1. ## Bounded Sets

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

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

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

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

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

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

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

3. Originally Posted by Failure

But it is impossible for $\displaystyle \inf(T) < \sup(S)$ to hold, I think.
Most likely you are right.

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

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

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

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

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

Then sup$\displaystyle (S)$$\displaystyle \leq inf\displaystyle (T) holds for \displaystyle S \cap T =\varnothing and \displaystyle S \cap \not = \varnothing Yah? 4. Originally Posted by novice Most likely you are right. \displaystyle s\leq t, \forall s\in S, \forall t \in T must mean no \displaystyle t \in T is smaller than all the \displaystyle s \in S, and no \displaystyle s \in S is larger than all the \displaystyle t \in T, which fit \displaystyle S \cap T= \varnothingeasily. No - \displaystyle \forall s \in S \ \forall t \in T \ s \le t means that if you take any \displaystyle s \in S, then it is smaller (or equal to) than all \displaystyle t \in T, and vice versa. Let's say \displaystyle S \cap T \not = \varnothing and \displaystyle s \leq t, \forall s \in S, \forall t \in T. I think this can only be true for two cases: Case 1: \displaystyle s \leq t, \forall s \in S, \forall t \in T/S Case 2: \displaystyle s \leq t, \forall s \in S/T, \forall t \in T If so inf\displaystyle (T) \displaystyle \not < sup\displaystyle (S) Then sup\displaystyle (S)$$\displaystyle \leq$ inf$\displaystyle (T)$ holds for $\displaystyle S \cap T =\varnothing$ and $\displaystyle S \cap \not = \varnothing$

Yah?

5. Originally Posted by Defunkt
No - $\displaystyle \forall s \in S \ \forall t \in T \ s \le t$ means that if you take any $\displaystyle s \in S$, then it is smaller (or equal to) than all $\displaystyle t \in T$, and vice versa.
Any $\displaystyle s \in S$ can easily be smaller than all the $\displaystyle t \in T$, but for any $\displaystyle s \in S$ to be equal to all the $\displaystyle t \in T$, then $\displaystyle |S|=|T|=1$.