1. ## Bounded Sets

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 $s \leq t, \exists s \in S, \exists t \in T$

2. 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.

3. 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?

4. 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?

5. 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 $|S|=|T|=1$.