Compactness of a Subspace of $\mathbb{R}$

Question

I am trying to determine if the following set is compact:

$$A=\bigg\{\frac{1}{n}: n\in \mathbb{N}, n>0\bigg\}\subset\mathbb{R}$$

When I consider the subspace topology induced by the standard topology of $\mathbb{R}$

then I figure that the results are the discrete topology on $A$ since for any $Y\subset A$ can find an interval $(a,b)$ such that $A\cap(a,b)=Y$

If I consider an open covering of elements of $\mathscr{T}_A$ of $A$ consisting of every element in $A$ then no finite sub covering exists so cannot be compact.

What if I consider the set $$A^{'}=\{0\}\cup\bigg\{\frac{1}{n}: n\in \mathbb{N}, n>0\bigg\}\subset\mathbb{R}$$ is this set considered compact. Im guessing it is compact since it contains only sequences that converge to $0$ which is now included in the set. Consider an arbitrary open covering $\{U_\alpha: U_\alpha\in \mathscr{T}_{A^{'}}, \alpha\in I\}$of $A$ If I consider a sequence $\{X\}_n^{\infty}$ with each $X_n=\frac{1}{n}\in A$ where $X_n\rightarrow 0$ then consider an open nbhd of $U_0$ of 0 then for $n\ge N$ , $X_n\in U_0$ then consider the open neighborhoods $U_1,U_2, ...,U_{N-1}$ of $X_1,X_2, ...,X_{N-1}$ then $U_0,U_1,U_2, ...,U_{N-1}$ is a finite subcovering of $A$. Is my understanding correct? Outside using the idea of the set containing all its limit points intuitively why is $A$ not compact but $A^{'}$ is?

Answer 1

Your understanding is correct, but your formulation could be improved.

Let $\mathcal U = \{U_\alpha\}_{\alpha \in I}$ by an open cover of $A'$. Then there exists $\alpha_0 \in I$ such that $0 \in U_{\alpha_0}$. Write $U_{\alpha_0} = V \cap A'$ with an open $V \subset \mathbb R$ (subspace topology!). There exists $\epsilon > 0$ such that $0 \in (-\epsilon,\epsilon) \subset V$. Choose $N$ such $1/N <\epsilon$. Then $U_{\alpha_0}$ contains all points $1/n$ with $n \ge N$. Choose $\alpha_n \in I$ such that $1/n \in U_{\alpha_n}$ for $n = 1,\ldots,N-1$. Then $U_{\alpha_0},U_{\alpha_1},\ldots,U_{\alpha_{N-1}}$ is a finite subcover of $\mathcal U$.

Answer 2

Munkres explains it in Example 2, section 26 on p-164 of his text: Topology 2E as follows.

EXAMPLE 2. The following subspace of $\mathbb{R}$ is compact: $$X=\{0\}∪\{1/ n\; |\; n∈\mathbb{Z}_+ \}. $$ Given an open covering $\mathcal{A}$ of $X$, there is an element $U$ of $\mathcal{A}$ containing $0$. The set $U$ contains all but finitely many of the points $1/ n$; choose, for each point of $X$ not in $U$, an element of $\mathcal{A}$ containing it.

The collection consisting of these elements of $\mathcal{A}$, along with the element $U$, is a finite subcollection of $\mathcal{A}$ that covers $X$. \qed

The following is my discussion for clarity.

Suppose there exists $N$ such that $1/ n_i∉U\; ∀ i≤N$, and suppose $\mathcal{A}_i (i=1,…,N)$ is the element of $\mathcal{A}$ that contains $1/ n_i $. Then the collection $U∪⋃_{i=1}^N\; \mathcal{A}_i $ is a finite subcollection of $\mathcal{A}$ that covers $X$.