In Munkres Theorem 20.4 it is shown that the (relative) uniform topology induced by:
$$
d(x,y)\triangleq \sup\_{n \in \mathbb{N}} d(x\_n,y\_n)
$$
is strictly finer than the product topology on $\prod\_{n \in \mathbb{N}} \mathbb{R}$ and its relative topology on $\ell^{\infty}(\mathbb{R})$. 


Now, for each positive integer $n$, let $K\_k\triangleq \left\{x \in \prod\_{n \in \mathbb{N}} \mathbb{R}:\,
|x\_n|\leq k
\right\}$. Clearly $K\_k \subseteq \ell^{\infty}(\mathbb{R})\subseteq \prod\_{n \in \mathbb{N}} \mathbb{R}$ and $K\_k \subseteq K\_{k+1}$ defines an inductive system in the category **Top**. 


How does the inductive limit topology on $\bigcup\_{k \in \mathbb{N}} K\_k$ compare to (restriction of) the *uniform topology* thereon? My intuition tells me that they coincide, is this correct?



Comparison of product topology and colimit topology in sequence spaces
As user131781 noted the correct answer depends on the topology of $K\_n$. If it is the norm topology the result is a special case of the following more general case:
Let $K$ be a metric space with topology $\cal{O}$ and $K\_n$, $n \in \mathbb{N}$ a sequence of subsets with $K\_n \subset K^o\_{n+1}$ ($X^0$ the interior of $X$) and with $K\_n \uparrow K$. Let $\cal{O}\_1$ be the inductive topology on $K$ induced by the injections $i\_n \colon K\_n \to K$. By definition $\cal{O}\_1$ is finer than $\cal{O}$. But since $(K,\cal{O})$ is metrizable to show that $\cal{O}$ is finer than $\cal{O}\_1$ it suffices to show that each convergent sequence $x\_n \to x$ w.r.t. $\cal {O}$ converges w.r.t. $\cal{O}\_1$. But this is a consequnce of the assumption $K\_n \subset K^o\_{n+1}$.


