Consider the sets [1,3) and  `NN`.Show that each subset of `RR` is not compact by describing an open cover for it that has no finite subcover.

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

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I edited your question because you should make separate posts for separate questions. However, the idea behind all of them is so similar that if I do the first two you might see how to do the others.

Let `S_n=(0,3-1/n)`   for `n=1,2,3...`

As `n` gets larger, `3-1/n` gets arbitrarily close to 3 (but never equals 3), so `uuu S_n=(0,3),` which covers `[1,3).` However, if we take a finite subset of this cover, let `N` be the largest number such that `S_N` is in this finite subset. No matter how large `N` is, `3-1/N<3-1/(2N)<3,` so there will be elements of `[1,3)` missing, in particular `3-1/(2N).` Thus we have found a cover such that no finite subset covers `[1,3),` so `[1,3)` isn't compact.

Now for the set `NN` , simply let `S_n=(-1,n),` so again

`uuuS_n=(-1,oo),` covering `NN.` But again, if `N` is the largest number such that `S_N` is in a finite subset of this union, then clearly any natural number greater than or equal to `N` isn't covered, so `NN` isn't compact.