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

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1. Jun 26, 2016

### 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.

2. Jun 26, 2016

### andrewkirk

This step is invalid.

3. Jun 26, 2016

### Utilite

why??

4. Jun 26, 2016

### andrewkirk

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. Jun 27, 2016

### Utilite

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. Jun 27, 2016

### pasmith

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

7. Jun 27, 2016

### Utilite

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. Jun 27, 2016

### The Bill

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

9. Jun 28, 2016

### Utilite

(a+r_1-r_2,a+r_1+r_2) is in an open set i am just ignoring the unnecessary part

10. Jun 28, 2016

### pasmith

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 $b$.