# connectedness

• Apr 28th 2009, 05:07 PM
chaoticmath
connectedness
Hi,
Could anyone help me prove that " an open interval A is connected".
I tried proving it by contradiction and had 3 cases for A and taking two open sets U=(a,b) and V=(c,d)
case 1: b<c and showed it was impossible for A to be in U union V
case 2: b=c similarly
case 3: b>c similarly

Thank you
• Apr 28th 2009, 06:28 PM
HallsofIvy
Quote:

Originally Posted by chaoticmath
Hi,
Could anyone help me prove that " an open interval A is connected".
I tried proving it by contradiction and had 3 cases for A and taking two open sets U=(a,b) and V=(c,d)
case 1: b<c and showed it was impossible for A to be in U union V
case 2: b=c similarly
case 3: b>c similarly

Thank you

Any interval whether open, closed, neither, or both is connected so you will need to use the definition of "interval". What is that definition?
• Apr 28th 2009, 08:00 PM
chaoticmath
Quote:

Originally Posted by HallsofIvy
Any interval whether open, closed, neither, or both is connected so you will need to use the definition of "interval". What is that definition?

Well I only have the definition of an open interval:
If a,b in X, then (a,b) denotes the set of points in X for which a<x and x<b. A set of the form (a,b) is called an open interval.
• Apr 28th 2009, 09:45 PM
redsoxfan325
Right, so if $\displaystyle a<x<b$, then $\displaystyle x\in(a,b)$. Thus it's a connected set.
• Apr 29th 2009, 03:00 AM
HallsofIvy
?? You are simply asserting that an interval is connected which is what was supposed to be proved!

Chaoticmath, if the interval (a, b) were NOT connected, there would exist two open sets, A, B, such that $\displaystyle A\cup B= (a, b)$ and $\displaystyle A\cap B= \phi$. Take x in A, y= B and show that there must exist points in (x, y) that are not in either A or B.
• Apr 29th 2009, 10:16 AM
redsoxfan325
Oh sorry. I have a theorem in my book that says "A subset E of the real line is connected if and only if it has the following propoerty: if $\displaystyle x\in E$, $\displaystyle y\in E$, and $\displaystyle x<z<y$, then $\displaystyle z\in E$ ". I had this theorem in mind when I posted that.
• Apr 29th 2009, 03:38 PM
chaoticmath
Thank you that helped!
Here is what I have, please let me know if it's correct:
Take x in U, y in V such that x/=y (/= is for different), x<y and x,y belong to (a,b).
By a theorem we saw, we know (x,y)/=0. So let z belong to (x,y).
Now consider, the set A1={m in V : m>n for all n in U}
A1 included in A and A1 /= 0.
Let z be the greatest lower bound of A1. Since V is open then z is not in V so z is in U.
By another theorem we saw, there exists an open interval A2 containing z so there exists z2 in A2 such that z2>z but this contradicts the fact that z is the greatest lower bound.

Someone else read it and told me I may still have to show that my A1 was in A because U could be the union of many open sets, but I'm not sure I understand that.

• Apr 29th 2009, 04:59 PM
Plato
Definition: A set is connected if and only if is not the union of two separated sets (R.L. Moore)
Definition: Two sets are separated if each contains neither a point nor a limit point of the other.
(Remember that Prof. Moore did not think that a point set could be empty).

Lemma: If $\displaystyle t=\text{lub or glb} (D)$ then $\displaystyle t$ is either in $\displaystyle D$ or is a limit point of $\displaystyle D$.

For the purpose of contradiction suppose that $\displaystyle \left( {a,b} \right) = H \cup K$ where $\displaystyle H~\&~K$ are separated sets.

Without loss of generality we can say that $\displaystyle \left( {\exists p \in H} \right)\left( {\exists q \in K} \right)\left[ {a < p < q < b} \right]$.
Now consider these: $\displaystyle d = \text{lub} \left[ {H \cap \left( {a,q} \right)} \right]\;\& \;e = \text{glb} \left[ {K \cap \left( {d,b} \right)} \right]$.

Can you find a contradiction in all of that?