Completeness proofs often begin with coordinatewise or pointwise limits, but metric convergence can be stronger. In finite-dimensional spaces, coordinatewise convergence and metric convergence coincide. In infinite-dimensional spaces, coordinatewise convergence may require tail control. In function spaces, convergence in the uniform metric is exactly uniform convergence. :::theorem[Coordinatewise Convergence in Finite-Dimensional p-Metrics] Let $X=\mathbb{R}^n$ with $d_p(x,y)=(\sum_{j=1}^{n}|x_j-y_j|^p)^{1/p}$, where $p\geq 1$. Then $x^{(k)}\to x$ in $d_p$ if and only if $x_j^{(k)}\to x_j$ for each $j$. ::: :::proof Given that $X=\mathbb{R}^n$ with metric $d_p$. To prove the equivalence. If $x^{(k)}\to x$ in $d_p$, then $|x_j^{(k)}-x_j|\leq d_p(x^{(k)},x)\to 0$ for each $j$. Conversely, if every coordinate converges, then the finite sum $\sum_{j=1}^{n}|x_j^{(k)}-x_j|^p$ tends to $0$. Hence $d_p(x^{(k)},x)\to 0$. $\square$ ::: :::corollary In $\mathbb{R}^n$ with the maximum metric, convergence is equivalent to coordinatewise convergence. ::: :::proof Given the maximum metric. To prove the equivalence. If the maximum distance tends to zero, then each coordinate difference tends to zero. Conversely, if finitely many coordinate differences tend to zero, then their maximum tends to zero. $\square$ ::: :::example In a discrete metric space, $x_n\to x$ if and only if $x_n=x$ for all sufficiently large $n$. ::: :::proof Given the discrete metric. To prove the characterization. If $x_n\to x$, take $\epsilon=1/2$. Then $d(x_n,x)<1/2$ eventually, so $d(x_n,x)=0$ eventually, and therefore $x_n=x$ eventually. The converse is direct. $\square$ ::: :::theorem[Coordinatewise Convergence in the Space of All Sequences] Let $X$ be the space of all scalar sequences with metric $$\begin{equation*} d(x,y)=\sum_{j=1}^{\infty}\frac{1}{2^j}\frac{|x_j-y_j|}{1+|x_j-y_j|}. \end{equation*}$$ Then $x^{(n)}\to x$ if and only if $x_j^{(n)}\to x_j$ for every fixed $j$. ::: :::proof Given the stated metric. To prove the equivalence. Metric convergence forces every fixed coordinate term to tend to zero. Conversely, given $\epsilon>0$, choose a finite number of coordinates whose complement contributes less than $\epsilon/2$. Coordinatewise convergence controls the finite part, and the weighted tail remains small. Hence the metric distance tends to zero. $\square$ ::: :::theorem[Tail Condition for Convergence in Summable Sequence Spaces] Let $p\geq 1$. If $x_k^{(n)}\to x_k$ coordinatewise, $x\in\ell_p$, and the tails of $x^{(n)}$ are uniformly small, then $x^{(n)}\to x$ in $\ell_p$. ::: :::proof Given coordinatewise convergence and uniform tail control. To prove convergence in $\ell_p$. Choose a finite cutoff so that both the tail of $x$ and all tails of $x^{(n)}$ are small. Coordinatewise convergence controls the finite part. Minkowski's inequality combines the finite part and the tails, giving $d_p(x^{(n)},x)\to 0$. $\square$ ::: :::definition[Pointwise Convergence of Functions] Let $\{f_n\}$ be a sequence of functions on a set $S$. Then $f_n$ converges $pointwise$ to $f$ on $S$ if $f_n(x)\to f(x)$ for every fixed $x\in S$. ::: :::definition[Uniform Convergence through the Supremum Metric] Let $B(S)$ have the uniform metric. Then $f_n\to f$ in this metric if and only if $$\begin{equation*} \sup_{x\in S}|f_n(x)-f(x)|\to 0. \end{equation*}$$ This is exactly uniform convergence. ::: :::exercise [1] State the finite-dimensional coordinatewise convergence criterion. [2] What does convergence in the uniform metric mean? [3] Why is tail control needed in $\ell_p$? ::: :::answer [1] Metric convergence is equivalent to convergence of every coordinate. [2] The supremum of $|f_n-f|$ tends to zero. [3] Coordinatewise convergence alone does not control infinitely many tail terms. ::: :::call-to-action[Continue to Complete Function Spaces] subtitle: Next, use these ideas in differentiable and bounded continuous function spaces. button: Next Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/complete-function-spaces-and-sequence-problems button-ghost: Previous Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/completeness-of-standard-metric-spaces :::