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

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The discussion centers on proving that the interval [0,1] is compact by demonstrating that the set E, defined as the collection of points t in [0,1] for which the interval [0,t] is covered by a finite number of open sets from an open cover, is non-empty. Participants confirm that E is indeed non-empty since it includes the point t=0. They also note that E is bounded above by 1. There is a critique of the clarity of the original argument and a suggestion to improve notation and rigor. Additionally, some users express frustration with the textbook "Rudin" and seek alternative resources for learning real analysis.
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Homework Statement


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

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

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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colstat said:

Homework Statement


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

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

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

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

In general it is indeed true that E is nonempty since 0\in E and that E is bounded above by 1. But you got to be careful with your notation.
 
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.
 
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.
 
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 \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...
 
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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.
 

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