Connected subset

• Oct 19th 2010, 12:35 PM
tarheelborn
Connected subset
If $\displaystyle S \subseteq \mathbb{R}$ is bounded below but not bounded above, then $\displaystyle S$ is either $\displaystyle \left( a,\infty \right]$ or $\displaystyle \left[ a,\infty \right)$ for some $\displaystyle a \in \mathbb{R}$.

I know from a previously proved theorem that if $\displaystyle S$ is a connected subset of $\displaystyle \mathbb{R}$ and $\displaystyle a<b$, $\displaystyle a,b \in S$, then $\displaystyle [a,b] \subseteq S$. And it seems intuitive that $\displaystyle a$ is the greatest lower bound for this set, but I am not sure how to begin to prove it.
• Oct 19th 2010, 12:55 PM
Plato
Surely you left something out.
Did you mean that $\displaystyle S$ is connected?
• Oct 19th 2010, 01:00 PM
tarheelborn
I thought the same thing, but I am not sure. I have rewritten the problem as it was presented to me, but I feel sure that it must also mean S is a connected subset.
• Oct 19th 2010, 01:02 PM
Plato
Quote:

Originally Posted by tarheelborn
I thought the same thing, but I am not sure. I have rewritten the problem as it was presented to me, but I feel sure that it must also mean S is a connected subset.

Surely. Otherwise it is false.

There is a well known theorem: The only nontrivial connected sets of real numbers are:
$\displaystyle ( - \infty ,a),( - \infty ,a],(a,b),[a,b),(a,b],[a,b],(a,\infty ),[a,\infty )$.
Only the last two fit the description of $\displaystyle S$.
• Oct 19th 2010, 01:07 PM
tarheelborn
OK, then. If $\displaystyle S$ is a connected subset of $\displaystyle \mathbb{R}$ and $\displaystyle S$ is bounded below, but not above, then $\displaystyle S$ is either $\displaystyle \left[ a, \infty \right)$ or $\displaystyle \left( a, \infty \right)$ for some $\displaystyle a \in \mathbb{R}$.
• Oct 20th 2010, 06:52 AM
tarheelborn
I am trying to prove the two parts of that theorem described in the last post. I believe I need to show that since $\displaystyle S$ is bounded below, it has a greatest lower bound and that said bound doesn't actually need to be included in the interval. But I am not sure how to get that started.
• Oct 21st 2010, 09:17 AM
tarheelborn
I have gotten this far with the proof:

Since $\displaystyle S$ is bounded below, $\displaystyle S$ has a greatest lower bound, say $\displaystyle a$. Since $\displaystyle S$ is not bounded above, I claim that $\displaystyle S=(a, \infty)$ or $\displaystyle S=[a, \infty)$.
Case 1: Suppose $\displaystyle a,x \in S$ such that $\displaystyle a \neq x$. Since $\displaystyle a=g.l.b.(S)$, $\displaystyle a<x$. Now since $\displaystyle S$ is unbounded, there is some $\displaystyle s \in S$ such that $\displaystyle s>x$. but since $\displaystyle a,s,x \in S$, by previously proved lemma, $\displaystyle [a,x] \in S$ and $\displaystyle [x,s] \in S$. Henc e $\displaystyle S=[a, \infty)$.
Case 2: Suppose $\displaystyle a \notin S$ and suppose $\displaystyle x \in S$, $\displaystyle x>a$.

Now I am not sure how to move on to say that everything approaching a is in S but a is not in S.
• Oct 21st 2010, 09:41 AM
Plato
I will help with this point.
If $\displaystyle A$ is connected then the closure, $\displaystyle \overline{A}$, is also connected.
So no limit point can be separated from the set.
Thus both $\displaystyle (a,\infty)~\&~[a,\infty)$ are connected.
• Oct 21st 2010, 09:45 AM
tarheelborn
Thank you, Plato!