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

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In summary: This can be shown by considering the sequence 1/n which converges to a point in [0,1] but not a point in (0,1). This means that (0,1) is not compact because it does not have a finite subcover.
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Mathsadness
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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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Mathsadness said:
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?
 
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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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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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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
Mathsadness said:
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 said:
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
Mathsadness said:
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
Mathsadness said:
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
Mathsadness said:
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.
 

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

1. Why is [0,1] compact?

The interval [0,1] is compact because it satisfies the definition of compactness, which states that every open cover of the set has a finite subcover. In other words, any collection of open sets that covers [0,1] can be reduced to a finite number of open sets that still cover [0,1].

2. How does compactness relate to open neighborhoods?

Compactness is a property of a set, while open neighborhoods are subsets of a topological space. In the case of [0,1], the set is compact because it can be covered by open neighborhoods in a finite manner. This means that any open neighborhood of [0,1] can be reduced to a finite number of open neighborhoods that still cover [0,1].

3. Can you provide an example of a collection of open neighborhoods for [0,1]?

One example of a collection of open neighborhoods for [0,1] is the set of all open intervals (a,b) where a and b are real numbers between 0 and 1. This collection of open sets covers [0,1] and can be reduced to a finite subcover, making [0,1] compact.

4. Is [0,1] compact in all topological spaces?

No, [0,1] is not necessarily compact in all topological spaces. Compactness is a property that depends on the specific topological space and the definition of open sets within that space. In some topological spaces, [0,1] may not satisfy the definition of compactness.

5. What are the practical applications of understanding compactness in the case of open neighborhoods?

Understanding compactness in the case of open neighborhoods is important in many areas of mathematics and science, including topology, analysis, and physics. It allows us to prove theorems and make predictions about the behavior of sets and spaces. Additionally, compactness is a useful concept in optimization problems and in understanding the convergence of sequences and series.

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