Results 1 to 7 of 7

Math Help - Show a nonempty finite subset of R has a max and min

  1. #1
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230

    Show a nonempty finite subset of R has a max and min

    Show that a nonempty finite subset of the Real Numbers has both a maximum and minimum.

    I know I can use induction to solve this but I seem to be having some problems getting started:


    Let P(n) be assertion: a nonempty finite subset of R has a max and min

    P(1) is the assertion: a set S={1} has a max and min, both of which are 1.

    Suppose P(k) is true for: a set S={k} containing a max and min.

    Show P(k+1) true for S={k+1}


    This is where I seem to be getting stuck. I am just not sure what to do from here (or even if my set up is correct)

    Any help would be greatly appreciated!

    Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,616
    Thanks
    1579
    Awards
    1

    Re: Show a nonempty finite subset of R has a max and min

    Quote Originally Posted by mybrohshi5 View Post
    Show that a nonempty finite subset of the Real Numbers has both a maximum and minimum.
    Let P(n) be assertion: a nonempty finite subset of R has a max and min
    P(1) is the assertion: a set S={1} has a max and min, both of which are 1.
    Suppose P(k) is true for: a set S={k} containing a max and min.
    Show P(k+1) true for S={k+1}
    I can see how this could be tricky depending of how picky an instructor can be.
    But in a set S of n+1 real numbers remove any one, say x.
    Now T=S\setminus\{x\} is set of n elements.
    So by the inductive step let \alpha=\max(T)~\&~\omega=\min(T).
    Now what? For example what if x<\alpha~?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230

    Re: Show a nonempty finite subset of R has a max and min

    Now there are a few cases to cover!

    Here's what I have come up with now,

    Let  \alpha = max(T) and  \beta = min(T)

    if  x < \alpha then  \alpha is a maximum for T

    if  x > \alpha then  x is a maximum for T

    if  x < \beta then  x is a minimum for T

    if  x > \beta then  \beta is a minimum for T
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,616
    Thanks
    1579
    Awards
    1

    Re: Show a nonempty finite subset of R has a max and min

    Quote Originally Posted by mybrohshi5 View Post
    Now there are a few cases to cover!

    Here's what I have come up with now,

    Let  \alpha = max(T) and  \beta = min(T)

    if  x < \alpha then  \alpha is a maximum for T

    if  x > \alpha then  x is a maximum for T

    if  x < \beta then  x is a minimum for T

    if  x > \beta then  \beta is a minimum for T
    You are trying to show that S has a max & min.
    T is the set in the inductive step and S=T\cup \{x\} has one more element.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230

    Re: Show a nonempty finite subset of R has a max and min

    Okay I may be getting a little confused now.

    so i have a set S which contains {n+1} elements. Now I let  x be an element of S and let T = S \ {x}.

    So now T is of size "n" and we know that T contains a max and min, namely  \alpha = max(T) and  \beta = min(T)

    if  x < \alpha then  \alpha is a maximum for S

    if  x > \alpha then  x is a maximum for S

    similar reasoning for minimum

    Thus we have shown that S will always contain a max and min (whether that be x,  \alpha or  \beta )

    Am I missing something?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,616
    Thanks
    1579
    Awards
    1

    Re: Show a nonempty finite subset of R has a max and min

    Quote Originally Posted by mybrohshi5 View Post
    so i have a set S which contains {n+1} elements. Now I let  x be an element of S and let T = S \ {x}.
    So now T is of size "n" and we know that T contains a max and min, namely  \alpha = max(T) and  \beta = min(T)
    if  x < \alpha then  \alpha is a maximum for S
    if  x > \alpha then  x is a maximum for S
    similar reasoning for minimum
    Thus we have shown that S will always contain a max and min (whether that be x,  \alpha or  \beta )
    Am I missing something?
    No, you have it now. You may want to add that \alpha\in T\subseteq S and \beta\in T\subseteq S
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230

    Re: Show a nonempty finite subset of R has a max and min

    Great! I will do that.

    Thanks again Plato. Your knowledge is greatly appreciated. Have a great day!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: November 17th 2011, 12:27 PM
  2. Nonempty Subset of Q
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: May 6th 2011, 07:12 PM
  3. Replies: 7
    Last Post: February 19th 2011, 03:29 PM
  4. Replies: 9
    Last Post: September 24th 2010, 06:27 AM
  5. Replies: 1
    Last Post: October 15th 2008, 11:34 AM

/mathhelpforum @mathhelpforum