Topologically, what distinguishes $\mathbb{R}^\omega_\text{prod}$ and $\mathbb{R}^J_\text{prod}$

Question

Let $\mathbb{R}^\omega_\text{prod}$ be countable product of $\mathbb{R}$ equipped with the product topology

Let $\mathbb{R}^J_\text{prod}$ be arbitrary product of $\mathbb{R}$ with the product topology.

Is there a topological invariant that distinguishes these spaces?

Any suggestions helps.

Answer 1

To reduce typing I will use $R$ for $\mathbb R$.

1. If $J$ is empty : Obvious.

2. If $J$ is finite,not empty: Then $R^J$ is homeomorphic to $R^n$ for some $n\in N,$ which is locally compact. But $R^{\omega}$ is not locally compact. PROOF: If $B$ is a base for a locally compact Hausdorff space then there is a base $B'\subset B$ such that $\bar c$ is compact for every $c\in B'.$ So it suffices to find a base $B$ for $R^{\omega}$ such that $\bar b$ is not compact for any $b\in B.$

   Let $B$ be the base of all non-empty sets of the form $\prod_{n\in \omega}A_n$ where each $A_n$ is open in $R$, and $A_n=R$ for all but finitely many $n.$ Now for $b=\prod_{n\in \omega}b_n \in B,$ take $m_0\in N$ such that $b_{m_0}=R.$ Let $c_{n,m}=R$ for $m\ne m_0$ and $c_{n,m_0}=(-n,n).$ Let $U_n(b)=\prod_{m\in N}c_{n,m}.$ Then $\{U_n(b):n\in N\}$ is an open cover of $\bar b$, and since $b \ne \emptyset,$ there is no finite subcover.

3. If $J$ is uncountable: Then $R^J$ has uncountable pseudo-character: If $f\in R^J$ and $V$ is a countable open family in $R^J$ then $\cap V\ne \{f\}.$ PROOF: Suppose $f\in \cap V.$ For each $v\in V$ take $v'=\prod_{j\in J}v'_j$ where (i). Each $v'_j$ is open in $R,$ and (ii) . $S(v')=\{j:v'_j\ne R\}$ is finite, and (iii). $f\in v'\subset v.$

   Then $J$ \ $\cup_{v\in V}S(v') \ne \emptyset$ (because $V$ is countable, each $S(v')$ is finite , and $J$ is uncountable.)

   Let $g(j)=f(j)$ for all $j\in \cup_{v\in V}S(v')$ and $g(j)=f(j)+1$ for $ j\in J$ \ $\cup_{j\in J}S(v').$ Then $f\ne g\in \cap V, $ so $\cap V\ne \{f\}.$

   But $R^{\omega}$ has countable pseudo-character: For $f\in R^{\omega}$ and $n\in \omega$ let $U_{j,n}$ be the interval $(f(j)-2^{-n},f(j)+2^{-n})$ for $j\leq n$ and let $ U_{j,n}=R$ for $j>n.$ Then $V_n=\prod_{j\in N}U_{j,n}$ is open in $R^{\omega}$ and $\cap_{n\in \omega}V_n=\{f\}.$

There are many other differences between $R^{\omega}$ and $R^J$ for uncountable $J$.

Answer 2

Let $J$ be any infinite set. Then the cardinality of $J$ is equal to the weight of the space $\mathbb{R}^J$ (the minimal cardinality of a basis for the topology). Indeed, taking a countable basis for $\mathbb{R}$, the induced basis on $\mathbb{R}^J$ has cardinality $|J|$. Conversely, if you have a basis, then for each $j\in J$ there must be some $U$ in your basis whose projection onto the $j$th coordinate is not all of $\mathbb{R}$. Since each $U$ can have this property for only finitely many $j$, there must be at least $|J|$ different elements of the basis.

Answer 3

There are a lot of topological properties that distinguish $\mathbb{R}^{\omega}$ from $\mathbb{R}^J$ where $J$ is a non- countable set. Several of then have already been mentioned by other users. I summarize them here:

1. The set $\mathbb{R}^{\omega}$ is second countable and $\mathbb{R}^J$ is not.

2. Indeed $\mathbb{R}^{\omega}$ is first countable and $\mathbb{R}^J$ is not.

3. You can prove that $\mathbb{R}^\omega$ is a metrizable space but since $\mathbb{R}^J$ is not first countable then you can not find any metric that induce the topology of $\mathbb{R}^J$.

All of these properties I mentioned are topological invariants.