Prove [0,1] is non-empty and bounded above

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In summary: Anyways, let \gamma=Sup E, how do you prove \gamma\inE. I know the definition of Sup, but how do you show something is a supremum?Sup(E) is the least upper bound of E. But 1 is an upper bound. So sup(E) must be smaller... Continue...great! I will get started, I am not sure if I will come back for more on this one.But one thing for sure, I am learning more on PF than in class.
  • #1
colstat
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Homework Statement


Want to prove that [0,1] in R is compact. Let [itex]\bigcup_{\alpha\in A}[/itex] I[itex]_{\alpha}[/itex] be an open cover of [0,1].

By open sets in R.
Let E={t[itex]\in[/itex][0,1] s.t. [0,t] is covered by a finite number of the open cover sets I[itex]_{\alpha}[/itex]}.
Prove that E[itex]\neq[/itex][itex]\emptyset[/itex].

The Attempt at a Solution


Let t=0, the set E=[0,0] has only one element, it is non-empty.
Is this ok for the non-empty part?
 
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  • #2
colstat said:

Homework Statement


Want to prove that [0,1] in R is compact. Let [itex]\bigcup_{\alpha\in A}[/itex] I[itex]_{\alpha}[/itex] be an open cover of [0,1].

By open sets in R.
Let E={t[itex]\in[/itex][0,1] s.t. [0,t] is covered by a finite number of the open cover sets I[itex]_{\alpha}[/itex]}.
Prove that E[itex]\neq[/itex][itex]\emptyset[/itex].

The Attempt at a Solution


Let t=0, the set E=[0,0] has only one element, it is non-empty.
Is this ok for the non-empty part?

Your argument makes no sense. You wish to prove E to be non-empty. Thus you wish to find a t in E. So you want to find a t such that [0,t] is covered by finitely many sets.
 
  • #3
I found t=0, there is only one element in that interval, namely {0}. What's wrong?

Let me continue from earlier.

[0,0]={0} [itex]\in[/itex][0,1]
There exists some I[itex]_{\alpha}[/itex] that covers {0}, and there is a finite number of open covers. So, E[itex]\neq[/itex][itex]\emptyset[/itex].
Since E={t|t[itex]\in[/itex][0,1] and [0,t]...}
E[itex]\subset[/itex][0,1]. So it is bounded above by 1.
 
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  • #4
Does anyone else know? someone is got to know this, chapter 2 of Rudin.

*patiently waiting*
 
  • #5
Clean up your argument first. Things like [itex]\{0\}\in [0,1][/itex] really make no sense.

In general it is indeed true that E is nonempty since [itex]0\in E[/itex] and that E is bounded above by 1. But you got to be careful with your notation.
 
  • #6
thanks, micromass. :)
So, I was right? I feel like Rudin is a little overated(no examples). How do I learn how to prove this? I know I am suppose to work hard, but you can't just beat me around the bush.

Anywho, does anyone know some good websites? I've been look at other course websites, but it's very frustrating.
 
  • #7
colstat said:
thanks, micromass. :)
So, I was right? I feel like Rudin is a little overated(no examples). How do I learn how to prove this? I know I am suppose to work hard, but you can't just beat me around the bush.

Anywho, does anyone know some good websites? I've been look at other course websites, but it's very frustrating.

Rudin is a very good book... if you already know the material :biggrin:
Yes, I consider it to be overrated and not really suitable for a first encounter with real analysis. There are better books out there though. Take a look at http://hbpms.blogspot.com/2008/05/stage-3-introductory-analysis.html for some good (and free) books.
 
  • #8
Thanks again! I just feel like banging my head against the wall. Actually I want to bang Rudin's head against the wall. So, frustrating!

Anyways, let [itex]\gamma[/itex]=Sup E, how do you prove [itex]\gamma[/itex][itex]\in[/itex]E. I know the definition of Sup, but how do you show something is a supremum?
 
  • #9
Sup(E) is the least upper bound of E. But 1 is an upper bound. So sup(E) must be smaller... Continue...
 
  • #10
great! I will get started, I am not sure if I will come back for more on this one.
But one thing for sure, I am learning more on PF than in class.
 

1. How do you prove that [0,1] is non-empty?

To prove that [0,1] is non-empty, we simply need to show that there exists at least one element within this interval. In this case, we can see that 0 is an element of [0,1], therefore the interval is non-empty.

2. Can you explain what it means for [0,1] to be bounded above?

When we say that [0,1] is bounded above, it means that there exists a number that serves as an upper limit for all elements within the interval. In this case, the number 1 serves as the upper limit, as all elements in [0,1] are less than or equal to 1.

3. How can we prove that [0,1] is bounded above?

To prove that [0,1] is bounded above, we can use the completeness axiom which states that every non-empty set of real numbers that is bounded above must have a least upper bound (supremum). In this case, the supremum of [0,1] is 1, which proves that the interval is bounded above.

4. Is [0,1] the only interval that is non-empty and bounded above?

No, there are many intervals that are non-empty and bounded above. Any interval of the form [a,b] where a < b, is non-empty and bounded above. For example, [2,5] is a non-empty and bounded above interval.

5. What is the significance of proving that [0,1] is non-empty and bounded above?

Proving that [0,1] is non-empty and bounded above is significant because it helps us understand the properties of real numbers and intervals. It also allows us to use this information in other mathematical proofs and calculations. Additionally, it serves as a building block for more complex mathematical concepts and theories.

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