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

**mybrohshi5** · August 31st 2011, 01:22 PM

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

**Plato** · August 31st 2011, 01:53 PM

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 $S$ 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 $x<\alpha~?$

**mybrohshi5** · August 31st 2011, 02:24 PM

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

**Plato** · August 31st 2011, 02:43 PM

Originally Posted by mybrohshi5

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.

**mybrohshi5** · August 31st 2011, 02:59 PM

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?

**Plato** · August 31st 2011, 03:09 PM

Originally Posted by mybrohshi5

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 $\beta\in T\subseteq S$

**mybrohshi5** · August 31st 2011, 03:17 PM

Great! I will do that.

Thanks again Plato. Your knowledge is greatly appreciated. Have a great day!

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