- #1

- 53

- 0

## Main Question or Discussion Point

Hello there, can someone help me with the proof? The proof in my text (Advanced Calculus, R Creighton Buck) is long and tedious and I hoped to be able to make it shorter

Let O be a non empty open connected set.

----------

Aside:

Then if O is broken up into Kn sets, then the intersection of some Kn with some other Kn is nonempty just by the definition of connectedness.

----------

So I guess my idea is that we can make all these Kn's into open balls(or some sets) and connect every point through the intersection of the Kn's.

Consider an open ball completely contained in O. Since any point 'p' of an open set must not be isolated, there exists another point 'q' such that p and q can be connected by a straight line inside the ball.

Then by my "idea" every point in O can be connected by a sequence of straight lines.

Could someone point me in the right direction if I'm off? The thing I'm having trouble wrapping my head around is filling up all of O with open balls. It seems this only works in some finite sense or something is mucked up.

Let O be a non empty open connected set.

----------

Aside:

Then if O is broken up into Kn sets, then the intersection of some Kn with some other Kn is nonempty just by the definition of connectedness.

----------

So I guess my idea is that we can make all these Kn's into open balls(or some sets) and connect every point through the intersection of the Kn's.

Consider an open ball completely contained in O. Since any point 'p' of an open set must not be isolated, there exists another point 'q' such that p and q can be connected by a straight line inside the ball.

Then by my "idea" every point in O can be connected by a sequence of straight lines.

Could someone point me in the right direction if I'm off? The thing I'm having trouble wrapping my head around is filling up all of O with open balls. It seems this only works in some finite sense or something is mucked up.