# Sergei Yakovenko's blog: on Math and Teaching

## 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 $\mathbf K$.