## 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 .

[…] 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 […]

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