📂Topology

Uniform Continuity Theorem

Definition

For metric spaces $(X, d)$ and $(Y, d’)$, let $f : X \to Y$. If for every $\varepsilon > 0$ and $x_{1}, x_{2} \in X$ there exists a $\delta > 0$ satisfying $$ d(x_{1}, x_{2}) < \delta \implies d’( f( x_{1} ) , f( x_{2} ) ) < \varepsilon $$ then $f$ is called uniformly Continuous.

Explanation

Just as the concept of continuity learned in analysis was generalized in topology, uniform continuity can likewise be generalized in topology. However, the point to be careful about here is that, unlike continuity, it is not defined on all topological spaces but is discussed only on metric spaces.

The following theorem is also a generalization of a fact already well known in analysis.

Theorem ¹

If $(X,d)$ is a compact metric space, $(Y,d’)$ is a metric space, and $f : X \to Y$ is a continuous function, then $f$ is uniformly continuous.

Proof

Suppose $\varepsilon > 0$ is given. Since $f$ is continuous, for each $x \in X$ there exists a $\delta_{x} > 0$ satisfying $$ d (x,y) < \delta_{x} \implies d’ \left( f(x) , f(y) \right) < {{ \varepsilon } \over { 2 }} \qquad , \forall y \in X $$ For such $\delta_{x}$, the collection $\left\{ B_{d} \left( x , {{ \delta_{x} } \over { 2 }} \right) : x \in X \right\}$ is an open cover of $X$, and since $X$ is compact, there exists a finite set $\left\{ x_{1} , \cdots , x_{n} \right\} \in X$ satisfying $$ X = \bigcup_{i=1}^{n} B_{d} \left( x_{i} , {{ \delta_{x_{i}} } \over { 2 }} \right) $$ Therefore, the existence of $\delta > 0$ can be guaranteed as $$ \delta := \min \left\{ {{ \delta_{x_{1}} } \over { 2 }} , \cdots , {{ \delta_{x_{n}} } \over { 2 }} \right\} $$ Since $ X = \bigcup_{i=1}^{n} B_{d} \left( x_{i} , {{ \delta_{x_{i}} } \over { 2 }} \right)$, any $x \in X$ must satisfy, for some $1 \le j \le n$, $$ x \in B_{d} \left( x_{j} , {{ \delta_{x_{j}} } \over { 2 }} \right) $$ Hence $$ d \left( x_{j} , y \right) \le d \left( x_{j} , x \right) + d (x,y) < {{ \delta_{x_{j}} } \over { 2 }} + \delta \le \delta_{x_{j}} \implies d’ \left( f(x_{j}) , f(y) \right) < {{ \varepsilon } \over { 2 }} \\ d \left( x_{j} , x \right) \le {{ \delta_{x_{j}} } \over { 2 }} < \delta_{x_{j}} \implies d’ \left( f(x_{j}) ,f( x) \right) < {{ \varepsilon } \over { 2 }} $$ so that whenever $d(x , y) < \delta$, $$ d’ \left( f(x), f(y) \right) < d’ \left( f(x) , f(x_{j}) \right) + d’ \left( f(x_{j}) , f(y) \right) < {{ \varepsilon } \over { 2 }} + {{ \varepsilon } \over { 2 }} = \varepsilon $$ there exists a $\delta > 0$ making this hold. In other words, $f$ is uniformly continuous.

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