# Nonempty set bounded from below

• Oct 28th 2008, 08:55 AM
kathrynmath
Nonempty set bounded from below
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.
• Oct 28th 2008, 09:12 AM
Plato
We don't know what you are to prove,
• Oct 28th 2008, 11:48 AM
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.
• Oct 28th 2008, 12:54 PM
Jhevon
Quote:

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 $x$
• Oct 28th 2008, 01:11 PM
kathrynmath
Quote:

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?
• Oct 29th 2008, 01:29 PM
kathrynmath
Quote:

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.
• Oct 29th 2008, 02:55 PM
Plato
Quote:

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 $S$.