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Prove if $n in mathbb{Z} - mathbb{N}$, then max$(nA)=n$ min$(A)$, for $nA = left{na left| right. a in A right}$

$left|{sin(pi alpha N)}/{sin(pi alpha)}right| leq {1}/{2 | alpha |}$real analysis on max min and boundednessAre sup, inf, max, min correct for the set ${frac{x}{x+1}: x in A}$?Prove that set is bounded but has no max/minEstablishing whether a set is open or closed and bounded or unboundedHow can I find the sup, inf, min, and max of $bigcupleft[frac{1}{n}, 2-frac{1}{n}right]$inf, sup, max, min for $bigcap_{n in mathbb N}left[-frac{1}{3^n},4+frac{1}{2n}right)$Prove the interval contains infinitely many points of AProving that $minleft { f,g right }$ and $maxleft { f,g right }$ are continuousIf $f,g$ are continuous at $x_0 in mathbb{R}$ then $min{f,g}$ and $max{f,g}$ are at $x_0 in mathbb{R}$ as well

0

$begingroup$

I'm having trouble with the following:

Let $A$ be a nonempty bounded subset of $mathbb{R}$ and let $n in mathbb{Z}$. Define $nA = left{na left| right. a in A right}$. Prove the following:
If $n in mathbb{Z} - mathbb{N}$, then max$(nA)=n$ min$(A)$.

I started out by writing:

Suppose $n in mathbb{Z} - mathbb{N}$, i.e. $n in left{...,-2,-1,0 right}$. If $n=0$, the result is trivial.

If $n<0$, and we let $m=$min$(A)$, then $m=A^l cap A$, where $A^l$ denotes the set of all lower bounds of $A$. Now $nm=n cdot A^l cap A$.

I'm not sure where to go from here?

real-analysis

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asked Mar 19 at 16:05

bb411bb411

16919

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0

$begingroup$

I'm having trouble with the following:

Let $A$ be a nonempty bounded subset of $mathbb{R}$ and let $n in mathbb{Z}$. Define $nA = left{na left| right. a in A right}$. Prove the following:
If $n in mathbb{Z} - mathbb{N}$, then max$(nA)=n$ min$(A)$.

I started out by writing:

Suppose $n in mathbb{Z} - mathbb{N}$, i.e. $n in left{...,-2,-1,0 right}$. If $n=0$, the result is trivial.

If $n<0$, and we let $m=$min$(A)$, then $m=A^l cap A$, where $A^l$ denotes the set of all lower bounds of $A$. Now $nm=n cdot A^l cap A$.

I'm not sure where to go from here?

real-analysis

share|cite|improve this question

asked Mar 19 at 16:05

bb411bb411

16919

$endgroup$

add a comment | 

$begingroup$

I'm having trouble with the following:

Let $A$ be a nonempty bounded subset of $mathbb{R}$ and let $n in mathbb{Z}$. Define $nA = left{na left| right. a in A right}$. Prove the following:
If $n in mathbb{Z} - mathbb{N}$, then max$(nA)=n$ min$(A)$.

I started out by writing:

Suppose $n in mathbb{Z} - mathbb{N}$, i.e. $n in left{...,-2,-1,0 right}$. If $n=0$, the result is trivial.

If $n<0$, and we let $m=$min$(A)$, then $m=A^l cap A$, where $A^l$ denotes the set of all lower bounds of $A$. Now $nm=n cdot A^l cap A$.

I'm not sure where to go from here?

real-analysis

share|cite|improve this question

asked Mar 19 at 16:05

bb411bb411

16919

$endgroup$

share|cite|improve this question

asked Mar 19 at 16:05

bb411bb411

16919

share|cite|improve this question

asked Mar 19 at 16:05

bb411bb411

16919

asked Mar 19 at 16:05

bb411bb411

16919

asked Mar 19 at 16:05

bb411bb411

16919

1 Answer
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1

$begingroup$

It's easier than that since $ forall a : a geq min(A)$. Then multiplying by $n$ changes the inequality and finally, you know that the maximum is in $A$ :)

share|cite|improve this answer

answered Mar 19 at 16:11

MarcMMarcM

362

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1

$begingroup$

It's easier than that since $ forall a : a geq min(A)$. Then multiplying by $n$ changes the inequality and finally, you know that the maximum is in $A$ :)

share|cite|improve this answer

answered Mar 19 at 16:11

MarcMMarcM

362

$endgroup$

add a comment | 

1

$begingroup$

It's easier than that since $ forall a : a geq min(A)$. Then multiplying by $n$ changes the inequality and finally, you know that the maximum is in $A$ :)

share|cite|improve this answer

answered Mar 19 at 16:11

MarcMMarcM

362

$endgroup$

add a comment | 

$begingroup$

It's easier than that since $ forall a : a geq min(A)$. Then multiplying by $n$ changes the inequality and finally, you know that the maximum is in $A$ :)

share|cite|improve this answer

answered Mar 19 at 16:11

MarcMMarcM

362

$endgroup$

share|cite|improve this answer

answered Mar 19 at 16:11

MarcMMarcM

362

answered Mar 19 at 16:11

MarcMMarcM

362

answered Mar 19 at 16:11

MarcMMarcM

362