Exercise 2.1.4 (Closure in discrete, indiscrete and cofinite topologies)

Question

Let $S$ be a subset of $X$. Describe the closure of $S$ when
\begin{enumerate}[label=(\roman*)]
\item $X$ has the discrete topology
\item $X$ has the indiscrete topology
\item $X$ has the cofinite topology.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(\roman*)]
\item Notice that every set in the discrete topology is both open and closed since its complement is a union of (open) singletons. Thus, the closure of $S$ in the discrete topology is $S$ itself.

\item Notice that the only closed sets in the indiscrete topology are $\emptyset = X \setminus X$ and $X = X \setminus \emptyset$. By Theorem 1.3, since $S \subseteq \bar{S}$, we only have two possibilities: the closure of $S$ is $\emptyset$ if $S = \emptyset$, or the closure of $S$ is $X$, if $S$ is larger than $\emptyset$.

\item Notice that the only closed sets in the cofinite topology are finite subsets of $X$ plus the indiscrete sets $\emptyset$ and $X$. Thus, if $S$ is finite, it is already closed, whereas if $S$ is infinite, then $S \subseteq \bar{S} = X$ (other closed sets are simply too small to contain $S$).
\end{enumerate}

Exercise 2.1.4 (Closure in discrete, indiscrete and cofinite topologies)

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Exercise 2.1.4 (Closure in discrete, indiscrete and cofinite topologies)

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