The open and closed set criteria allow continuity arguments to be made structurally. The focus keyword composition of continuous mappings describes the first and most important permanence property: composing continuous maps gives a continuous map. Closure tests also show how continuous functions behave with closures of sets and inverse images. These tools are frequently used in functional analysis, where complicated mappings are built by combining simpler mappings. :::theorem[Composition of Continuous Mappings] Let $(X,d_X)$, $(Y,d_Y)$, and $(Z,d_Z)$ be metric spaces. If $f:X\to Y$ and $g:Y\to Z$ are continuous, then $g\circ f:X\to Z$ is continuous. ::: :::proof Given that $f:X\to Y$ and $g:Y\to Z$ are continuous. To prove that $g\circ f$ is continuous. Let $G$ be an open subset of $Z$. Since $g$ is continuous, $g^{-1}(G)$ is open in $Y$. Since $f$ is continuous, $f^{-1}(g^{-1}(G))$ is open in $X$. But $f^{-1}(g^{-1}(G))=(g\circ f)^{-1}(G)$. Therefore inverse images of open sets under $g\circ f$ are open. Hence, $g\circ f$ is continuous. $\square$ ::: :::theorem[Closure Characterization of Continuity] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let $f:X\to Y$. The following statements are equivalent: (i) $f$ is continuous on $X$. (ii) For every subset $B\subseteq Y$, $\overline{f^{-1}(B)}\subseteq f^{-1}(\overline B)$. (iii) For every subset $A\subseteq X$, $f(\overline A)\subseteq \overline{f(A)}$. ::: :::proof Given that $f:X\to Y$. To prove the equivalence. If $f$ is continuous and $B\subseteq Y$, then $f^{-1}(\overline B)$ is closed and contains $f^{-1}(B)$. Hence $\overline{f^{-1}(B)}\subseteq f^{-1}(\overline B)$. Now assume the second condition. Let $A\subseteq X$ and put $B=f(A)$. Since $A\subseteq f^{-1}(f(A))$, we get $\overline A\subseteq \overline{f^{-1}(f(A))}\subseteq f^{-1}(\overline{f(A)})$. Therefore $f(\overline A)\subseteq\overline{f(A)}$. Finally assume the third condition. If $F$ is closed in $Y$ and $A=f^{-1}(F)$, then $f(\overline A)\subseteq\overline{f(A)}\subseteq F$. Hence $\overline A\subseteq A$, so $A$ is closed. Thus inverse images of closed sets are closed, and $f$ is continuous. $\square$ ::: :::solved-problem Let $Y_1,Y_2\subseteq X$ satisfy $Y_1\cup Y_2=X$. Suppose $f_1:Y_1\to Z$ and $f_2:Y_2\to Z$ are continuous and agree on $Y_1\cap Y_2$. Show that the pasted map is continuous if $Y_1,Y_2$ are both open or both closed, but need not be continuous in general. ::: :::solution For a counterexample, take $Y_1=\mathbb Q$, $Y_2=\mathbb R\setminus\mathbb Q$, $f_1=1$, and $f_2=0$. The agreement condition is vacuous, but the pasted Dirichlet-type function is discontinuous everywhere. If $Y_1,Y_2$ are open, then inverse images of open sets under the pasted map are unions of open sets in the pieces, hence open in $X$. If $Y_1,Y_2$ are closed, use inverse images of closed sets instead. Hence, in either case the pasted map is continuous. ::: :::theorem[Continuous Image of a Separable Metric Space] Let $(X,d)$ be a separable metric space, and let $f:X\to Y$ be continuous. Then $f(X)$ is separable. ::: :::proof Given that $X$ is separable. To prove that $f(X)$ is separable. Let $D$ be a countable dense subset of $X$. We claim that $f(D)$ is dense in $f(X)$. If $y\in f(X)$, then $y=f(x)$ for some $x\in X$. Since $D$ is dense, choose $x_n\in D$ with $x_n\to x$. By continuity, $f(x_n)\to f(x)=y$. Thus $f(D)$ is dense in $f(X)$. Since $f(D)$ is countable, $f(X)$ is separable. $\square$ ::: :::exercise 1. Prove composition using sequences. 2. Give a continuous image of a closed set that is not closed. 3. Explain why the pasting theorem needs hypotheses on the pieces. ::: :::answer 1. If $x_n\to x$, then $f(x_n)\to f(x)$ and $g(f(x_n))\to g(f(x))$. 2. $x\mapsto 1/(1+x^2)$ sends $\mathbb R$ to $(0,1]$. 3. Without open or closed pieces, inverse images may fail to be open or closed in the full space. ::: :::faq[Frequently Asked Questions] Q: Why are closure tests useful? A: They express continuity by the behaviour of closures and limit points. Q: Is a continuous image of a separable space separable? A: Yes, the image of a countable dense subset is countable and dense in the image. :::