Basic question regarind archimedian property

[semidevil]

I need to prove that if I := [0, 1/n], the element 0 belongs to all I(n) and the archimedian property to show that 0 is the only common point.

so basically, I need to prove the intersection from 1 to infinite of I(n) = {0}.

the book says to use the archimedian property that if t > 0, then tehre exsists n(t) such that 0 < 1/n(t) < t.

i've been thinking all day, but still can't go anywhere.

ok, so I guess to start off,

let t > 0, this implies that that there is a n in N, such that t < n(t).

so this means 1/n(t) < t.

and I guess by looking at the archimedian property that 0 < 1/n(t) < t, this means that 1/n(t) is always less then any given t. so I guess if it is less then t, then obviously, t is not included with any 1/(t + 1), and it is always greater then 0, so I guess that's why 0 is the only thing in common w/ all of them.

notice I said "I guess" a lot, because I really don't believe my proof...so can someone give me some tips? am I close?

and then, part 2:

same question, but with (0, 1/n) and need to prove that the intersection is the empty set. both are confusing

 

[Muzza]

Seems like you were on the right track.

(I'll use )0, 1/n( to denote [0, 1/n], since the latter seems to mess up Latex).

You want to show that $\bigcap^{\infty}_{n = 1} )0, 1/n( = \{0 \} .$

The hard part is $\bigcap^{\infty}_{n = 1} )0, 1/n( \subseteq \{ 0 \}$, the other inclusion is trivial.

So, suppose $x \in \bigcap^{\infty}_{n = 1} )0, 1/n($ (that set will be referred to as "the set"). x can't be negative, so we have x >= 0. If x = 0, we are done. If x > 0, take an integer n such that nx > 1. Since x was in the set, we must have $x \in )0, 1/n($, so that 0 < x < 1/n. But then nx < 1, which contradicts nx > 1. Thus, there are no non-zero numbers in the set.

 

[matt grime]

Hmm, review the definitions etc.

1. the intersection is the set of all elements common to all the sets. Obviously 0 is in all the sets, and the intersection contains only positive numbers.

2. Given any number, t, strictly greater than 0 there is an integer n, dependent on t, such that 0<1/n<t, ie t isn't in the inteval [0,1/n]

3. If the intersection contained anything other then 0, it contains a strictly positive t, so by 2. we see no such t exists. Hence the intersection only contains 0.

So, where's the problem? I just wrote out the information in your question and it's the proof you require.

 

[semidevil]

Hmm, review the definitions etc.

1. the intersection is the set of all elements common to all the sets. Obviously 0 is in all the sets, and the intersection contains only positive numbers.

2. Given any number, t, strictly greater than 0 there is an integer n, dependent on t, such that 0<1/n<t, ie t isn't in the inteval [0,1/n]

3. If the intersection contained anything other then 0, it contains a strictly positive t, so by 2. we see no such t exists. Hence the intersection only contains 0.

So, where's the problem? I just wrote out the information in your question and it's the proof you require.

hm...still not understanding quite yet.

I understand number 1..
on number 2, is ok
but I still don't quite follow you on number 3. what do you mean that t doesn't exist if it is in the intersection, and again...can you expalin why??

 

[matt grime]

Suppose that t is strictly positive and in the intersection. Got it? ASSUME that is true...

then since by the archimedian principle there is an n in N suc that 1/n <= t t is not in [0,1/n]

This is a contradicion, so something we assumed is wrong. Let's look, what did we ASSUME...

(apologies if the idea of proof by contradiction is alien, it can be done without contradiction)