Topology on the real line
Definition. A homeomorphism between two subsets is a one-to-one map such that both and its inverse are continuous. If such a map exists, the two subsets are called homeomorphic (or topologically equivalent) and we write .
- Homeomorphisms between and .
- The property “Every continuous function is bounded” is invariant by homeomorphism: if it is true for a set and , then this property also holds for . Non-homeomorphy of and .
- The properties “Closedness” and “Openness” are invariant by homeomorphism: if is closed/open and , then is closed/open respectively.
- Is the property “Boundedness” invariant by homeomorphism?
- The disjoint union is homeomorphic to yet non-homeomorphic to or .
Finite point sets are compact. Union of two compacts is a compact.
Theorem. The segment is compact.
Proof. Assume that it is not compact, and is an open covering that does not admit a finite subcovering. Then out of the two halves and 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 of lengths decreasing to 0, which has a (unique) common point : none of these segments by construction admits a finite subcovering by the sets . But the open interval covering , covers also all sufficiently small segments . The resulting contradiction proves that the initial segment was compact. ♦
Proposition. A function continuous on a compact set, is bounded (and achieves its extremal values).
Proof. For each point there exists a neighborhood such that is bounded on (e.g., takes values on the interval . Choose a finite sub-covering and pick the maximal out of the (finitely many!!) local bounds. ♦
Definition. A subset is called connected, if together with any pair of points it contains the entire segment .
The property “Being connected” is invariant by homeomorphisms (what does this mean?)
There are only four non-empty connected subsets on , not homeomorphic to each other: a point, an open interval, a closed segment and a semi-interval .
The Cantor set
From the closed segment delete the (open) middle third . To each of the two remaining closed segments apply the same procedure (deleting the middle third), and repeat it.
As a result, we obtain an infinite collection (family) of nested closed subsets , each of which is a finite union of closed segments of total length going to zero.
The intersection is the Cantor set.
- is closed. For any it can be covered by finitely many intervals of total length less than .
- contains no open subset (everywhere holes!), yet no point of $latex\mathbf K$ is isolated.
- is in one-to-one correspondence with the entire segment .
These properties can be seen from the alternative description of as points on the segment , 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 is not isolated, we show that each interval of length centered at , contains another point of . For this, it is sufficient to change a “ternary digit” with the number greater than 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 and correspond to the same point! how does this affect the above arguments?
Difficult question. Construct a continuous function such that the preimage of any point always consists of only one or two points from .