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

• Aug 31st 2011, 01:22 PM
mybrohshi5
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 :D
• Aug 31st 2011, 01:53 PM
Plato
Re: Show a nonempty finite subset of R has a max and min
Quote:

Originally Posted by mybrohshi5
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 $\displaystyle S$ of $\displaystyle n+1$ real numbers remove any one, say $\displaystyle x$.
Now $\displaystyle T=S\setminus\{x\}$ is set of $\displaystyle n$ elements.
So by the inductive step let $\displaystyle \alpha=\max(T)~\&~\omega=\min(T)$.
Now what? For example what if $\displaystyle x<\alpha~?$
• Aug 31st 2011, 02:24 PM
mybrohshi5
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 $\displaystyle \alpha = max(T)$ and $\displaystyle \beta = min(T)$

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

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

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

if $\displaystyle x > \beta$ then $\displaystyle \beta$ is a minimum for T
• Aug 31st 2011, 02:43 PM
Plato
Re: Show a nonempty finite subset of R has a max and min
Quote:

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

Here's what I have come up with now,

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

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

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

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

if $\displaystyle x > \beta$ then $\displaystyle \beta$ is a minimum for T

You are trying to show that $\displaystyle S$ has a max & min.
$\displaystyle T$ is the set in the inductive step and $\displaystyle S=T\cup \{x\}$ has one more element.
• Aug 31st 2011, 02:59 PM
mybrohshi5
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 $\displaystyle x$ 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 $\displaystyle \alpha = max(T)$ and $\displaystyle \beta = min(T)$

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

if $\displaystyle x > \alpha$ then $\displaystyle x$ 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, $\displaystyle \alpha$ or $\displaystyle \beta$)

Am I missing something?
• Aug 31st 2011, 03:09 PM
Plato
Re: Show a nonempty finite subset of R has a max and min
Quote:

Originally Posted by mybrohshi5
so i have a set S which contains {n+1} elements. Now I let $\displaystyle x$ 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 $\displaystyle \alpha = max(T)$ and $\displaystyle \beta = min(T)$
if $\displaystyle x < \alpha$ then $\displaystyle \alpha$ is a maximum for S
if $\displaystyle x > \alpha$ then $\displaystyle x$ 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, $\displaystyle \alpha$ or $\displaystyle \beta$)
Am I missing something?

No, you have it now. You may want to add that $\displaystyle \alpha\in T\subseteq S$ and $\displaystyle \beta\in T\subseteq S$
• Aug 31st 2011, 03:17 PM
mybrohshi5
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!