Compactness of $\{0,1\} \cup (\bigcup_{n \in \mathbb{N}} \frac{n}{n+1})$

Question

Compactness of $\{0,1\} \cup (\bigcup_{n \in \mathbb{N}} \frac{n}{n+1})$

Let $A = \{0,1\} \cup (\bigcup_{n \in \mathbb{N}} \frac{n}{n+1})$

The definition of compactness that I'm working with is that every open cover of a set $X$ has a finite subcover.

First off, we have to show this for every open cover, if there are infinitely many then this won't be possible. Secondly, what do open sets in $A$ look like? (Subspace topology with $A \subseteq \mathbb{R}$). It looks like open sets in $A$ are just every singleton except for $\{1\}$ because each element can be intersected by some open neighborhood about that element in $\mathbb{R}$.

Answer 1

Hint If you pick any open cover of $A$, then one of the open sets, lets call it $U$ contains $1 \in A$. Show that there exists some $N$ so that $$\frac{n}{n+1} \in U \forall n>N$$

To find the finite subcover, use this $U$ and any finite cover of the finitely many remaining terms.

Answer 2

Let $O=\{A_\lambda\mid\lambda\in\Lambda\}$ be an open cover of $\{0,1\}\cup\left\{\frac n{n+1}\,\middle|\,n\in\mathbb N\right\}$. Then $1\in A_{\lambda_0}$ for some $\lambda_0\in\Lambda$. Since $\lim_{n\to\infty}\frac n{n+1}=1$, and since $A_{\lambda_0}$ is open, $A_{\lambda_0}$ has every number of the form $\frac n{n+1}$ in it, with finitely many exceptions. Let $\frac{n_1}{n_1+1},\frac{n_2}{n_2+1},\ldots,\frac{n_k}{n_k+1}$ be those exceptions. Each $\frac{n_j}{n_j+1}$ belongs to some $A_{\lambda_j}$. Take $\lambda_{k+1}\in\Lambda$ such that $0\in A_{\lambda_{k+1}}$. So$$\{0,1\}\cup\left\{\frac n{n+1}\,\middle|\,n\in\mathbb N\right\}\subset\bigcup_{j=0}^{k+1}A_{\lambda_j}.$$

Answer 3

I guess that you're talking about the set $$ A=\{0,1\}\cup\Bigl\{\frac{n}{n+1}:n\in\mathbb{N}\Bigr\} $$

Let $(a_n)_{n\in\mathbb{N}}$ be a convergent sequence, with limit $l$. Then the set $$ A=\{a_n:n\in\mathbb{N}\}\cup\{l\} $$ is compact. Indeed, let $(U_\lambda)_{\lambda\in I}$ be an open cover. Then there is $\lambda_0$ such that $l\in U_{\lambda_0}$.

Since the sequence is convergent, there exists $k$ such that, for $n>k$, $a_n\in U_{\lambda_0}$.

Now pick a finite number of sets in the family that contain $a_1,a_2,\dots,a_k$.

Adding a finite number of points to a compact set obviously yields a compact set.