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

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?

Mathematics Asked by Happy Turn on December 28, 2020

4 Answers

4 Answers

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 $A$

Answered by GraduateStudent on December 28, 2020

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

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