Connected vs. Path Connected Sets

[lmedin02]

In general, if S is a connected set, can I conclude that S must be path connected?

Definition 1: S is connected if it is not disconnected. A set S is disconnected if it can be written as the union of two mutually separated sets, where mutually separated sets are two nonempty sets that do not contain any of each others boundary points.

Definition 2: S is path (or polygon) connected if any two points in S can be joined by a chain of line segments contain in S which abut (joined end to end), starting from one point and ending at the other.

 

[rasmhop]

Those definitions seem somewhat non-standard (especially 2). What is a line segment? If it has the elementary meaning consider the unitcircle $\{(x,y) | x^2+y^2 =1\}$ which doesn't seem to be path connected according to your definition 2 (but is path connected according to the standard definition of path connectedness which says that points can be joined by a continuous function $[0,1]\to S$).

A standard example for the normal definitions is the http://mathworld.wolfram.com/TopologistsSineCurve.html" .

 

[HallsofIvy]

Here's a cute problem connected with the "topologist's sine curve" that rasmhop refers to:

Find two set, P and Q satisfying:

1) Both P and Q are contained in the closed square in R² with vertices (1, 1), (-1, 1), (-1, -1), and (1, -1).

2) P contains the diagonally opposite points (-1, -1) and (1, 1) and Q contains the diagonally opposite points (-1, 1) and (1, -1).

3) P and Q are disjoint.

4) P and Q are both connected sets.

The solution involves using the "topologist's sine function" to construct two connected but NOT path connected sets that satisfy these conditions.

By the way, if a set is path connected, then it is connected. If a set is either open or closed and connected, then it is path connected. But there exist sets, neither open nor closed, that are connected but not path connected so "path connected" is a stronger condition than just "connected".

 

[lmedin02]

I was trying to understand the topologists sine curve and see how this forms a connected set, but I am having a difficult time.
First the function sin(1/x) has an essential singularity at x=0, thus it cannot be extended to a continuous function in a neighborhood of the origin.
How can I convince myself that the topologists sine curve forms a connected set?
Intuitively this set looks disjoint in a neighborhood of the origin.

 

[rasmhop]

I was trying to understand the topologists sine curve and see how this forms a connected set, but I am having a difficult time.
First the function sin(1/x) has an essential singularity at x=0, thus it cannot be extended to a continuous function in a neighborhood of the origin.
How can I convince myself that the topologists sine curve forms a connected set?
Intuitively this set looks disjoint in a neighborhood of the origin.

If it was disconnected you could find a separation A, B of it. Let A be the component containing (0,0). Clearly the ray y=0,x<=0 is connected so it must be in A. The graph of (x,sin(1/x)) for x>0 is also connected and so it must be B (since if it were in A, B would be empty). Now define a sequence,
$$a_n = (1/(n\pi),\sin(n\pi)) = (1/(n\pi),0)$$
This sequence is in B, but converges to (0,0) so (0,0) can't be in A (since A is open and disjoint from B).

 

[lmedin02]

Ok, I think I'm getting somewhere. I followed your argument and define A as you stated. From B we can see that the origin is a boundary point of B contained in A. From the definition then A and B are not mutually separated. What confuses me is that A and B are disjoint, hence giving me an intuitive idea that A and B must be mutually separated; in other words, if two sets are disjoint they are not necessarily mutually separated.

Also, I do not agree with A being open. It looks like a half open interval; it is closed at the origin.

 

[rasmhop]

Ok, I think I'm getting somewhere. I followed your argument and define A as you stated. From B we can see that the origin is a boundary point of B contained in A. From the definition then A and B are not mutually separated. What confuses me is that A and B are disjoint, hence giving me an intuitive idea that A and B must be mutually separated; in other words, if two sets are disjoint they are not necessarily mutually separated.

Also, I do not agree with A being open. It looks like a half open interval; it is closed at the origin.

Sorry about stating that A is open. I switched back to an alternative definition of a separation that says that A and B are disjoint and both open which is equivalent to your definition. Anyway you don't need it by the argument you presented as you show that the limit of the sequence is in A and therefore a boundary point of B.

Also yes mutually separated and disjoint are different. Consider for instance (-infty,0], (0,infty) which are two disjoint subsets of the real line, but they are not mutually separated since 0 is in the first interval and a boundary point of the second.