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

[Mathsadness]

TL;DR 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. 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.

 

[PeroK]

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?

 

[PeroK]

PS an example of an open cover for $(0, 1)$ that does not have a finite subcover is $\{(\frac 1 n, 1 - \frac 1 n): n = 3, 4, 5 \dots \}$.

 

[Office_Shredder]

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).

 

[Mathsadness]

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]?

 

[PeroK]

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.

 

[PeroK]

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.

 

[Office_Shredder]

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)

 

[pasmith]

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 $\epsilon > 0$ the open cover $\{ B(x,\epsilon) : x \in (0,1)\}$ does in fact admit a finite subcover of $(0,1)$: $$\left\{ B\left(\tfrac{n\epsilon}2, \epsilon\right) : 1 \leq n \leq \left\lceil \tfrac 2\epsilon \right\rceil - 1 \right\}.$$

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 $(0,1)$.

 

[AndreasC]

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.