# Proof

• Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces.
• Let $$X$$ be compact.
• Let $$f:X\mapsto Y$$ be a continuous function.
• It is to be shown that $$f:X\mapsto Y$$ is uniformly continuous, i.e. for every $\epsilon > 0$ there is a $\delta > 0$ such that $d_Y(f(x),f(y)) < \epsilon$ for all $x,y\in X$ with $d_X(x,y) < \delta.$
• Choose a fixed $\epsilon > 0$.
• According to the definition of continuity using open sets, $$f$$ is continuous at a point $x\in X$, if and only if for any neighborhood $V\subset Y$ of $f(x)$ there exists a neighborhood $U$ of $x$ such that $f(U)\subset V.$ In the following argument, for every $x\in X$ we will choose $V$ to be a neighborhood of $f(x)$ with a diameter $\operatorname{diam}(V) < \epsilon / 2$ and construct $U$ to be an open ball $B(x, r(x))$ with an appropriate radius $r(x) > 0$ depending on $$x$$ and $\epsilon$.
• Since $f$ is continuous at any point $x\in X$, for every $x\in X$ there is a radius $r(x) > 0$ such that $$d_Y(f(x),f(y)) < \frac \epsilon2$$ for all $$y\in B(x,r(x)).$$
• Note that $$\bigcup_{x\in X} B\left(x,\frac {r(x)}2 \right)$$ is an open cover of $X$.
• Since $X$ is compact, there exist finitely many $$x_1,\ldots,x_n\in X$$ with $$X\subset \bigcup_{k=1}^n B\left(x_k,\frac {r(x_k)}2 \right).$$
• Set the minimum $\delta:=\frac 12\min(r(x_1),\ldots,r(x_n))$.
• For this $\delta$, we have that if we take $x,y\in X$ with $d_X(x,y) < \delta$, we can find an index $k\in \{1, \ldots,n\}$ with $x,y\in B(x_k,r(x_k)/2)$.
• For these $$x,y\in X$$ with $d_X(x,y) < \delta$ we have that $$d_Y(f(x),f(y))\le d_Y(f(x),f(x_k)) + d_Y(f(x_k),f(y))=\frac \epsilon2 + \frac \epsilon2 = \epsilon.$$

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### References

#### Bibliography

1. Forster Otto: "Analysis 2, Differentialrechnung im $$\mathbb R^n$$, Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984