Exercise 2.4.3 (Continuous functions and bases)

Question

Let $X$ and $Y$ be topological spaces and let $\mathscr{B}$ be a base of open sets for $Y$. Show that a function $f: X \rightarrow Y$ is continuous if and only if $f^{-1}(U)$ is an open subset of $X$ for every $U \in \mathscr{B}$.

Elu Thingol · Accepted Answer

\begin{proof}
If $f$ is continuous, then $f^{-1}(U)$ is open in $X$ for any open $U$ in $Y$; in particular, for any open $U \in \mathscr{B}$. Conversely, suppose that $f^{-1}(U)$ is open for any $U \in \mathscr{B}$. Let $V$ be an open set in $Y$. We can write $V$ as the union of the base sets, i.e., $V=\bigcup_{U \in B} U$ for some $B \subseteq \mathscr{B}$. We then have
$$f^{-1}\left( V \right) = f^{-1}\left( \bigcup_{U \in B} U \right)
= \bigcup_{U \in B} f^{-1}\left( U \right),$$
which is open as the union of open sets. Since our choice of an open set $V$ was arbitrary, $f$ must be continuous.
\end{proof}

Exercise 2.4.3 (Continuous functions and bases)

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Exercise 2.4.3 (Continuous functions and bases)

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