# How to show $A$ is compact in $Bbb{R}$ with standard topology?

Mathematics Asked by Happy Turn on December 28, 2020

Let $$A = {0} cup { frac{1}{n}| n in Bbb{N}} subseteq Bbb{R}$$. I am struggling with showing whether $$A$$ is compact or not. I do not even know how to start.
Is $$u={0} cup left{left(frac{1}{n+1}, frac{1}{n-1}right)| n in Bbb{N}right} cap {(1/2, 2)}$$ an open cover of A with no finite subcover? If so, what is the next step?

If you can use the Heine-Borel theorem is easy to see that the set of acumulation points $$A' ={0} subseteq A$$ therefore $$A$$ is closed and observe that $$forall x in A$$ we have that $$|x - 1| leq 1$$ hence $$A$$ is bounded. So $$A$$ is a closed and bounded set, by Heine-Borel theorem we can conclude that $$A$$ is compact.

Answered by Iñaki Mendieta on December 28, 2020

The given set is $$A={0} cup big{ frac1n : n in mathbb{N} big}$$

To show that $$A$$ is compact set in $$mathbb{R}$$, we need to show that any open cover of $$A$$ has a finite subcover.

Consider any open cover of $$A= bigcup_{i in I} U_i$$. Then since $$0 in A$$, then for some $$i_0 in I, 0 in U_{i_0}$$. since $$U_{i_0}$$ is an open set. Hence there exists $$epsilon>0$$ such that $$(-epsilon,epsilon ) subset U_{i_0}$$.

Archimedean Property

For any $$epsilon >0$$, there exists $$n in mathbb{N}$$ such that $$frac1n < epsilon$$

Using the Archimedean Property, choose an $$n_0$$ such that $$frac{1}{n_0} < epsilon$$. Thus for $$n ge n_0$$ we have $$frac1n le frac{1}{n_0} < epsilon$$. Thus all but finite number of elements of $$A$$ are contained in $$(-epsilon,epsilon)$$. Thus they are all contained in $$U_{i_0}$$. Now choose a finite number of open sets from $$U_i$$ to create a finite subcover of $$A$$

This is a nice exercice for practicing with definitions.

So $$X$$ is compact if EVERY open cover has a finite sub cover. In order tho prove that $$A$$ is compact, you then have to start with a generic open cover of $$A$$ (you cannot choose the cover because the proof has to work for every cover). Say $$U={U_i}_{iin I}$$ is an open cover. Then there is at least one of the $$U_i$$, say $$U_{i_0}$$ which contains $$0$$. Since $$U$$ is open for the standard topology, and contains $$0$$, it must also contain an interval $$(-epsilon,epsilon)$$, which contains all but finitely many points of $$A$$. (Precisely, it contains all points $$1/n$$ for wich $$n>1/epsilon$$). For any of those finitely many points $$1/n$$ left outside $$U_{i_0}$$ there is an $$U_{i_n}$$ containing it. So the family formed by $$U_0$$ and those finitely many $$U_{i_n}$$ is a finite cover of $$A$$.

Answered by user126154 on December 28, 2020

Hint : Compacts in $$mathbb{R}$$ are precisely the closed and bounded sets. $$A$$ is clearly bounded. What about $$L=lim_{n}frac{1}{n}$$ ? Does $$Lin A$$ ?

Answered by JPA on December 28, 2020