Comparison of product topology and colimit topology in sequence spaces

Question

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?

Answer 1

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 $K_n \subset K^o_{n+1}$.