Proof of Euclidean, Metric and Sequential Continuity of a Real Function of a Real Variable

Each result cited below is universally quantified over the data appearing in its own statement, and is applied here to the data named in the step in question. Throughout, $s^{2}$ denotes $s\,s$ for $s\in\mathbb{R}$.

A preliminary remark on absolute values. For every $s\in\mathbb{R}$ one has $|s|\,|s|=s\,s$. Indeed, by claim 1 of Properties of the Absolute Value in an Ordered Field either $|s|=s$, in which case $|s|\,|s|=s\,s$, or $|s|=-s$, in which case $|s|\,|s|=(-s)(-s)=s\,s$ by the sign rules of the field $\mathbb{R}$. Consequently $0\le s^{2}$: by claim 1 of Properties of the Absolute Value in an Ordered Field one has $0\le|s|$, so claim 5 of Elementary Arithmetic in an Ordered Field, applied with $0\le|s|$ and the nonnegative multiplier $|s|$, gives $0\,|s|\le|s|\,|s|$, and $0\,|s|=0$ by claim 1 of Zero Products and Elementary Identities in a Field.

A preliminary remark on squares and order. Let $t,\delta\in\mathbb{R}$ with $0\le t$ and $0<\delta$. Then $t^{2}<\delta^{2}$ if and only if $t<\delta$. Suppose first $t<\delta$. By claim 5 of Elementary Arithmetic in an Ordered Field, applied with $t\le\delta$ and the nonnegative multiplier $t$, we get $t\,t\le\delta\,t$; by claim 10 of Elementary Order Arithmetic in an Ordered Field, applied with $t<\delta$ and the positive multiplier $\delta$, we get $\delta\,t<\delta\,\delta$; and claim 2 of Elementary Order Arithmetic in an Ordered Field combines these into $t^{2}<\delta^{2}$. Suppose conversely that $t<\delta$ fails. The order of $\mathbb{R}$ is a total order by Ordered Field, so $\delta\le t$; the same two multiplication steps, with the roles of $t$ and $\delta$ exchanged, give $\delta^{2}\le t^{2}$, so $t^{2}<\delta^{2}$ fails.

Claim 1. Let $s,t\in\mathbb{R}$ and regard them as points of $\mathbb{R}^{1}$, so that $s-t$ is the point of $\mathbb{R}^{1}$ whose single coordinate is $s-t$. By claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ one has $d_{E}(s,t)=\lVert s-t\rVert$, and by claim 1 of that lemma $\lVert s-t\rVert$ is the unique nonnegative real number $r$ with $r^{2}=\sum_{i=1}^{1}(s-t)^{2}$; the sum over the single index equals $(s-t)^{2}$ by claim 1 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set. Now $|s-t|$ is nonnegative by claim 1 of Properties of the Absolute Value in an Ordered Field and satisfies $|s-t|\,|s-t|=(s-t)(s-t)$ by the first preliminary remark. By the asserted uniqueness, $\lVert s-t\rVert=|s-t|$. Finally $d_{\mathbb{R}}(s,t)=|s-t|$ by The Absolute Value Metric on the Real Line.

Claim 2. Regarded as a map from $\mathbb{R}^{1}$ to $\mathbb{R}^{1}$, $f$ has the single coordinate function $f$ itself, and a point $x\in\mathbb{R}^{1}$ has the single coordinate $x$. So, by Continuity at a Point for Maps Between Euclidean Spaces and claim 1 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set, $f$ is Euclidean continuous at $a$ exactly when for every positive $\eta\in\mathbb{R}$ there is a positive $\delta\in\mathbb{R}$ such that every $x\in\mathbb{R}$ with $(x-a)^{2}<\delta^{2}$ satisfies $(f(x)-f(a))^{2}<\eta^{2}$. By the first preliminary remark, $(x-a)^{2}=|x-a|\,|x-a|$ and $(f(x)-f(a))^{2}=|f(x)-f(a)|\,|f(x)-f(a)|$, and both $|x-a|$ and $|f(x)-f(a)|$ are nonnegative; so by the second preliminary remark the two displayed conditions are equivalent, for the same $\delta$ and $\eta$, to $|x-a|<\delta$ and $|f(x)-f(a)|<\eta$ respectively. By claim 1 these read $d_{\mathbb{R}}(x,a)<\delta$ and $d_{\mathbb{R}}(f(x),f(a))<\eta$. Hence the condition just displayed is, verbatim, the condition of Continuous Map Between Metric Spaces for $f$ to be continuous at $a$ relative to $\mathbb{R}$ as a map from $(\mathbb{R},d_{\mathbb{R}})$ to $(\mathbb{R},d_{\mathbb{R}})$. The two readings therefore agree.

Claim 3. By claim 2 it suffices to prove metric continuity at $a$. Let $\eta\in\mathbb{R}$ be positive. The number $L+1$ is positive, since $0\le L$ and $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field give $0<L+1$ by claim 3 of Elementary Order Arithmetic in an Ordered Field; hence $\eta\,(L+1)^{-1}$ is positive by claims 7 and 5 of that lemma. Let $\delta$ be the lesser of $\rho$ and $\eta\,(L+1)^{-1}$, as in claim 9 of Elementary Order Arithmetic in an Ordered Field; it is one of the two and so is positive. Let $x\in\mathbb{R}$ satisfy $|x-a|<\delta$. Then $|x-a|<\rho$, so the hypothesis gives $|f(x)-f(a)|\le L\,|x-a|$. Moreover $L\le L+1$, since $0\le1$ by claim 1 of Elementary Arithmetic in an Ordered Field and claim 3 of that lemma turns $0\le(L+1)-L$ into $L\le L+1$; so claim 5 of Elementary Arithmetic in an Ordered Field, applied with the nonnegative multiplier $|x-a|$, gives $L\,|x-a|\le(L+1)\,|x-a|$; claim 10 of Elementary Order Arithmetic in an Ordered Field, applied with $|x-a|<\delta$ and the positive multiplier $L+1$, gives $(L+1)\,|x-a|<(L+1)\,\delta$; and $\delta\le\eta\,(L+1)^{-1}$ gives $(L+1)\,\delta\le(L+1)\,\eta\,(L+1)^{-1}=\eta$ by claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative multiplier $L+1$. Chaining these by repeated application of claim 2 of Elementary Order Arithmetic in an Ordered Field yields $|f(x)-f(a)|<\eta$, which by claim 1 is $d_{\mathbb{R}}(f(x),f(a))<\eta$. As $\eta$ was an arbitrary positive real number, $f$ is metrically continuous at $a$.

Claim 4. Let $(x_{k})_{k\in\mathbb{N}}$ be a sequence in $\mathbb{R}$ and let $x\in\mathbb{R}$ be such that $(d_{E}(x_{k},x))_{k\in\mathbb{N}}$ converges to $0$; by claim 1 this says that $(|x_{k}-x|)_{k\in\mathbb{N}}$ converges to $0$. Let $\eta\in\mathbb{R}$ be positive. By hypothesis $f$ is Euclidean continuous at $x$, hence metrically continuous at $x$ by claim 2, so there is a positive $\delta\in\mathbb{R}$ such that every $y\in\mathbb{R}$ with $|y-x|<\delta$ satisfies $|f(y)-f(x)|<\eta$. By Limit of a Sequence of Real Numbers, applied to the convergent sequence $(|x_{k}-x|)_{k}$ with the positive number $\delta$, there is a natural number $K$ such that every natural number $k$ with $K\le k$ satisfies $\bigl||x_{k}-x|-0\bigr|<\delta$; and $\bigl||x_{k}-x|-0\bigr|=\bigl||x_{k}-x|\bigr|=|x_{k}-x|$. For the last step, write $t=|x_{k}-x|$, which is nonnegative by claim 1 of Properties of the Absolute Value in an Ordered Field; by that same claim $|t|$ is $t$ or $-t$, and in the second case $0\le|t|=-t$ forces $t\le0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field, whence $t=0$ by antisymmetry of the order and $|t|=-0=0=t$ by claim 4 of Additive Cancellation and Elementary Additive Identities in a Field. Hence $|f(x_{k})-f(x)|<\eta$ for every $k$ with $K\le k$. As $\eta$ was an arbitrary positive real number, Limit of a Sequence of Real Numbers gives that $(f(x_{k}))_{k\in\mathbb{N}}$ converges to $f(x)$.