Exercise 2.1.12 (Cocountable topology)

Question

Let $X$ be a set and let $\mathcal{T}$ be the family of subsets $U$ of $X$ such that $X \setminus U$ is
at most countable, together with the empty set $\emptyset$.
\begin{enumerate}[label=(\alph*)]
\item Prove that $\mathcal{T}$ is a topology for $X$.
\item Describe the convergent sequences in X with respect to this topology.
\item Prove that if $X$ is uncountable, then there is a subset $S$ of $X$ whose closure contains points that are not limits of convergent sequences in $S$.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item We use the same line of reasoning as in the proof that the cofinite topology is a topology.
\begin{enumerate}
\item $\emptyset \in \mathcal{T}$ by construction and $X \in \mathcal{T}$ since its complement $X \setminus X = \emptyset$ is trivially at most countable.

\item Let $(A_j)_{j \in J}$ be a family of sets in $\mathcal{T}$. To see whether $\bigcup_{j \in J} A_j \in \mathcal{T}$, consider the cardinality of its complement. By \href{../../terence_tao/analysis_i/exercise_3-4-11}{de Morgan's laws} and by \href{../../terence_tao/analysis_i/exercise_3-6-4}{cardinal arithmetic},
\begin{align*}
\# \left(X \setminus \bigcup_{j \in J} A_j ight)
&= \# \left(\bigcap_{j \in J} X \setminus A_j ight)\
&\leq \# \left(X \setminus A_1 ight).
\end{align*}
Since $A_1 \in \mathcal{T}$, its complement must be at most countable.

\item Let $(A_n)_{n=1}^{N}$ be a finite family of sets in $\mathcal{T}$. We look at the complement of their intersection. Using \href{../../terence_tao/analysis_i/exercise_3-4-11}{de Morgan's laws} and basic \href{../../terence_tao/analysis_i/exercise_3-6-4}{cardinal arithmetic} again, we obtain
\begin{align*}
\# \left(X \setminus \bigcap_{1 \leq n \leq N} A_n ight)
&= \# \left(\bigcap_{1 \leq n \leq N} X \setminus A_n ight)\
&\leq \sum_{1 \leq n \leq N} \# \left(X \setminus A_n ight) \leq \aleph_0.
\end{align*}
In other words, $\bigcap_{1 \leq n \leq N} A_n$ is contained in $\mathcal{T}$.
\end{enumerate}

\item We argue that the only convergent sequences in cocountable topologies are those who eventually become constant.
\begin{description}
\item[$\implies$] Let $\{x_n\}$ be a convergent sequence in $X$, and let $x \in X$ be its limit. Define $U := X \setminus \{x_n : x_n 
eq x\}$. Notice that (1) $U$ is open and (2) that $\{x_n\}$ is eventually contained in $U$, i.e., $\exists N \in \mathbb{N}$ $\forall n \geq N:$ $x_n \in U$. Since $\{x_n\}$ is convergent and since $U$ is an open set containing the limit $x$, the latter assertion means that $\{x_n\}$ becomes $x$ eventually, by construction of $U$.
\item[$\impliedby$] Let $\{x_n\}$ be an eventually constant sequence, i.e., there is a limit $x$ and a bound $N \in \mathbb{N}$ such that $\forall n \geq N:$ $x_n = x$. Then any open neighbourhood of $x$ will contain $x_n$ eventually, proving trivially that $\{x_n\}$ converges to $x$.
\end{description}

\item The only closed sets in the cocountable topology are the at most countable sets, the empty set and the $X$ itself. Thus, if $S$ is any proper uncountable set, then the only closed set big enough to contain it, and thus is its closure, is $X$.
\end{enumerate}

Exercise 2.1.12 (Cocountable topology)

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Exercise 2.1.12 (Cocountable topology)

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