Why is [0,1] compact in the case of a collection of open neighborhoods?

  • #1

Summary:

This is not an assignment question but has been bugging me for a while. (0,1) is not compact as it has at least one open cover which does not have a finite sub cover. [0,1] is different as every open cover has a finite subcover for this.
For (0,1), the collection of neighborhoods N_e of q from (0,1) is an open cover. However, there exists e>0 such that it will not have a finite sub cover. Let us take e=0.5*min{|p-q|}, where p=/=q and both are from (0,1). I am not sure if the construction of e here is right, please correct me if not. Then every neighborhood will only cover one point from (0,1) which is q itself, hence a finite subcollection cannot cover (0,1) as there are infinitely many points. Then, similarly the collection neighborhoods for [0,1] is also an open cover given the neighborhood is taken from a point, say x from [0,1]. my question is then how can there be a finite sub covers for THIS particular open cover which consists of the collections of the neighborhood of x, where x is from [0,1]. By definition there has to be as the set is compact but I fail to see one for this particular open cover which is the collection of the neighborhoods of its points.

Thank you for any suggestion. i am new to this and would be obliged if the math shown is simplified.
 

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  • #2
PeroK
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Summary:: my question is then how can there be a finite sub covers for THIS particular open cover which consists of the collections of the neighborhood of x, where x is from [0,1].
Can you define this open cover for ##[0, 1]## more precisely?
 
  • #3
PeroK
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PS an example of an open cover for ##(0, 1)## that does not have a finite subcover is ##\{(\frac 1 n, 1 - \frac 1 n): n = 3, 4, 5 \dots \}##.
 
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  • #4
Office_Shredder
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Being compact is equivalent to sequences having convergent subsequences. Do you know why every sequence in [0,1] has a convergent subsequence? And why that's not true for (0,1)?

I think it's easier to think of it from this point of view, and consider the following:

Consider an open cover of sets ##U_i##. If there is no finite subcover, then for each k, you can find ##x_k## such that ##x_k \notin \bigcup_{i=1}^{k} U_k##. Some subsequence of ##x_k##s converges to some point ##y \in [0,1]##. Let's call this sequence ##y_k## - ##y_k## is the ##k##th ##x_i## in the subsequence converging to ##y##.
Then there must be some ##j## such that ##y\in U_j## since the Us form an open cover. But since ##U_j## is open, and ##y_k## converges to ##y##, there is some ##K## such that if ##k>K##, then ##y_k \in U_j##. But this contradicts the definition of the ##y_k##s in particular ##y_k \notin U_j## for ##k > j##.

Perok of course has posted the classic example of a cover with no finite subcover on (0,1) that fails to cover [0,1]. Under the above equivalence (well, I only proved one way but you can also do it the other way) this is basically the same thing as observing that the sequence 1/n converges to a point in [0,1] but not a point in (0,1).
 
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  • #5
Can you explain the part why 1/n does not converge to a point in (0,1)? Is it because we define limit as |1/n-0|<e, which boils down to 1/n<e which can be made arbitrarily small by archimedean property? Essentially saying that 1/n=0. As 0 is not in (0,1) it converges in [0,1]?
 
  • #6
PeroK
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Can you explain the part why 1/n does not converge to a point in (0,1)? Is it because we define limit as |1/n-0|<e, which boils down to 1/n<e which can be made arbitrarily small by archimedean property? Essentially saying that 1/n=0. As 0 is not in (0,1) it converges in [0,1]?
It has nothing to do with limits. I have specified an open cover with no finite subcover.
 
  • #7
PeroK
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PS an example of an open cover for ##(0, 1)## that does not have a finite subcover is ##\{(\frac 1 n, 1 - \frac 1 n): n = 3, 4, 5 \dots \}##.
What I have specified here is an infinite family of open sets. One for every natural number starting from ##3##. It's an open cover for ##(0, 1)## because every member of that set is in at least one of the sets in the open cover.

But, it has no finite subcover (exercise) hence ##(0, 1)## is not compact.
 
  • #8
Office_Shredder
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Can you explain the part why 1/n does not converge to a point in (0,1)? Is it because we define limit as |1/n-0|<e, which boils down to 1/n<e which can be made arbitrarily small by archimedean property? Essentially saying that 1/n=0. As 0 is not in (0,1) it converges in [0,1]?
Yes, 1/n converges to 0, which is in [0,1] but not in (0,1)
 
  • #9
pasmith
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Summary:: This is not an assignment question but has been bugging me for a while. (0,1) is not compact as it has at least one open cover which does not have a finite sub cover. [0,1] is different as every open cover has a finite subcover for this.

For (0,1), the collection of neighborhoods N_e of q from (0,1) is an open cover. However, there exists e>0 such that it will not have a finite sub cover.
This is false.

For any [itex]\epsilon > 0[/itex] the open cover [itex]\{ B(x,\epsilon) : x \in (0,1)\} [/itex] does in fact admit a finite subcover of [itex](0,1)[/itex]: [tex]
\left\{ B\left(\tfrac{n\epsilon}2, \epsilon\right) : 1 \leq n \leq \left\lceil \tfrac 2\epsilon \right\rceil - 1 \right\}.[/tex]

Non-compactness does not require that every open cover not admit a finite subcover, but only that at least one open cover does not admit a finite subcover. @PeroK has exhibited such a cover for [itex](0,1)[/itex].
 
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  • #10
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Summary:: This is not an assignment question but has been bugging me for a while. (0,1) is not compact as it has at least one open cover which does not have a finite sub cover. [0,1] is different as every open cover has a finite subcover for this.

Then every neighborhood will only cover one point from (0,1) which is q itself
That wouldn't be an open neighbourhood if it only contained a single real point, so I'm not sure what you mean.
 

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