Sergei Yakovenko's blog: on Math and Teaching

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.

Cantor set: first several steps

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.


1 Comment »

  1. […] the function is squeezed between and .  The union of all these small intervals covers which is compact. Hence there exists a finite covering of by these intervals. The endpoints of these intervals […]

    Pingback by Lecture 13 (Feb 2, 2010) « Sergei Yakovenko’s Weblog — Monday, February 8, 2010 @ 5:06 | Reply

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