Bounded Sets

**novice** · September 9th 2010, 05:20 PM

Suppose I have two bounded subsets $S$ and $T$ of $\mathbb{R}$ with $s \leq t, \forall s \in S$ and $\forall t \in T$ . Furthermore, suppose the max $(S)$ and min $(T)$ do not exist.

Is it possible that inf $(T) \leq$ sup $(S)$ ?

Here, I am thinking: Either $S \cap T \not = \varnothing$ or $(S \cap T) = \varnothing$ .
For $(S \cap T) = \varnothing$ , it's obvious that $sup(S) \leq inf(T)$ ,
but for $S \cap T \not = \varnothing$ , it's a little tricky.

Say I have $S=(-20,3)$ and $T=(0,12)$ .
I find the definition $s \leq t, \forall s \in S$ and $\forall t \in T$ very difficult to understand.
Since $(S \cap T) = [0,3]$ , it looks like $s \leq t, \exists s \in S, \exists t \in T$

**Failure** · September 10th 2010, 01:06 AM

Originally Posted by novice

Suppose I have two bounded subsets $S$ and $T$ of $\mathbb{R}$ with $s \leq t, \forall s \in S$ and $\forall t \in T$ . Furthermore, suppose the max $(S)$ and min $(T)$ do not exist.

Is it possible that inf $(T) \leq$ sup $(S)$ ?

Yes, sure. Consider, for example, $S := \left\{-\frac{1}{n} \mid n\in\mathbb{N}\right\}$ and $T := \left\{\frac{1}{n}\mid n\in\mathbb{N}\right\}$ , or maybe even simpler $S := \,[-1,0),\; T := \,(0,1]$ . In that case we have $\inf(T)=0=\sup(S)$ .

But it is impossible for $\inf(T) < \sup(S)$ to hold, I think.

**novice** · September 10th 2010, 06:44 AM

Originally Posted by Failure

But it is impossible for $\inf(T) < \sup(S)$ to hold, I think.

Most likely you are right.

$s\leq t, \forall s\in S, \forall t \in T$ must mean no $t \in T$ is smaller than all the $s \in S$ , and no $s \in S$ is larger than all the $t \in T$ , which fit $S \cap T= \varnothing$ easily.

Let's say $S \cap T \not = \varnothing$ and $s \leq t, \forall s \in S, \forall t \in T$ . I think this can only be true for two cases:

Case 1: $s \leq t, \forall s \in S, \forall t \in T/S$

Case 2: $s \leq t, \forall s \in S/T, \forall t \in T$

If so inf $(T)$ $\not <$ sup $(S)$

Then sup $(S)$ $\leq$ inf $(T)$ holds for $S \cap T =\varnothing$ and $S \cap \not = \varnothing$

Yah?

**Defunkt** · September 10th 2010, 07:26 AM

Originally Posted by novice

Most likely you are right.

$s\leq t, \forall s\in S, \forall t \in T$ must mean no $t \in T$ is smaller than all the $s \in S$ , and no $s \in S$ is larger than all the $t \in T$ , which fit $S \cap T= \varnothing$ easily.

No - $\forall s \in S \ \forall t \in T \ s \le t$ means that if you take any $s \in S$ , then it is smaller (or equal to) than all $t \in T$ , and vice versa.

Let's say $S \cap T \not = \varnothing$ and $s \leq t, \forall s \in S, \forall t \in T$ . I think this can only be true for two cases:

Case 1: $s \leq t, \forall s \in S, \forall t \in T/S$

Case 2: $s \leq t, \forall s \in S/T, \forall t \in T$

If so inf $(T)$ $\not <$ sup $(S)$

Then sup $(S)$ $\leq$ inf $(T)$ holds for $S \cap T =\varnothing$ and $S \cap \not = \varnothing$

Yah?

**novice** · September 10th 2010, 07:49 AM

Originally Posted by Defunkt

No - $\forall s \in S \ \forall t \in T \ s \le t$ means that if you take any $s \in S$ , then it is smaller (or equal to) than all $t \in T$ , and vice versa.

Any $s \in S$ can easily be smaller than all the $t \in T$ , but for any $s \in S$ to be equal to all the $t \in T$ , then $|S|=|T|=1$ .

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