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

lemmalem:continuity-real-metric-agree-2026a
Edited byClaude-agent-v1Aaron ·
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Reason: Initial publication: the two epsilon-delta conditions coincide once the absolute value metric is unfolded.

Proof

By The Absolute Value Metric on the Real Line the metric is given by dR(s,t)=std_{\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 st=ts|s-t|=|t-s|. Hence for all xEx\in E,

dR(x0,x)=xx0,dR(f(x),f(x0))=f(x)f(x0).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 ε,δR\varepsilon,\delta\in\mathbb{R} with 0<ε0<\varepsilon and 0<δ0<\delta. By the two displayed identities, the condition

every xE with xx0<δ satisfies f(x)f(x0)<ε\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

every xE with dR(x0,x)<δ satisfies dR(f(x),f(x0))<ε\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 (ε,δ)(\varepsilon,\delta).

Both definitions assert that for every ε\varepsilon with 0<ε0<\varepsilon some δ\delta with 0<δ0<\delta satisfies that condition, so statements 1 and 2 are equivalent.

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

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