Prove that:-- A set is said to be pathwise connected if any two points in can be

Question

Prove that:--

A set is said to be pathwise connected if any two points in can be joined by a (piecewise-smooth) curve entirely contained in . The purpose of this exercise is to prove that an open set is pathwise connected if and only if is connected.

While the hint is--(a) Suppose first that is open and pathwise connected, and that it can be written as = 1 2 where 1 and 2 are disjoint non-empty open sets. Choose two points w1 1 and w2 2 and let denote a curve in joining w1 to w2 . Consider a parametrization z : [0, 1] of this curve with z(0) = w1 and z(1) = w2, and let t = sup {t:z(s)1 for all 0s<t}. 0t1. Arrive at a contradiction by considering the point z(t). (b) Conversely, suppose that is open and connected. Fix a point w and let 1 denote the set of all points that can be joined to w by a curve contained in . Also, let 2 denote the set of all points that cannot be joined to w by a curve in . Prove that both 1 and 2 are open, disjoint and their union is . Finally, since 1 is non-empty (why?) conclude that = 1 as desired.

Part a, what does t* mean? Is it a function or a point in 1?;

and how to prove the problem strictly?

Explanation / Answer

Background and motivation:

To start with , think of open sets in R. Any open set in R is a disjoint union of open intervals. For example, the open set (1,2) U (3,4) is an open set . Note that it is is neither connected, nor path connected. (No path from a point (say 1.5) in (1,2) to any point (say 2.3) in (3,4) can exist as it would violate the continuity of the path--- A path being a continuous image of {0,1], is itself a connected space.

Thus , it is intuitively clear that an open set (in the Euclidean space) is path connected if and only if it is connected.

Here is a detailed proof for the points (a) and (b) in the question.

t* is a point in the interval [0,1] (Note that is a continuous map from the interval [0,1]. (It is neither a function nor a point in 1.

(a) An open path connected set is connected

the set {t:z(s)1} is an open set as it is the inverse image of the open set 1 under a continuous map. So z(t) will belong to both 1 as well as 2, which is a contradiction. In other words, the existence of z(t*) will establish a disconnection of the path from w1 to w2 , which is not possible , as a path is a connected space. This establishes the first part--a path connected open set has to be connected. To put in an another way, there cannot be a disconnection of a path connected open set .

(b) Here we prove the converse--an open connected set is path connected.

Let P be a point in 1. This implies there exists a curve C starting with w and ending with P.

As is open, there exists an open set U containing P. Let Q belong to U. There exists a path from P to Q. Combining this with C, we get a path from w to Q. Thus every point in U belongs to 1--in other words , 1 is an open set. Similary one can prove that 2 is an open set.

Now 1 is non-empty as w belongs to it. If 1 is non-empty as w belongs to it. If 2 is non-empty , 1 and 2 will constitute a disconnection of the connected set .

This forces 2 to be empty. Thus all points can be connected to w--this means is path connected, as desired.

This is clearly a contradiction and it establishes (b).

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