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

• I
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.

andrewkirk
Homework Helper
Gold Member
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.

This step is invalid.
why??

andrewkirk
Homework Helper
Gold Member
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)##.

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
Homework Helper
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 $b$.

$b$ 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].

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

[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
Homework Helper
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 $b$.

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