## September 2001 #5

Problem Statement: Use the open cover definition of compact to show that $E=\{0\}\cup\{1,1/2,1/3,1/4,\dots\}$ is a compact subset of $\mathbb{R}$.

Proof: Let $S$ be an open cover of $E$, $S=\bigcup_{j\in\mathcal{J}}S_j$ for some indexing set $\mathcal{J}$. If $S$ is a finite cover then we’re done, so suppose that $S$ is not a finite open cover of $E$.

Since $0\in E$ there is at least one $S_k\in S$ such that $0\in S_k$. Since $\{1/n\}$ converges to $0$ there is an $N\in \mathbb{N}$ such that $1/n\in S_k$ for every $n>N$. Furthermore, since $S$ is an open cover of $E$ there is at least one $S_j\in S$ such that $1\in S_j$. Pick one of these and call it $S_1$. We may do this for $1/2, 1/3,\dots,1/N$ and find $S_2,S_3,\dots,S_N$.

Let $T=S_k\cup S_1\cup S_2\cup\dots\cup S_N$ which is contained in $S$. We claim that $T$ is a finite open cover of $E$. First note that each of the $S_n$ are open and so their union is open. Let $x=\dfrac{1}{n}\in E$. If $n>N$ then $x\in S_k$, otherwise, $x\in S_n$. Either way, $x\in T$. Thus, $T$ is a finite open cover of $E$.

Since such a $T$ can be constructed from any open cover $S$ of $E$ it follows that every open cover of $E$ in $\mathbb{R}$ has a finite subcover in $\mathbb{R}$, thus, $E$ is compact in $\mathbb{R}$.

$\Box$

Reflection: The most important part of this proof rests in the fact that $0$ is the limit of the sequence $\{\dfrac{1}{n}\}$. This allows us to cover infinitely many of the points in $E$ with a single open set $S_k$ and leave only finitely many left to cover. After that, the rest is a snap 🙂

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### 4 Responses to September 2001 #5

1. j2kun says:

Ooooooooh snap!