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The displayed metric on a countable Cartesian product of metric spaces combines the bounded metric transform with summable positive weights. It is finite even when individual factor metrics are unbounded. The triangle inequality follows from subadditivity of $t/(1+t)$ and summation. Zero distance forces equality in each coordinate.

Convergence in this metric is exactly coordinatewise convergence. One term is bounded by the sum, so $D(x^{(n)},x)\to 0$ implies each $d_i(x_i^{(n)},x_i)\to 0$. Conversely, choose a finite head whose omitted weight sum is small; convergence in those finitely many coordinates controls the head uniformly, while boundedness of the transformed distances controls the entire tail. Thus it induces the countable product topology. Arbitrary positive summable weights work by the same proof.

The function $\varphi (t)=t/(1+t)$ increases for $t\ge 0$, andThus $a\le b+c$ implies $\boxed{\varphi (a)\le \varphi (b)+\varphi (c)}$. Applying this to the triangle inequality for a metric proves the triangle inequality for the bounded metric transform $d_b=\varphi \circ d$. Symmetry and nonnegativity are immediate, and $d_b(x,y)=0$ exactly when $d(x,y)=0$, so $\boxed{d_b\text{ is a metric}}$.

Each $K_n$ is compact and lies inside $D$, so the supremum norm $\Vert f-g{\Vert }_n$ is finite for continuous $f,g$. Every summand of $\sigma$ lies between zero and $2^{-n}$, giving absolute convergence and $0\le \sigma \le 1/2$. Symmetry is inherited from these norms. If $\sigma (f,g)=0$, all the nonnegative summands vanish, hence $f=g$ on every $K_n$; their union is $D$, so $f=g$ everywhere. Finallyand the subadditivity already proved give the triangle inequality term by term and then after summation. Therefore $\boxed{\sigma \text{ is a metric on }\mathcal{F}}$. This is a weighted metric for local uniform convergence, representing the compact-open topology.

The geometric-series sum is $s(z)=1/(1-z)$ and its remainder is $s-s_k=z^{k+1}/(1-z)$. Writing $r_n=1-1/n$, the triangle inequality givesEquality occurs at the positive real point $z=r_n$. ThusFor every integer $N\ge 2$, split the defining sum into a finite head and its tail. Since $\varphi (t)\le t$ and $\varphi (t)\le 1$,Given $\epsilon >0$, choose $N$ with $2^{-N}<\epsilon /2$. For that fixed $N$, each of the finitely many displayed norms tends to zero, so their sum is below $\epsilon /2$ for sufficiently large $k$. Hence $\boxed{\sigma (s_k,s)\to 0}$, even though convergence is not uniform on the whole open disk.

Put $\varphi (t)=t/(1+t)$. Every summand is nonnegative and at most $2^{-i}$, so the series defining $d$ converges and $d\le 1$. Symmetry follows from each $d_i$. If $d(x,y)=0$, every nonnegative summand is zero, hence every $d_i(x_i,y_i)=0$ and $x_i=y_i$. The converse is immediate.

For the triangle inequality, first observe that for $a,b\ge 0$,Monotonicity of $\varphi$ and the triangle inequality for $d_i$ therefore giveMultiply by $2^{-i}$ and sum the convergent nonnegative series. This proves the required triangle inequality for $d$, so $d$ is a metric. The construction combines a bounded metric transform with a countable product metric.

Let $X$ have nested compact sets $K_n$ covering it, with $K_n\subseteq \mathrm{int}{K}_{n+1}$. Then every point has a neighborhood in some $K_n$, and every compact subset is contained in one $K_n$. For continuous scalar functions, take positive weights with ${\sum }_n{w}_n<\infty$ and defineThe bounded metric transform makes the sum finite and gives its triangle inequality; the covering property gives definiteness. On $\mathbb{R}$, one may use $K_n=[-n,n]$ and $w_n=2^{-n}$.

If $\sigma (f_j,f)\to 0$, each fixed compact supremum norm tends to zero, since its summand is bounded by the whole metric. Conversely, control finitely many compact supremum norms and then the summable weight tail. This proves that $\sigma$ realizes locally uniform convergence under the stated exhaustion conditions.

Merely writing a space as a union of compact sets is insufficient. Enumerate the points of the compact space $\{0\}\cup \{1/n:n\ge 1\}$ and take finite initial subsets as $K_n$. The resulting metric detects pointwise control on those finite sets, not uniform control on the entire compact space. The neighborhood and compact-containment conditions explain the difference.