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I recently began self-studying real analysis on a whim and I have run into some trouble understanding the idea of compactness. I have no accessible living sources near me and I have yet to find a source that has explained the matter clearly, so I shall turn here for help.

From my primitive understanding of classifying a set as compact, I understand that every open cover of said set has to have a finite sub-cover that still covers that set. So, in order to declare a set as non-compact, it must be shown that there exists some open cover that has no finite sub-cover. Take, for instance, the closed interval [0, 1]: from what I've read, this is a compact set. A possible open cover for this set would simply be the ball centered at x = 1/2 with a radius greater than 1/2. This is an open cover that has no finite sub-cover - it is a single set so it has no sub-covers! How can [0, 1] be compact if it has an open cover that has no finite sub-cover?

I realize that the previous example might be special since it consists of a singular set as the open cover, and maybe that is what tripped me up; but, here is another example: consider the finite set {1, 2, 3}. From what I've read, all finite sets are compact, so therefore this set is compact. But take the open cover {B_{1}, B_{2}, B_{3}}, where B_{i}is an open ball with radius 0.1 centered at x = i. If you consider any sub-cover of this open cover, you must remove one of the sets in the open cover. Once you do, your finite sub-cover does not cover the entire set! Again, how can any finite set be compact if it can be shown that there exists an open cover that does not have a finite sub-cover?

I'm either missing something or drawing some wildly incorrect conclusions - if anyone could offer any insight to clear up my muddled thinking I would greatly appreciate it.

Thanks!

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# Question About Compact Sets

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