2
$\begingroup$

I wish to show that the uniform topology is finer than the product topology on countable Cartesian products on $\mathbb{R}$ denoted $\mathbb{R}^\mathbb{N}$

We know both spaces are metrizable:

  • The metric on the product topology on $\mathbb{R}^\mathbb{N}$ is assumed to be:

$$d_p(x,y) = \sup_{n\in \mathbb{N}}(\dfrac{1}{n}\min\{1, |x_n-y_n|\})$$

  • The metric on the uniform topology on $\mathbb{R}^\mathbb{N}$ is assumed to be:

$$d_u(x,y) = \sup_{n\in \mathbb{N}}(\min\{1, |x_n-y_n|\})$$

I am not sure how to proceed exactly, let me try compare their metric balls:

$$B_\epsilon^p(x) =\{y \in \mathbb{R}^\mathbb{N}| \sup_{n\in \mathbb{N}}(\dfrac{1}{n}\min\{1, |x_n-y_n|\})< \epsilon\}$$

$$B_\epsilon^u(x) =\{y \in \mathbb{R}^\mathbb{N}| \sup_{n\in \mathbb{N}}(\min\{1, |x_n-y_n|\})< \epsilon\}$$

(These metric balls are frightening to say the least...)

Then, the uniform topology is finer than the product topology if for every basic open set in the uniform topology, we can fit a basic open set in the product topology i.e. $B^p \subseteq B^u \Leftrightarrow \mathcal{T}^u \subseteq \mathcal{T}^p$

The metric balls are exactly the basic open sets.

So the claim is $B_\epsilon^p(x) \subseteq B_\epsilon^u(x)$

Attempt:

Fix $x$. Take $y \in B_\epsilon^p(x)$, then $\sup_{n\in \mathbb{N}}(\dfrac{1}{n}\min\{1, |x_n-y_n|\})< \epsilon$ ...does this necessarily imply $\sup_{n\in \mathbb{N}}(\min\{1, |x_n-y_n|\})< \epsilon$? It seems it would be the other way around, if $ \sup_{n\in \mathbb{N}}(\min\{1, |x_n-y_n|\})< \epsilon$ (the bigger guy is less than $\epsilon$) then $ \sup_{n\in \mathbb{N}}(\dfrac{1}{n}\min\{1, |x_n-y_n|\})< \epsilon$ (the smaller guy is less than $\epsilon$)

I am confused, can anyone help out?

$\endgroup$
1
$\begingroup$

Your interpretation of "finer" is backwards: for the uniform topology to be finer, each product-open set must be uniform-open, meaning that given any product-open neighborhood of a point we can find a uniform-open neighborhood inside it. So it would suffice to show that $B_\epsilon^u(x) \subseteq B_\epsilon^p(x)$, which you have correctly observed is true since $d_p(x,y)\leq d_u(x,y)$ for any $x$ and $y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.