Why would a topologist's sine curve be connected but is not path connected in general topology?

In general topology, a space is said to be connected if it cannot be expressed as the union of two disjoint non-empty open sets. A space is said to be path connected if there exists a continuous path between any two points in the space.

The sine curve is a continuous function that takes values in the interval [-1, 1]. As a function, it is connected. However, the sine curve is not path connected in the usual topology on the real line. This is because there is no continuous path from the point (-1, 0) to the point (1, 0) that stays within the sine curve.

One way to see this is to consider the following sets:

A = {(x, y) | y > 0} B = {(x, y) | y < 0}

These sets are both open in the usual topology on the real line, and they cover the entire sine curve. However, they are disjoint, and there is no continuous path from a point in A to a point in B that stays within the sine curve.

Therefore, the sine curve is connected but not path connected in the usual topology on the real line.

Related Questions

  1. What is the difference between a connected space and a path connected space?
    • A connected space cannot be expressed as the union of two disjoint non-empty open sets, while a path connected space has a continuous path between any two points in the space.
  2. Why is the sine curve not path connected in the usual topology on the real line?
    • Because there is no continuous path from the point (-1, 0) to the point (1, 0) that stays within the sine curve.
  3. What is a topological space?
    • A set equipped with a collection of subsets, called open sets, which satisfies certain axioms.
  4. What is a continuous function?
    • A function that preserves the topology of the space, meaning that the pre-image of an open set is open.
  5. What is a homeomorphism?
    • A continuous function that is also a bijection, meaning that it has an inverse that is also continuous.

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