion. All these spectacular things are discussed on the level accessible to high school students in the most fascinating animation (21 min.) discovered on the web by Dmitry Novikov (thanks!). A much shorter animation (mere 22 sec.) does not easily reveal the mistery, so the longer one is really worth its time!

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