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

[Utilite]

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.

 

[andrewkirk]

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.

 

[Utilite]

This step is invalid.

why??

 

[andrewkirk]

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 $r_k=2^{-k}(b-a)$.

 

[Utilite]

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.

 

[pasmith]

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.

b is not a member of [a,b), so an open cover of [a,b) doesn't have to contain any sets which contain b.

 

[Utilite]

b is not a member of [a,b), so an open cover of [a,b) doesn't have to contain any sets which contain b.

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].

 

[The Bill]

[a+r_1,a+r_1+r_2) isn't an open set with respect to the topology of [a,b].

 

[Utilite]

[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

 

[pasmith]

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 [a,b] must contain a set U such that (b - \epsilon, b] \subset U for some \epsilon > 0.

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

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

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

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

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

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

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