Proof of (0,100) Cover Not Finite: Title Under 65 Characters

[QuarkCharmer]

Homework Statement

(0,100) has a cover that consists of a finite number of closed interval subsets.


I'm really lost with this one. I can clearly understand why the statement is false, but I'm not sure my proof is good.

Homework Equations

The Attempt at a Solution

Clearly this is false, so I am trying to disprove it.

Proof:
Let S =(0,100) and let C be a cover of S.
Since C contains finitely many closed interval subsets of S,
C has a least element.
Let X_i =[a,b]$\in$C be a subset of S, $\forall$a,b$\in$R^+
such that 0<a<b<100

Since there is no smallest positive real number, (which i have proved before),
there is an infinite number of X_i's in C.
But C has a finite number of elements.
This is a contradiction.
Q.E.D.

 

[verty]

What does it mean for C to have a least element? Can one have a least set?

Stuff like this is the problem you have, things that are meaningless or not properly defined.

 

[QuarkCharmer]

It certainly seems like it. If C is a set containing a finite number of sets, then you could number each set in C, and since there is a finite number, one of them must be the least?

 

[QuarkCharmer]

Well, after considering what you said a little more...

I was thinking that I could let C be the cover of (0,100), then come up with some sort of closed interval like

[50-v(n), 50+v(n)]

to represent a generic closed interval subset in C.
v(n) in this case, would be a function of real number n, such that as n approaches infinity,
the interval converges on [0,100]
(If this is even possible to write, I haven't came up with a v(n) yet).

If the cover is a set of many such [50-v(n), 50+v(n)] 's with different n's, then could I show that there is always another n value (and thus, another set in C), which would bring the interval closer to (0,1).

Which means there would NEED to be an infinite number of closed-interval subsets in C, and thus, there is a contradiction.

My thought is, that this is not valid because then it only shows that C contains infinitely many closed interval subsets of (0,100) for that specific type of closed interval, and not "any". Since there could be many different covers...

Am I even on the right path here? When I try to look up more information that would be helpful, like that "The union of finitely many closed intervals is closed", I only find information on Topology which is far outside the scope of my class.

I feel that there is just something really obvious that I am missing here.

 

[Dick]

Well, after considering what you said a little more...

I was thinking that I could let C be the cover of (0,100), then come up with some sort of closed interval like

[50-v(n), 50+v(n)]

to represent a generic closed interval subset in C.
v(n) in this case, would be a function of real number n, such that as n approaches infinity,
the interval converges on [0,100]
(If this is even possible to write, I haven't came up with a v(n) yet).

If the cover is a set of many such [50-v(n), 50+v(n)] 's with different n's, then could I show that there is always another n value (and thus, another set in C), which would bring the interval closer to (0,1).

Which means there would NEED to be an infinite number of closed-interval subsets in C, and thus, there is a contradiction.

My thought is, that this is not valid because then it only shows that C contains infinitely many closed interval subsets of (0,100) for that specific type of closed interval, and not "any". Since there could be many different covers...

Am I even on the right path here? When I try to look up more information that would be helpful, like that "The union of finitely many closed intervals is closed", I only find information on Topology which is far outside the scope of my class.

I feel that there is just something really obvious that I am missing here.

If by 'cover' you mean there are a finite number of sets whose union equals (0,100), then I think your original idea for the proof works fine. If $X_i=[a_i,b_i]$ take the minimum of the $a_i$.

 

[QuarkCharmer]

I think I ended up doing something like that.

My proof was basically this (I'm paraphrasing from memory)Proof:
Assume that (0,100) has a cover consisting of a finite number of closed interval subsets.
Let S be a collection of closed interval subsets of (0,100), such that
$(0,100)\in\bigcup_{i=1}^{n}S_{i}$ So, it's a cover of (0,100)...

We organize the closed intervals in S first by left bound values in ascending order, then by right bounds in ascending order. Resulting in the ordered S having the form:
$S = {[a_1,b_1],[a_1,b_2],...,[a_1,b_j],[a_2,b_1],[a_2,b_2],...,[a_2,b_j],...,[a_i,b_{j-i}],[a_i,b_j]}$ for i,j are natural numbers.

So each closed interval subset of (0,100) in S, is represented as $[a_1, b_j]$

Then I noted for clarification, that i and j are simply indexes, and that each a_i and b_j were in fact real numbers.

Now I defined a value that I called L, the lowest value of S. (This is just my convention for this).
L is the smallest of either $a_1$ or $b_1$,
if they are equal (then the interval is a closed singleton?), then I just said "take a_1 then"...

so L is now basically defined as the smallest number like Dick mentioned above.

Then I claimed that for any L in (0,1), there is a K in (0,1) such that K = L/2.

This shows that S has infinitely many elements, and thus, is contradictory to my assumption that it had a finite number of elements.
Q.E.D.
How does that look?

 

[Dick]

I think I ended up doing something like that.

My proof was basically this (I'm paraphrasing from memory)


Proof:
Assume that (0,100) has a cover consisting of a finite number of closed interval subsets.
Let S be a collection of closed interval subsets of (0,100), such that
$(0,100)\in\bigcup_{i=1}^{n}S_{i}$ So, it's a cover of (0,100)...

We organize the closed intervals in S first by left bound values in ascending order, then by right bounds in ascending order. Resulting in the ordered S having the form:
$S = {[a_1,b_1],[a_1,b_2],...,[a_1,b_j],[a_2,b_1],[a_2,b_2],...,[a_2,b_j],...,[a_i,b_{j-i}],[a_i,b_j]}$ for i,j are natural numbers.

So each closed interval subset of (0,100) in S, is represented as $[a_1, b_j]$

Then I noted for clarification, that i and j are simply indexes, and that each a_i and b_j were in fact real numbers.

Now I defined a value that I called L, the lowest value of S. (This is just my convention for this).
L is the smallest of either $a_1$ or $b_1$,
if they are equal (then the interval is a closed singleton?), then I just said "take a_1 then"...

so L is now basically defined as the smallest number like Dick mentioned above.

Then I claimed that for any L in (0,1), there is a K in (0,1) such that K = L/2.

This shows that S has infinitely many elements, and thus, is contradictory to my assumption that it had a finite number of elements.
Q.E.D.



How does that look?

Your set of closed intervals is {[a1,b1],[a2,b2],[a3,b3],...,[an,bn]}. That doesn't include intervals like [a1,b2]. And in the notation [ai,bi] it's always the case that ai<=bi. If you clear up some confusion about the notation the proof is a lot simpler.