Connected subset

**tarheelborn** · October 19th 2010, 12:35 PM

If $S \subseteq \mathbb{R}$ is bounded below but not bounded above, then $S$ is either $\left( a,\infty \right]$ or $\left[ a,\infty \right)$ for some $a \in \mathbb{R}$ .

I know from a previously proved theorem that if $S$ is a connected subset of $\mathbb{R}$ and $a<b$ , $a,b \in S$ , then $[a,b] \subseteq S$ . And it seems intuitive that $a$ is the greatest lower bound for this set, but I am not sure how to begin to prove it.

**Plato** · October 19th 2010, 12:55 PM

Surely you left something out.
Did you mean that $S$ is connected?

**tarheelborn** · October 19th 2010, 01:00 PM

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.

**Plato** · October 19th 2010, 01:02 PM

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:
$( - \infty ,a),( - \infty ,a],(a,b),[a,b),(a,b],[a,b],(a,\infty ),[a,\infty )$ .
Only the last two fit the description of $S$ .

**tarheelborn** · October 19th 2010, 01:07 PM

OK, then. If $S$ is a connected subset of $\mathbb{R}$ and $S$ is bounded below, but not above, then $S$ is either $\left[ a, \infty \right)$ or $\left( a, \infty \right)$ for some $a \in \mathbb{R}$ .

**tarheelborn** · October 20th 2010, 06:52 AM

I am trying to prove the two parts of that theorem described in the last post. I believe I need to show that since $S$ 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.

**tarheelborn** · October 21st 2010, 09:17 AM

I have gotten this far with the proof:

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

Now I am not sure how to move on to say that everything approaching a is in S but a is not in S.

**Plato** · October 21st 2010, 09:41 AM

I will help with this point.
If $A$ is connected then the closure, $\overline{A}$ , is also connected.
So no limit point can be separated from the set.
Thus both $(a,\infty)~\&~[a,\infty)$ are connected.

**tarheelborn** · October 21st 2010, 09:45 AM

Thank you, Plato!

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