# Comparison of product topology and colimit topology in sequence spaces

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?

• In the definition of $K_n$, what are $t$ and $\|\cdot\|$? Does it mean something like $\forall t \in \mathbb N, \lvert x_t\rvert \le n$, where $\lvert\cdot\rvert$ is the usual absolute value? (Also, you currently ask for the inductive-limit topology on $\bigcup_{n \in \mathbb N}$; I guess you mean $\bigcup_{n \in \mathbb N} K_n$?) Commented Mar 27, 2020 at 13:00
• Exactly, thanks for pointing that out.
– ABIM
Commented Mar 27, 2020 at 13:06
• It is not clear to me what topology you are considering on $K_k$, the norm or product topology. If the former, your conjecture is correct, if the latter you get the so-called weak $\ast$-topology as the dual of $\ell^1$.(I deleted my previous comment since I then clocked on this ambivalence) Commented Mar 27, 2020 at 13:40
• The quickest way is via Google—“weak topology” will give you an article on Wikipedia which starts with the weak topology on a Banach space, then goes on to the weak $\ast$-topology on a dual Banach space. Any text on Banach spaces will discuss this (sorry, I don’t have access to a library to give a more precise reference for the obvious reason) Commented Mar 27, 2020 at 13:49
• Sorry, I messed up my first comment when I deleted and repaced it. As was correctly stated there, the topology you get is not the weak $\ast$-topology but the finest one on the whole space which agrees with the latter on bounded sets—commonly known as the bounded weak $\ast$-topology. Sorry for confusing the issue—mea culpa. This is a complete lc topology and its convergent sequences are the norm bounded ones which converge in the product topology. You could check out the Banach-Steinhaus theorem. Commented Mar 27, 2020 at 18:46

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