Let \(X\subseteq\RR\) and let \(f\colon X\to \RR\).
The function \(f\) is continuous on \(X\) if \[ \forall x_0\in X \quad \forall \varepsilon>0 \quad \exists \delta=\delta(\varepsilon, x_0)>0 \quad \forall y\in X: \quad \abs{x_0 - y} < \delta \implies \abs{f(x_0) - f(y)} < \varepsilon. \]
The function \(f\) is uniformly continuous on \(X\) if \[ \forall \varepsilon>0 \quad \exists \delta=\delta(\varepsilon)>0 \quad \forall x, y \in X: \quad \abs{x - y} < \delta \implies \abs{f(x) - f(y)} < \varepsilon. \]
Substituting \(y = x_0 + h\), the two conditions read \[ \begin{aligned} &\forall x_0\in X \quad \forall \varepsilon>0 \quad \exists \delta(\varepsilon, x_0)>0 \quad \forall h \text{ with } x_0+h\in X: && \abs{h} < \delta \implies \abs{f(x_0) - f(x_0 + h)} < \varepsilon, \\ &\forall \varepsilon>0 \quad \exists \delta(\varepsilon)>0 \quad \forall x\in X \quad \forall h \text{ with } x+h\in X: && \abs{h} < \delta \implies \abs{f(x) - f(x + h)} < \varepsilon. \end{aligned} \] In the first, \(\delta\) may depend on the point \(x_0\); in the second, one \(\delta\) serves every point of \(X\).