Results 1 to 5 of 5

Thread: Bounded Sets

  1. #1
    Banned
    Joined
    Sep 2009
    Posts
    502

    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 $\displaystyle s \leq t, \exists s \in S, \exists t \in T$
    Last edited by novice; Sep 9th 2010 at 08:19 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by novice View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by Failure View Post

    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by novice View Post
    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.
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by Defunkt View Post
    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 $\displaystyle |S|=|T|=1$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Polynomial integrable on bounded sets
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Jul 22nd 2011, 12:09 PM
  2. [SOLVED] compact sets are closed and bounded
    Posted in the Differential Geometry Forum
    Replies: 8
    Last Post: May 23rd 2011, 01:34 AM
  3. closed and bounded sets
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Dec 17th 2010, 01:00 AM
  4. Closed, bounded and perfect sets
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Apr 8th 2010, 12:45 AM
  5. Closed, non-empty and bounded sets.
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Oct 22nd 2009, 02:42 AM

Search Tags


/mathhelpforum @mathhelpforum