Let $\mathbb{R}^\mathbb{N}_\text{prod}$ be countable product of $\mathbb{R}$ with the product topology, $\mathbb{R}^\mathbb{N}_\text{prod}$ be countable product of $\mathbb{R}$ with the uniform topology. I wish to show that:
Show that $\mathbb{R}^\mathbb{N}_\text{prod}$ has the countable chain condition, and $\mathbb{R}^\mathbb{N}_\text{unif}$ does not.
Recall: A topological space $(X,\mathcal{T})$ has the countable chain condition (CCC) if there are no uncountable collections of pairwise disjoint open sets (PDNO sets)
I am confused why the proof that $\mathbb{R}^\mathbb{N}_\text{unif}$ is not CCC doesn't work for $\mathbb{R}^\mathbb{N}_\text{prod}$
Proof 1: $\mathbb{R}^\mathbb{N}_\text{unif}$ is not CCC
Take the collection of sequences that maps from the naturals to the integers $\mathbb{Z}^\mathbb{N} = \{x: x: \mathbb{N} \to \mathbb{Z} \}$
Since $\mathbb{R}^\mathbb{N}_\text{unif}$ is metrizable with the uniform metric $$d_u(x,y) = \sup\limits_{n \in \mathbb{N}}\{\min\{d(x_n,y_n),1\}\} =\sup\limits_{n \in \mathbb{N}}\{d(x_n,y_n) \wedge 1\} $$
Define the open ball $B_\epsilon(x) = \{y \in \mathbb{R}^\mathbb{N}_\text{unif} | d_u(x,y) < \epsilon\}$
- Then take $\epsilon = \dfrac{1}{2}$, we have $x \in B_\frac{1}{2}(x) \subset \mathbb{Z}^\mathbb{N}$ such that $B_\frac{1}{2}(x) \cap B_\frac{1}{2}(y) = \varnothing, \forall x \neq y$
- Since $|\mathbb{Z}^\mathbb{N}| = \aleph_1$ is uncountable, therefore the set of balls defined above is uncountable. Specifically, $\{B_\frac{1}{2}(x)| x \in \mathbb{Z}^\mathbb{N}\}$ is uncountable.
- This consititutes a collection of uncountable PDNO sets, hence $\mathbb{R}^\mathbb{N}_\text{unif}$ is not CCC.
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \nwarrow$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad $ This proof is unverified
My question is why doesn't the above imply that $\mathbb{R}^\mathbb{N}_\text{prod}$ is not CCC?
Recall that $\mathbb{R}^\mathbb{N}_\text{prod}$ is metrizable via $$d_p(x,y) = \sup\limits_{n \in \mathbb{N}}\{\dfrac{\min\{d(x_n,y_n),1\}}{n}\} =\sup\limits_{n \in \mathbb{N}}\{\dfrac{d(x_n,y_n)\wedge 1}{n} \} $$
Then fix $\epsilon$, $B_\epsilon^p(x) \subseteq B\epsilon^u(x)$.
Doesn't this mean we can put even open smaller ball around a given sequence $x : \mathbb{Z} \to \mathbb{N}$?
This means we still have an uncountable collection of PDNO set in the product topology on $\mathbb{R}^\mathbb{N}$.
Can someone provide a quick reason as to why this is? Is there a quick proof showing that $\mathbb{R}^\mathbb{N}_\text{prod}$ is CCC?
The proof that $\mathbb R^{\mathbb N}_{\text{unif}}$ is not c.c.c. doesn't carry over to $\mathbb R^{\mathbb N}_{\text{prod}}$ because of the structure of the open balls in $\mathbb R^{\mathbb N}_{\text{prod}}$. What you have essentially done is construct, for each $\mathbf x \in \mathbb Z^{\mathbb N}$, a uniform open ball about $\mathbf x$ that does not contain any other element of $\mathbb Z^{\mathbb N}$. The same cannot be done for the product topology.
For suppose that $\mathbf x = (x_1,x_2,\ldots) \in \mathbb Z^{\mathbb N}$ and let $\varepsilon > 0$. Then there must be an $n$ such that $\frac 1n < \varepsilon$. Now consider $\mathbf y = (x_1,x_2,\ldots,x_n,x_{n+1}+1,x_{n+2}+1,\ldots) \in \mathbb Z^{\mathbb N}$. Note that for $i \leq n$ we have that $\min(|x_i-y_i|,1) = \min(|x_1-x_i|,1) = \min(0,1) = 0$, and for $i > n$ we have that $\min(|x_i-y_i|,1) = \min(|x_i-(x_i+1)|,1) = \min(1,1) = 1$, and so it follows that $$\begin{align} d_p(\mathbf x , \mathbf y) &= \sup_{i \in \mathbb N} \{ \frac{\min ( d(x_i,y_i),1 )}n \} \\ %&= \sup \{ %\frac{\min(|x_1-x_1|,1)}1, \frac{\min(|x_2-x_2|,1)}2 , \ldots , %\frac{\min(|x_n-x_n|,1)}n , \frac{\min(|x_{n+1}-(x_{n+1}+1)|,1)}{n+1}, %\frac{\min(|x_{n+2}-(x_{n+2}+1)|,1)}{n+2}, \ldots \} \\ &= \sup \{ 0, 0, \ldots, 0 , \tfrac{1}{n+1}, \tfrac{1}{n+2}, \ldots \} \\ &= \frac{1}{n+1} \\ &< \varepsilon \end{align}$$
So it follows that every $d_p$-ball about an element of $\mathbb Z^{\mathbb N}$ contains another element of $\mathbb Z^{\mathbb N}$.
(This, of course, does not prove that $\mathbb R^{\mathbb N}_{\text{prod}}$ is c.c.c., but shows that the same argument cannot be used to show that it isn't c.c.c.)
Quick proofs that $\mathbb R^{\mathbb N}_{\text{prod}}$ is c.c.c. include the following:
By the Hewitt-Marczewski-Pondiczery Theorem $\mathbb R^{\mathbb N}_{\text{prod}}$ is actually separable, and so it is c.c.c. since all separable spaces are c.c.c.
It can be shown that a product space $\prod_{i \in I} X_i$ is c.c.c. iff for each finite $F \subseteq I$ the product $\prod_{i \in F} X_i$ is c.c.c. (See Theorem 2 on Product of Spaces with Countable Chain Condition from Dan Ma's Topology Blog.) In this case we have $I = \mathbb N$ and $X_i = \mathbb R$ for each $i$. For each finite $F \subseteq I$ the product $\prod_{i \in F} X_i$ is just the Euclidean $|F|$-space, which is clearly c.c.c. (again, use separability if you must).