Exercise 2.3.3 (Continuous functions on discrete and indiscrete spaces)

Question

Prove the following statements about continuous functions and discrete and indiscrete topological spaces.
\begin{enumerate}[label=(\alph*)]
\item If $X$ is discrete, then every function $f$ from $X$ to a topological space $Y$ is continuous.
\item If $X$ is not discrete, then there is a topological space $Y$ and a function $f: X ightarrow Y$ that is not continuous.
ewline
	extit{Hint:} Let $Y$ be the set $X$ with the discrete topology.
\item If $Y$ is an indiscrete topological space, then every function $f$ from a topological space $X$ to $Y$ is continuous.
\item If $Y$ is not indiscrete, then there is a topological space $X$ and a function $f: X ightarrow Y$ that is not continuous.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item 	extit{If $X$ is discrete, then every function $f$ from $X$ to a topological space $Y$ is continuous.}
ewline
\begin{proof}
Since every subset of $X$ is open, the set $f^{-1}(S)$ must be open for any subset $S$ of $Y$, particularly for every open subset $S$ of $Y$.
\end{proof}

\item 	extit{If $X$ is not discrete, then there is a topological space $Y$ and a function $f: X ightarrow Y$ that is not continuous.}
ewline
\begin{proof}
The trick is to make the topology of $Y$ as fine as possible so that $f^{-1}$ cannot translate $Y$ to $X$ without losing some information. 
Thus, let the range $Y$ be the same set $X$ but equipped with the discrete topology; let $f$ be the (one-to-one) identity function. Then, every point $x$ in the discrete range $X$ is open; but since the domain $X$ is non-discrete, there must be at least one element $x$ which does not constitute an open set. Taking $f^{-1}(\{x\})$ will then map the open set $\{x\}$ in discrete topology of $X$ to the non-open set $\{x\}$ in the non-discrete domain $X$.
\end{proof}

\item 	extit{If $Y$ is an indiscrete topological space, then every function $f$ from a topological space $X$ to $Y$ is continuous.}
ewline
\begin{proof}
The only open subsets of the indiscrete space are the underlying space $Y$ itself and the empty set $\emptyset$. Brute-force case verification shows that $f^{-1}(Y) = X$ and $f^{-1}(\emptyset) = \emptyset$, both open in $X$ by the axioms of topology.
\end{proof}

\item 	extit{If $Y$ is not indiscrete, then there is a topological space $X$ and a function $f: X ightarrow Y$ that is not continuous.}
\begin{proof}
Following the reverse of the strategy described in part (b), we make the domain $X$ as coarse as possible. A natural choice is taking $Y$ itself equipped with the indiscrete topology and looking at the identity transformation. Then there is an open set $S$ of $Y$ distinct from $X$ and $\emptyset$; thus, its inverse $f^{-1}(S) = S$ cannot be open in the indiscrete topology.
\end{proof}

\end{enumerate}

Exercise 2.3.3 (Continuous functions on discrete and indiscrete spaces)

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Exercise 2.3.3 (Continuous functions on discrete and indiscrete spaces)

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