Heine-Borel Theorem shouldn't work for open intervals?

  • #1
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Okay, I am studying Baby Rudin and I am in a lot of trouble.
I want to show that a closed interval [a,b] is compact in R. The book gives a proof for R^n but I am trying a different proof like thing.
Since a is in some open set of an infinite open cover, the interval [a,a+r_1) is in that open set for an r_1. Similarly a+r_1 is in the closed interval, [a+r_1,a+r_1+r_2) is in an open set. Going like this we can reach an r_k such that a+r_1+r_2...+r_k is greater than or equal to b. Therefore an interval [c,b] is an open set and those open sets form a subcover of an infinite open cover. M=min{r_1,r_2....r_k}>0 Therefore k needs to be less than or equal to (b-a)/M which is clearly finite. Even if every interval [a+r_1...r_i,a+r_1....r_(i+1)) is contained in different open sets the number of open sets is finite.
Every infinite open cover of an interval [a,b] has a finite subcover then it is compact.
I think I am making a mistake because this should work for [a,b). But [0,1) is not compact.
I am very confused, please help me out.
 

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  • #2
andrewkirk
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Going like this we can reach an r_k such that a+r_1+r_2...+r_k is greater than or equal to b.
This step is invalid.
 
  • #4
andrewkirk
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why??
We have no reason to expect that any sum of ##r_k##s will reach ##b-a##. That would be the case for instance if we had ##r_k=2^{-k}(b-a)##.
 
  • #5
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We have no reason to expect that any sum of ##r_k##s will reach ##b-a##. That would be the case for instance if we had ##r_k=2^{-k}(b-a)##.
But since our interval is covered by open sets some neighbourhood of each point is a subset of an open set. [a,c) is a subset of O_1, [c,d) is a subset of O_2 etc. Since B is contained in an open set B should we contained in a neighbourhood of some point.
 
  • #6
pasmith
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But since our interval is covered by open sets some neighbourhood of each point is a subset of an open set. [a,c) is a subset of O_1, [c,d) is a subset of O_2 etc. Since B is contained in an open set B should we contained in a neighbourhood of some point.
[itex]b[/itex] is not a member of [itex][a,b)[/itex], so an open cover of [itex][a,b)[/itex] doesn't have to contain any sets which contain [itex]b[/itex].
 
  • #7
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[itex]b[/itex] is not a member of [itex][a,b)[/itex], so an open cover of [itex][a,b)[/itex] doesn't have to contain any sets which contain [itex]b[/itex].
No but i am talking about [a,b] b is a member of an open set on an open cover of [a,b]. I don't know if this proves something but my argument seems valid, i am just finding a union of neighbourhoods that cover [a,b].
 
  • #8
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[a+r_1,a+r_1+r_2) isn't an open set with respect to the topology of [a,b].
 
  • #9
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[a+r_1,a+r_1+r_2) isn't an open set with respect to the topology of [a,b].
(a+r_1-r_2,a+r_1+r_2) is in an open set i am just ignoring the unnecessary part
 
  • #10
pasmith
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But since our interval is covered by open sets some neighbourhood of each point is a subset of an open set. [a,c) is a subset of O_1, [c,d) is a subset of O_2 etc. Since B is contained in an open set B should we contained in a neighbourhood of some point.
No but i am talking about [a,b] b is a member of an open set on an open cover of [a,b]. I don't know if this proves something but my argument seems valid, i am just finding a union of neighbourhoods that cover [a,b].
Well yes: an open cover of [itex][a,b][/itex] must contain a set [itex]U[/itex] such that [itex](b - \epsilon, b] \subset U[/itex] for some [itex]\epsilon > 0[/itex].

Here is a proof of compactness of [itex][a,b][/itex] which explicitly uses the least upper bound axiom:

Let [itex]\mathcal{U}[/itex] be an open cover of [itex][a,b][/itex] and define [itex]P \subset [a,b][/itex] such that [itex]x \in P[/itex] if and only if [itex][a,x][/itex] is covered by a finite subcollection of [itex]\mathcal{U}[/itex]. Our aim is to show that [itex]b \in P[/itex].

Observe that if [itex]x \in P[/itex] then [itex][a,x]\subset P[/itex].

Now [itex]a \in P[/itex] as [itex][a,a] = \{a\}[/itex] is covered by any set in [itex]\mathcal{U}[/itex] which contains [itex]a[/itex], and by definition [itex]b[/itex] is an upper bound for [itex]P[/itex]. Hence [itex]s = \sup P[/itex] exists and [itex]a < s \leq b[/itex]. Note that [itex][a, s) \subset P[/itex].

Now [itex]s \in [a,b][/itex] so there exists a [itex]U \in \mathcal{U}[/itex] such that [itex]s \in U[/itex]. If [itex]s < b[/itex] then by openness of [itex]U[/itex] there exists a [itex]\delta > 0[/itex] such that [itex](s - \delta,s + \delta) \subset U[/itex]. But then [itex][a,s + \frac12\delta] = [a,s - \frac12\delta] \cup (s-\delta,s+\frac12\delta][/itex] is covered by a finite subcollection of [itex]\mathcal{U}[/itex] so [itex]s < s + \frac12 \delta \in P[/itex]. This is a contradiction.

Thus [itex]s = b[/itex] and by openness of [itex]U[/itex] there exists an [itex]\epsilon > 0[/itex] such that [itex](b - \epsilon,b] \subset U[/itex]. This together with the finite subcollection which covers [itex][a, b - \frac12 \epsilon][/itex] forms a finite subcover of [itex][a,b][/itex] as required.

Note that starting from [itex]a[/itex] we could get arbitrarily close to [itex]b[/itex] using only a finite number of open sets, but to actually reach [itex]b[/itex] we had to use a set which contained [itex]b[/itex].
 
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