Proof of Composition of Continuous Euclidean Maps

Proof

Let $\varepsilon>0$ . Since $g$ is continuous at $f(a)$ , there exists $\eta>0$ such that whenever $y\in F$ and

\sum_{j=1}^m (y_j-f_j(a))^2<\eta^2,

one has

\sum_{\alpha=1}^p \bigl(g_\alpha(y)-g_\alpha(f(a))\bigr)^2<\varepsilon^2.

Since $f$ is continuous at $a$ , there exists $\delta>0$ such that whenever $x\in E$ and

\sum_{i=1}^n (x_i-a_i)^2<\delta^2,

one has

\sum_{j=1}^m \bigl(f_j(x)-f_j(a)\bigr)^2<\eta^2.

Now let $x\in E$ satisfy

\sum_{i=1}^n (x_i-a_i)^2<\delta^2.

Then $f(x)\in F$ and

\sum_{j=1}^m \bigl(f_j(x)-f_j(a)\bigr)^2<\eta^2.

Applying the choice of $\eta$ with $y=f(x)$ , we obtain

\sum_{\alpha=1}^p \bigl(g_\alpha(f(x))-g_\alpha(f(a))\bigr)^2<\varepsilon^2.

Since $(g\circ f)(x)=g(f(x))$ and $(g\circ f)(a)=g(f(a))$ , this is exactly the continuity condition for $g\circ f$ at $a$ in the sense of Continuity at a Point for Maps Between Euclidean Spaces.

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