Proof of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions

Throughout, $a=(a_1,\dots,a_n)$.

Two elementary equivalences.

(i) For every $x=(x_1,\dots,x_n)\in\mathbb{R}^n$ and every $\delta\in\mathbb{R}$ with $0<\delta$,

$$\sum_{i=1}^n (x_i-a_i)^2<\delta^2\quad\text{if and only if}\quad d_E(x,a)<\delta .$$

Indeed, by the definition of the Euclidean distance, $d_E(x,a)$ is the nonnegative square root of $\sum_{i=1}^n(x_i-a_i)^2$, so by Existence and Uniqueness of the Nonnegative Square Root we have $0\le d_E(x,a)$ and $d_E(x,a)^2=\sum_{i=1}^n(x_i-a_i)^2$. From $0<\delta$ we get $0\le\delta$, so claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, applied with the two nonnegative elements $d_E(x,a)$ and $\delta$, gives $d_E(x,a)<\delta$ if and only if $d_E(x,a)^2<\delta^2$. Substituting the displayed identity for $d_E(x,a)^2$ yields the stated equivalence.

(ii) For all $s,t\in\mathbb{R}$ and every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$,

$$(s-t)^2<\varepsilon^2\quad\text{if and only if}\quad d_{\mathbb{R}}(s,t)<\varepsilon .$$

Indeed, $d_{\mathbb{R}}(s,t)=|s-t|$ by The Absolute Value Metric on the Real Line, where $|\cdot|$ is the absolute value on $\mathbb{R}$. By claim 1 of Properties of the Absolute Value in an Ordered Field we have $0\le|s-t|$, and $|s-t|$ equals either $s-t$ or $-(s-t)$; in the first case $|s-t|^2=(s-t)^2$, and in the second case $|s-t|^2=(-(s-t))\cdot(-(s-t))=(s-t)\cdot(s-t)=(s-t)^2$, using the field arithmetic of $\mathbb{R}$. From $0<\varepsilon$ we get $0\le\varepsilon$, so claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $|s-t|<\varepsilon$ if and only if $|s-t|^2<\varepsilon^2$, which is the stated equivalence.

Claim 1. By the Euclidean definition of continuity at a point, taken with $m=1$ and single coordinate function $f$, the function $f$ is continuous at $a$ in the Euclidean sense exactly when for every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there exists $\delta\in\mathbb{R}$ with $0<\delta$ such that every $x\in E$ satisfying $\sum_{i=1}^n(x_i-a_i)^2<\delta^2$ satisfies $(f(x)-f(a))^2<\varepsilon^2$.

By the metric definition of continuity at a point, applied with $X=\mathbb{R}^n$, $d_X=d_E$, $A=E$, $Y=\mathbb{R}$ and $d_Y=d_{\mathbb{R}}$, the function $f$ is continuous at $a$ relative to $E$ exactly when for every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there exists $\delta\in\mathbb{R}$ with $0<\delta$ such that every $x\in E$ satisfying $d_E(a,x)<\delta$ satisfies $d_{\mathbb{R}}(f(x),f(a))<\varepsilon$. Since $d_E$ is a metric by Euclidean Distance is a Metric on $\mathbb{R}^n$, it is symmetric by the definition of a metric space, so $d_E(a,x)=d_E(x,a)$.

Fix $\varepsilon$ with $0<\varepsilon$ and $\delta$ with $0<\delta$. By (i), the hypotheses $\sum_{i=1}^n(x_i-a_i)^2<\delta^2$ and $d_E(x,a)<\delta$ are satisfied by exactly the same points $x\in E$; by (ii), applied with $s=f(x)$ and $t=f(a)$, the conclusions $(f(x)-f(a))^2<\varepsilon^2$ and $d_{\mathbb{R}}(f(x),f(a))<\varepsilon$ hold for exactly the same points $x\in E$. Hence, for this $\varepsilon$ and $\delta$, the implication required by the Euclidean definition holds if and only if the implication required by the metric definition holds. As the two definitions quantify over $\varepsilon$ and $\delta$ in the same way, the two notions of continuity at $a$ are equivalent. This proves claim 1.

Claim 2. Let $x\in U$. By claim 1 of the definition of a function of class $C^2$, $u$ is of class $C^1$ on $U$, regarded as a map into $\mathbb{R}^m$ with $m=1$ and single coordinate function $u$. The definition of a map of class $C^1$ requires in particular that each coordinate function be continuous at every point of $U$ in the Euclidean sense; hence $u$ is continuous at $x$ in the Euclidean sense.

Applying claim 1 with $E=U$ and with $x$ in place of $a$, we conclude that $u$ is continuous at $x$ relative to $U$ as a map into $(\mathbb{R},d_{\mathbb{R}})$. As $x\in U$ was arbitrary, $u$ is continuous on $U$ in the sense of Continuous Map Between Metric Spaces.

By claim 2 of Semicontinuity Under Negation and Characterization of Continuity, applied with the metric space $(\mathbb{R}^n,d_E)$, with $A=U$ and with $u$ in place of the function there, continuity of $u$ at each $x\in U$ relative to $U$ implies that $u$ is both upper semicontinuous at $x$ relative to $U$ and lower semicontinuous at $x$ relative to $U$. Since this holds for every $x\in U$, the function $u$ is upper semicontinuous on $U$ and lower semicontinuous on $U$. This proves claim 2.