Proof of Continuity of a Real Function Agrees with Metric Continuity on the Real Line

By The Absolute Value Metric on the Real Line the metric is given by $d_{\mathbb{R}}(s,t)=|s-t|$, with $|\,\cdot\,|$ the absolute value, and by claim 2 of Properties of the Absolute Value in an Ordered Field we have $|s-t|=|t-s|$. Hence for all $x\in E$,

$$d_{\mathbb{R}}(x_0,x)=|x-x_0|,\qquad d_{\mathbb{R}}\bigl(f(x),f(x_0)\bigr)=|f(x)-f(x_0)| .$$

Fix $\varepsilon,\delta\in\mathbb{R}$ with $0<\varepsilon$ and $0<\delta$. By the two displayed identities, the condition

$$\text{every }x\in E\text{ with }|x-x_0|<\delta\text{ satisfies }|f(x)-f(x_0)|<\varepsilon$$

used in Continuity at a Point and the condition

$$\text{every }x\in E\text{ with }d_{\mathbb{R}}(x_0,x)<\delta\text{ satisfies }d_{\mathbb{R}}\bigl(f(x),f(x_0)\bigr)<\varepsilon$$

used in Continuous Map Between Metric Spaces are the same condition on the pair $(\varepsilon,\delta)$.

Both definitions assert that for every $\varepsilon$ with $0<\varepsilon$ some $\delta$ with $0<\delta$ satisfies that condition, so statements 1 and 2 are equivalent.

For the final assertion, $f$ is continuous on $E$ in the sense of Continuous Map Between Metric Spaces exactly when statement 2 holds at every point of $E$, which by the equivalence just proved happens exactly when statement 1 holds at every point of $E$.