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Following a reference from "General Topology" by Ryszard Engelking

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Now usinge the above proposition we prove the following theorem

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Well I desire to discuss the proof of the theorem $2.3.13$, or rather I want understad better why it follow immediately from the proposition $2.3.1$. Anyway from $2.3.1$ we know that if we pick a base $\mathcal{B}_s$ of $X_s$ for any $s\in S$ such that $|\mathcal{B}_s|=w(X_s)$ then the set

$$\mathcal{B}=\{\pi^{-1}_{s_1}(B_1)\cap...\cap\pi^{-1}_{s_n}(B_n):B_i\in\mathcal{B}_{s_i}\land i=1,....,n\land n\in\Bbb{N}\}$$

is a base of $\prod_{s\in S}X_s$ and it result that $$ |\mathcal{B}| \le \left|\prod_{s\in S} \mathcal{B}_s\right| = \prod_{s\in S} |\mathcal{B}_s| = \prod_{s\in S} w(X_s) \le \prod_{s\in S} m = m\cdot |S| = m $$

since $m$ is an infinite cardinal such that $m>|S|$.

Could be this a right observation? Could someone help me, please?

gt6989b
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Antonio Maria Di Mauro
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1 Answers1

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No, you have to use in an essential way that the base elements have a finite support, and count them more accurately, as I did in this recent answer.

Each $X_s$ has a base of size $\le \mathbf{m}$, and the index set has size $\le \mathbf{m}$ too, so it has $\le \mathbf{m}$ many subsets of size $n \in \omega$ and for each such set we have $\mathbf{m}^n = \mathbf{m}$ many base element choices, so still $\le \mathbf{m}$ choices for basic sets that depend on $n$ coordinates. This holds for each $n$ so in total the base size does not exceed $\mathbf{m}$ too, as $\aleph_0 \cdot \mathbf{m}=\mathbf{m}$. (We're only concerned with upper bounds here).

In your formula $$\prod_{s \in S} w(X_s)$$ could well be equal to $$\mathbf{m}^\mathbf{m} = 2^{\mathbf{m}} > \mathbf{m}$$

so naive products won't work (then you get to te weight of box products instead of the usual product topology).

Henno Brandsma
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