# Compact sets and coverings

## Homework Statement

Show that S = [0,1) is not compact by giving an closed cover of S that has no finite subcover.

## The Attempt at a Solution

I know that S is not compact because it is [STRIKE]an open[/STRIKE] not a closed set even though it is bounded.

But I am completely lost on the open cover part.

I understand an open cover is a union of open sets where S is a subset of the union...

But I appear to missing something very fundamental. If I picked (-1, 2) for the cover that is an open set and S is a subset of it's "trivial" union. Why doesn't that work?

<edit> OK, I understand it needs to work for every cover, that's why,
but is (-1,2) a cover? <end edit>

If instead I had [0,1], that is closed and bounded so it is compact.
What sort of sets would go into a cover for it?

Does anyone know some extremely elementary references for this?

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Dick
Homework Helper
(-1,2) is a cover. And it has a finite subcover. No problems there. Really big hint. What about (-1,1-1/n) for all n. Can you tell me why i) it is a cover and ii) why it has no finite subcover? And S ISN'T open.

Thanks, I believe I understand.

[STRIKE]I said it backwards ... S is closed and bounded on the Reals so it is compact.[/STRIKE]

The union of the open intervals U = (-1, 1-1/n) for all positive n, covers S.

That is, S is a subset of U.

But if one takes a finite subset of u, then the finite union F is the interval (-1, 1-1/f) where f is the last interval in F.

Since 1/f < 1 for every positive integer f S is not a subset of F.

Does that sound good?

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Dick