The distance definition of continuity is local, but many metric space arguments become easier when continuity is translated into set language. The focus keyword open set continuity refers to the fact that continuity is characterized by inverse images of open sets. A common mistake is to think that a continuous mapping must send open sets to open sets. The correct general statement is that inverse images of open sets are open, and equivalently inverse images of closed sets are closed. :::lemma Let $f:X\to Y$ be a function, let $A\subseteq X$, and let $B\subseteq Y$. Then $f(A)\subseteq B$ if and only if $A\subseteq f^{-1}(B)$. ::: :::proof Given that $f:X\to Y$, $A\subseteq X$, and $B\subseteq Y$. To prove the equivalence. Suppose first that $f(A)\subseteq B$. If $x\in A$, then $f(x)\in f(A)\subseteq B$. Therefore $x\in f^{-1}(B)$, and $A\subseteq f^{-1}(B)$. Conversely, suppose $A\subseteq f^{-1}(B)$. If $y\in f(A)$, then $y=f(x)$ for some $x\in A$. Since $x\in f^{-1}(B)$, $f(x)\in B$. Therefore $y\in B$. Hence, $f(A)\subseteq B$. $\square$ ::: :::theorem[Neighbourhood Criterion for Continuity] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let $f:X\to Y$. Then $f$ is continuous at $a\in X$ if and only if for every $\epsilon>0$, there exists $\delta>0$ such that $$\begin{equation*} S(a,\delta)\subseteq f^{-1}(S(f(a),\epsilon)). \end{equation*}$$ ::: :::proof Given that $f:X\to Y$ and $a\in X$. To prove the equivalence. By definition, $f$ is continuous at $a$ if and only if for every $\epsilon>0$, there exists $\delta>0$ such that $d_Y(f(x),f(a))<\epsilon$ whenever $d_X(x,a)<\delta$. This means precisely that $f(S(a,\delta))\subseteq S(f(a),\epsilon)$. By the inverse image inclusion lemma, this is equivalent to $S(a,\delta)\subseteq f^{-1}(S(f(a),\epsilon))$. $\square$ ::: :::theorem[Open Set Criterion for Continuity] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let $f:X\to Y$. Then $f$ is continuous on $X$ if and only if $f^{-1}(G)$ is open in $X$ for every open set $G$ in $Y$. ::: :::proof Given that $f:X\to Y$. To prove the open set continuity criterion. Suppose first that $f$ is continuous on $X$. Let $G$ be open in $Y$ and let $a\in f^{-1}(G)$. Since $f(a)\in G$, there exists $\epsilon>0$ such that $S(f(a),\epsilon)\subseteq G$. By continuity at $a$, choose $\delta>0$ such that $S(a,\delta)\subseteq f^{-1}(S(f(a),\epsilon))$. Hence $S(a,\delta)\subseteq f^{-1}(G)$, so $f^{-1}(G)$ is open. Conversely, suppose inverse images of open sets are open. Let $a\in X$ and $\epsilon>0$. Since $S(f(a),\epsilon)$ is open in $Y$, its inverse image is open in $X$ and contains $a$. Hence, for some $\delta>0$, $S(a,\delta)\subseteq f^{-1}(S(f(a),\epsilon))$. Thus $f$ is continuous at $a$. Since $a$ was arbitrary, $f$ is continuous on $X$. $\square$ ::: :::theorem[Closed Set Criterion for Continuity] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let $f:X\to Y$. Then $f$ is continuous on $X$ if and only if $f^{-1}(F)$ is closed in $X$ for every closed set $F$ in $Y$. ::: :::proof Given that $f:X\to Y$. To prove the closed set criterion using complements. If $f$ is continuous and $F$ is closed in $Y$, then $Y\setminus F$ is open. Hence $f^{-1}(Y\setminus F)=X\setminus f^{-1}(F)$ is open. Therefore $f^{-1}(F)$ is closed. Conversely, suppose inverse images of closed sets are closed. Let $G$ be open in $Y$. Then $Y\setminus G$ is closed, so $f^{-1}(Y\setminus G)=X\setminus f^{-1}(G)$ is closed. Therefore $f^{-1}(G)$ is open. Hence, by the open set criterion, $f$ is continuous. $\square$ ::: :::solved-problem Let $f:X_1\to X_2$ be a mapping. Prove that $f$ is continuous if and only if $$\begin{equation*} f^{-1}(Y_2^\circ)\subseteq (f^{-1}(Y_2))^\circ \end{equation*}$$ for every $Y_2\subseteq X_2$. ::: :::solution If $f$ is continuous, then $f^{-1}(Y_2^\circ)$ is open and is contained in $f^{-1}(Y_2)$. Hence it is contained in the interior $(f^{-1}(Y_2))^\circ$. Conversely, if the condition holds for all $Y_2$, take $Y_2=G$ open. Then $G^\circ=G$, so $f^{-1}(G)\subseteq(f^{-1}(G))^\circ$. Since the reverse inclusion always holds, $f^{-1}(G)$ is open. Hence, $f$ is continuous. ::: :::solved-problem Let $\mathcal B$ be a base for the open sets of $Y$. Prove that $f:X\to Y$ is continuous if and only if $f^{-1}(B)$ is open in $X$ for every $B\in\mathcal B$. ::: :::solution If $f$ is continuous, inverse images of base elements are open. Conversely, suppose $f^{-1}(B)$ is open for every $B\in\mathcal B$. If $G$ is open in $Y$, then $G=\bigcup_\alpha B_\alpha$ for suitable $B_\alpha\in\mathcal B$. Therefore $f^{-1}(G)=\bigcup_\alpha f^{-1}(B_\alpha)$, which is open in $X$. Hence, $f$ is continuous. ::: :::faq[Frequently Asked Questions] Q: Does a continuous function always map open sets to open sets? A: No. Continuity concerns inverse images of open sets. Q: Why are closed sets also valid tests? A: Closed sets are complements of open sets, and inverse images commute with complements. :::