# Example of cover (of a set) having finite sub-covers in collection.

• experimentX
If it were finite, it would have a greatest member, and you could use that to show that (0,1] is compact. So you've shown that (0,1] is compact, but you didn't mean to; you meant to show that [0,1] is compact.)In summary, the concept of compactness can be understood as every open cover having a finite subcover. In the case of sets in R^n, compactness is equivalent to being closed and bounded. However, in metric spaces, there are related results that can be used to prove compactness. For example, every sequence in a compact set has a convergent subsequence. An example of a compact set is X = {0,
experimentX
Gold Member
I think I am not understanding the concept of compactness. Can anyone give me an example of a cover that contains finite sub-covers? for example:- G = {S1,S2, ... }, Sn={(1/n,2/n): n ≥ 2} is an example of cover of set (0,1) but it is an infinite collection.

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As you have shown, (0,1) is not compact.
It is not interesting to have some finite subcover - just add T=(0,1) to your list, and there is a finite subcover (any finite subset with T). Compactness means that every cover has a finite subcover. For compact sets, you cannot create examples like the one on your post.

Then how could we say that cover of compact sets have finite sub-cover if we can't construct them. Shouldn't there be at least one example? I understand the proof given in my book but I can't visualize it.

It depends. In R^n , there is a result that compact is equivalent to being closed and bounded. Maybe if you go over this theorem (Heine-Borel) , you can see how being closed and bounded implies that every cover has a finite subcover. In metric spaces there are related results you can use, like that every sequence has a convergent subsequence. So you can try to see the connection between these two by going over the proofs. But I don't know how to do this more generally.

experimentX said:
I think I am not understanding the concept of compactness. Can anyone give me an example of a cover that contains finite sub-covers? for example:- G = {S1,S2, ... }, Sn={(1/n,2/n): n ≥ 2} is an example of cover of set (0,1) but it is an infinite collection.
The set of all open intervals with length 1 is an open cover of the closed interval [0,1]. The set ##\left\{\left(\frac{-1}{3},\frac{2}{3}\right),\left(\frac{1}{3},\frac{4}{3}\right)\right\}## is a finite sub-cover.

but this also covers (0,1), this would also be a finite sub cover of (0,1) making (0,1) compact. What I am asking is an example of sub covers of [0,1] that makes [0,1] compact while it does not make (0,1) compact

experimentX said:
but this also covers (0,1), this would also be a finite sub cover of (0,1) making (0,1) compact. What I am asking is an example of sub covers of [0,1] that makes [0,1] compact while it does not make (0,1) compact
It covers (0,1), but it doesn't make (0,1) compact (or [0,1] for that matter...I mean, [0,1] is compact, but it takes more than this to prove it). A set is compact if every open cover has a finite subcover. I only mentioned one open cover that has an open subcover.

Also, every cover of [0,1] will of course also cover (0,1).

So I don't understand what you want. I'm not even sure I understand what you mean by an "example".

The set X= {0, 1, 1/2, 1/3, ..., 1/n,...}, that is, the set containing the sequence 1/n for n any positive integer and 0, is both closed (because it contains the limit of the sequence) and bounded (it obviously lies in [0, 1]) and so, as a subset of the real numbers, is compact. To prove that directly from the definition of "compact" I would do the following:

Suppose $\{U_n\}$ is a open cover for X. Then 0 is in one of those sets- rename it U0. Since U0 is open, there exist some r such that the interval (-r, r) is a subset of U0. There exis some integer N, such that N> 1/r and so 1/N< r. That means that for all $n\ge N$, 1/n< r and so is contained in U0. We then choose a set from the open cover containing 1/n for every n< N, a finite number of sets, and those, together with U0 form a finite sub-cover.

Notice that is essential that 0 be in this set. The set {1, 1/2, ..., 1/n, ...} itself is bounded but not closed and so not compact. For every integer n, Let Un be the interval from (2n+1)/(2n(n+1)) to (2n-1)/(2n(n-1)). The point of that is that (2n+1)/(2n(n+1)) is exactly half way between 1/n and 1/(n+1) while (2n-1)/(2n(n-1)) is exactly half way between 1/n and 1/(n-1). That is, that interval includes ONLY the point 1/n from this set. Since each set in the cover includes only one point of the set, we have to have all of them to cover the set- and we certainly can't reduce to a finite cover.

would this set be finite sub-cover of [0, 1] Sn={(1/n -ε, 2/n + ε): n ≥ 2} ? Choosing n > N ≥ 1/ε where ε > 0 would make it cover [0, 1]?

experimentX said:
would this set be finite sub-cover of [0, 1] Sn={(1/n -ε, 2/n + ε): n ≥ 2} ? Choosing n > N ≥ 1/ε where ε > 0 would make it cover [0, 1]?
The set ##\{(1/n-\varepsilon,2/n+\varepsilon):n\geq 2\}## has infinitely many members. n is just a dummy variable. The set is the same if you use another dummy variable. So it doesn't really make sense to call the set Sn. Also, (this is a bit nitpicky) the notation should specify that n is an integer.

I would just call the set S. It's an open cover of both [0,1] and (0,1). The subset of S that's defined by the condition n≥N contains all but a finite number of members of S. It's another open cover of both [0,1] and (0,1), and it's not finite.

## 1. What is a cover of a set?

A cover of a set is a collection of sets that completely covers all elements of the original set. In other words, every element in the original set is also an element of at least one set in the cover.

## 2. What does it mean for a cover to have finite sub-covers?

A cover having finite sub-covers means that the cover can be broken down into a finite number of smaller covers, each of which still completely covers the original set. This is useful for proving certain properties or theorems about the original set.

## 3. Can you provide an example of a cover having finite sub-covers?

One example would be a cover of the set of real numbers by the intervals (0,1), (1,2), (2,3), and so on. This cover has infinite sub-covers, but if we only consider the intervals (0, n) for n=1, 2, 3..., then we have a finite sub-cover.

## 4. Why is it important for a cover to have finite sub-covers?

Having finite sub-covers is important because it allows us to use techniques such as induction to prove properties about the original set. It also helps to simplify proofs and make them more manageable.

## 5. How is the concept of finite sub-covers related to topology?

In topology, the concept of finite sub-covers is important for understanding compactness. A set is compact if every open cover has a finite sub-cover. This is a fundamental concept in topology and is used in many proofs and theorems.

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