# 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.