Exercise 3.3.11

Question

Consider each of the sets listed in Exercise 3.3.2. For each one that is not compact, find an open cover for which there is no finite subcover.

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $\mathbf{N}$ and $\{V_1(n) : n \in \mathbf{N}\}$ has no finite subcover since each $V_1(n)$ covers exactly one $n \in \mathbf{N}$, meaning there are no subcovers at all!
  \item $\mathbf{Q} \cap [0,1]$: Choose some $y \in \mathbf{Q}^c \cap [0, 1]$, for example $y = \sqrt{2}/2$. Consider the open cover $\{(-1, y)\} \cup \{(y + 1/n, 2) : n \in \mathbf{N}\}$. Since $\mathbf{Q}$ is dense in $\mathbf{R}$, for any finite subcover there must be some rational number $q \in (y, y + 1/n)$ where $n$ is finite.

\item The Cantor is compact
  \item $K = \{1 + 1/2^2 + 1/3^2 + \dots + 1/n^2 : n \in \mathbf N\}$ and $\left\{V_{\frac 12 |x-L|}(x) : x \in K\right\}$ for $L = \pi^2/6$ since any finite cover $\left\{V_{\frac 12 |x_1-L|}(x_1), \dots, V_{\frac 12 |x_n-L|}(x_n)\right\}$, letting$\epsilon = \min\{\frac 12|x_i - L||\}$ will make $V_{\epsilon}(L)$ not in the finite cover, meaning there exists an $x \in V_\epsilon(L)$ with $x \in K$ (since $K$ gets arbitrarily close to $L$) but $x$ not in the finite cover.
  \item \{1, 1/2, 2/3, 3/4, 4/5, \dots\} is compact
   \end{enumerate}

Exercise 3.3.11

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Exercise 3.3.11

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