Nonempty set bounded from below

**kathrynmath** · October 28th 2008, 07:55 AM

S is a nonempty set of real numbers bounded from below.

Here is what I have done so far:
x=inf S=glb S
Since x=glb S, there cannot be an x+epsilon that is a lower bound. This implies there exists a elemt of S such that x-epsilon<a<x<x+epsilon.

My problem is that I think I went wrong somewhere in the 2nd sentence. I can do the rest of my problem if I can get this cleared up, but I'm not entirely sure where I went wrong.

**Plato** · October 28th 2008, 08:12 AM

Please state the problem!
We don't know what you are to prove,

**kathrynmath** · October 28th 2008, 10:48 AM

Let S be a nonempty set of real numbers that is bounded from below and let x=inf S. prove that either x belongs to S or x is an accumulation point of S.

**Jhevon** · October 28th 2008, 11:54 AM

Originally Posted by kathrynmath

Let S be a nonempty set of real numbers that is bounded from below and let x=inf S. prove that either x belongs to S or x is an accumulation point of S.

here is a sketch of the proof

if S has a minimum, then inf(S) = min{S} and so is in the set

now suppose S does not have a minimum, and let x = inf(s)

then there is some $\epsilon > 0$ so that $(x + \epsilon )\in S$ . let $s_1 = x + \epsilon$ . then by the denseness of $\mathbb{R}$ , there is some $s_2 \in \mathbb{R}$ , so that $x < s_2 < s_1$ and $s_2 \in S$ . continue this logic by induction to create a sequence $\{ s_n \}$ that converges to $x$

**kathrynmath** · October 28th 2008, 12:11 PM

Originally Posted by Jhevon

here is a sketch of the proof

if S has a minimum, then inf(S) = min{S} and so is in the set

now suppose S does not have a minimum, and let x = inf(s)

then there is some $\epsilon > 0$ so that $(x + \epsilon )\in S$ . let $s_1 = x + \epsilon$ . then by the denseness of $\mathbb{R}$ , there is some $s_2 \in \mathbb{R}$ , so that $x < s_2 < s_1$ and $s_2 \in S$ . continue this logic by induction to create a sequence $\{ s_n \}$ that converges to $x$

Ok, but was I on the right track at all in what I first wrote?

**kathrynmath** · October 29th 2008, 12:29 PM

Originally Posted by Jhevon

here is a sketch of the proof

if S has a minimum, then inf(S) = min{S} and so is in the set

now suppose S does not have a minimum, and let x = inf(s)

then there is some $\epsilon > 0$ so that $(x + \epsilon )\in S$ . let $s_1 = x + \epsilon$ . then by the denseness of $\mathbb{R}$ , there is some $s_2 \in \mathbb{R}$ , so that $x < s_2 < s_1$ and $s_2 \in S$ . continue this logic by induction to create a sequence $\{ s_n \}$ that converges to $x$

Ok I got that, but now I'm trying to figure out how to create a sequence that converges to x.

**Plato** · October 29th 2008, 01:55 PM

Originally Posted by kathrynmath

Ok I got that, but now I'm trying to figure out how to create a sequence that converges to x.

You don’t need a sequence.
Say that $\sigma = \inf (S)$ . If $\sigma \in S$ you are done.
So suppose that $\sigma \notin S$ .
Hence $\left( {\forall z \in S} \right)\left[ {\sigma < z} \right]$ .
If $\varepsilon > 0 \Rightarrow \left( {\exists x \in S} \right)\left[ {\sigma < x < \sigma + \varepsilon } \right]\quad$ .
That means that $\sigma$ is an accumulation point of $S$ .

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