Exercise 1.5.10

Question

\begin{enumerate}[label=(\alph*)] 
  \item Let $C \subseteq[0,1]$ be uncountable. Show that there exists $a \in(0,1)$ such that $C \cap[a, 1]$ is uncountable.
  \item Now let $A$ be the set of all $a \in(0,1)$ such that $C \cap[a, 1]$ is uncountable, and set $\alpha=\sup A$. Is $C \cap[\alpha, 1]$ an uncountable set?
  \item Does the statement in (a) remain true if ``uncountable'' is replaced by ``infinite''?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Suppose $a$ does not exist, then $C \cap [a, 1]$ is countable for all $a \in (0, 1)$ meaning
    $$
    \bigcup_{n=1}^\infty C \cap [1/n, 1] = C \cap [0, 1]
    $$
    Is countable (by \ref{ex:countable_union}), contradicting our assumption that $C \cap [0, 1]$ is uncountable.
  \item If $\alpha = 1$ then $C \cap [\alpha, 1]$ is finite. Now if $\alpha < 1$ we have $C \cap [\alpha+\epsilon, 1]$ countable for $\epsilon>0$ (otherwise the set would be in $A$, and hence $\alpha$ would not be an upper bound). Take
    $$
    \bigcup_{n=1}^\infty C \cap [\alpha+1/n, 1] = C \cap [\alpha, 1]
    $$
    Which is countable by \ref{ex:countable_union}.
  \item No, consider the set $C = \{1/n : n \in \mathbf{N}\}$ it has $C \cap [\alpha, 1]$ finite for every $\alpha$, but $C \cap [0, 1]$ is infinite.
   \end{enumerate}

Exercise 1.5.10

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Exercise 1.5.10

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